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Chapter 1 Even Answers 2. 4. 6. 8. 10. 12. 16. 18. 20. 22. 24. 26. 28. 30. 32. 34. 36. 38. 40. 42. 44. 46. 48. 50. 52. 54. 56. 58. 60. 62. 64. 66. 68. 70. 623 kg/m3 3 3 4 πρ (r2 – r1 ) 3 7.69 cm 8.72 × 1011 atoms/s (a) 72.6 kg (b) 7.82 × 1026 atoms equation is dimensionally consistent The units of G are: m3/kg ⋅ s2 9.19 nm/s (a) 3.39 × 105 ft3 (b) 2.54 × 104 lb 8.32 × 10–4 m/s 9.82 cm (a) 6.31 × 104 AU (b) 1.33 × 1011 AU (a) 1.609 km/h (b) 88.5 km/h (c) 16.1 km/h (a) 3.16 × 107 s/yr (b) 6.05 × 1010 yr 2.57 × 106 m3 1.32 × 1021 kg (a) 2.07 mm (b) 8.62 × 1013 times as large (a) 13.4 (b) 49.1 3 rAl = rFe (ρFe/ρAl) ~ 106 km ~ 109 drops time required ≅ 50 years or more; advise against accepting the offer ~ 105 tons (a) 2 (b) 4 (c) 3 (d) 2 (a) 797 (b) 1.1 (c) 17.66 (a) 3 (b) 4 (c) 3 (d) 2 5.2 m3, 2.7% 1.79 × 10–9 m 24.6° (b) Acylinder = πR2, Arectangular solid = lw 0.141 nm 289 µm (a) 1000 kg (b) 5.2 × 10–16 kg 0.27 kg (d) 1.3 × 10–5 kg g Aluminum: 2.75 3 (table value is 2% smaller) cm g Copper: 9.36 (table value is 5% smaller) cm3 g Brass: 8.91 cm3 g Tin: 7.68 cm3 g Iron: 7.88 (table value is 0.3% smaller) cm3 © 2000 by Harcourt College Publishers. All rights reserved. 2 Chapter 1 Even Answers © 2000 by Harcourt College Publishers. All rights reserved. Chapter 1 Solutions *1.1 With V = (base area) · (height) V = π r2 · h and ρ = m , we have V ρ= 9 3 m 1 kg 10 mm = π r2 h π (19.5 mm)2 39.0 mm 1 m3 ρ = 2.15 × 104 kg/m3 1.2 ρ= ρ= 1.3 M M = V 4 πR3 3 3(5.64 × 1026 kg) = 623 kg/m3 4 π (6.00 × 107 m) 3 VCu = V0 − Vi = VCu = 4 3 3 π (ro – ri ) 3 4 π [(5.75 cm)3 – (5.70 cm)3] = 20.6 cm3 3 5.7cm 0.05 cm ρ=m V m = ρV = (8.92 g/cm3)(20.6 cm3) = 184 g 1.4 V = Vo – Vi = 4 3 3 π (r2 – r1 ) 3 3 ρ= *1.5 (a) 3 4 πρ (r2 – r1 ) m 4 3 3 , so m = ρV = ρ π (r2 – r1) = V 3 3 The number of moles is n = m/M, and the density is ρ = m/V. Noting that we have 1 mole, V1 mol = mFe nFe MFe (1 mol)(55.8 g/mol) = = = 7.10 cm3 ρFe ρFe 7.86 g/cm3 © 2000 by Harcourt College Publishers. All rights reserved. 2 Chapter 1 Solutions (b) In 1 mole of iron are NA atoms: V1 atom = V1 mol 7.10 cm3 = = 1.18 × 10–23 cm3 NA 6.02 × 1023 atoms/mol = 1.18 × 10-29 m3 3 1.18 × 10–29 m3 = 2.28 × 10–10 m = 0.228 nm (c) datom = (d) V1 mol U = (1 mol)(238 g/mol) = 12.7 cm3 18.7 g/cm3 V1 atom U = V1 mol U 12.7 cm3 = = 2.11 × 10–23 cm3 NA 6.02 × 1023 atoms/mol = 2.11 × 10-29 m3 datom U = *1.6 1.7 r2 = r1 3 3 V1 atom U = 3 2.11 × 10–29 m3 = 2.77 × 10–10 m = 0.277 nm 5 = (4.50 cm)(1.71) = 7.69 cm Use m = molar mass/NA and 1 u = 1.66 × 10-24 g 4.00 g/mol = 6.64 × 10-24 g = 4.00 u 6.02 × 1023 mol-1 (a) For He, m = (b) For Fe, m = 55.9 g/mol = 9.29 × 10-23 g = 55.9 u 6.02 × 1023 mol-1 (c) For Pb, m = 207 g/mol -22 g = 207 u 23 -1 = 3.44 × 10 6.02 × 10 mol © 2000 by Harcourt College Publishers. All rights reserved. Chapter 1 Solutions Goal Solution Calculate the mass of an atom of (a) helium, (b) iron, and (c) lead. Give your answers in atomic mass units and in grams. The molar masses are 4.00, 55.9, and 207 g/mol, respectively, for the atoms given. Gather information: The mass of an atom of any element is essentially the mass of the protons and neutrons that make up its nucleus since the mass of the electrons is negligible (less than a 0.05% contribution). Since most atoms have about the same number of neutrons as protons, the atomic mass is approximately double the atomic number (the number of protons). We should also expect that the mass of a single atom is a very small fraction of a gram (~10–23 g) since one mole (6.02 × 1023) of atoms has a mass on the order of several grams. Organize: An atomic mass unit is defined as 1/12 of the mass of a carbon-12 atom (which has a molar mass of 12.0 g/mol), so the mass of any atom in atomic mass units is simply the numerical value of the molar mass. The mass in grams can be found by multiplying the molar mass by the mass of one atomic mass unit (u): 1 u = 1.66 × 10–24 g. Analyze: For He, m = 4.00 u = (4.00 u)(1.66 × 10–24 g/u) = 6.64 × 10–24 g For Fe, m = 55.9 u = (55.9 u)(1.66 × 10–24g/u) = 9.28 × 10–23 g For Pb, m = 207 u = (207 u)(1.66 × 10–24 g/u) = 3.44 × 10–22 g Learn: As expected, the mass of the atoms is larger for bigger atomic numbers. If we did not know the conversion factor for atomic mass units, we could use the mass of a proton as a close approximation: 1u ≈ mp = 1.67 × 10–24 g. *1.8 ∆n = ∆m 3.80 g – 3.35 g = = 0.00228 mol M 197 g/mol ∆N = (∆n)NA = (0.00228 mol)(6.02 × 1023 atoms/mol) = 1.38 × 1021 atoms ∆t = (50.0 yr)(365 d/yr)(24.0 hr/d)(3600 s/hr) = 1.58 × 109 s ∆N 1.38 × 1021 atoms = = 8.72 × 1011 atoms/s ∆t 1.58 × 109 s 1.9 (a) m = ρL3 = (7.86 g/cm3)(5.00 × 10-6 cm)3 = 9.83 × 10-16 g (b) N=m NA = (9.83 × 10-16 g)(6.02 × 1023 atoms/mol) Molar mass 55.9 g/mol = 1.06 × 107 atoms © 2000 by Harcourt College Publishers. All rights reserved. 3 4 1.10 Chapter 1 Solutions (a) The cross-sectional area is 15.0 cm A = 2(0.150 m)(0.010 m) + (0.340 m)(0.010 m) 1.00 cm = 6.40 × 10-3 m2 36.0 cm The volume of the beam is V = AL = (6.40 × 10-3 m2)(1.50 m) = 9.60 × 10-3 m3 1.00 cm Thus, its mass is m = ρV = (7.56 × 103 kg/m3)(9.60 × 10-3 m3) = 72.6 kg (b) Presuming that most of the atoms are of iron, we estimate the molar mass as M = 55.9 g/mol = 55.9 × 10-3 kg/mol. The number of moles is then n= m 72.6 kg = = 1.30 × 103 mol M 55.9 × 10-3 kg/mol The number of atoms is N = nNA = (1.30 × 103 mol)(6.02 × 1023 atoms/mol) = 7.82 × 1026 atoms *1.11 (a) n= m 1.20 × 103 g = = 66.7 mol, and M 18.0 g/mol 23 Npail = nNA = (66.7 mol)(6.02 × 10 molecules/mol) = 4.01 × 1025 molecules (b) Suppose that enough time has elapsed for thorough mixing of the hydrosphere. Nboth = Npail mpail = (4.01 × 1025 molecules) 1.20 kg , or Mtotal 1.32 × 1021 kg Nboth = 3.65 × 104 molecules 1.12 r, a, b, c and s all have units of L. (s – a)(s – b)(s – c) s = L×L×L = L L2 = L Thus, the equation is dimensionally consistent. © 2000 by Harcourt College Publishers. All rights reserved. Chapter 1 Solutions 1.13 The term s has dimensions of L, a has dimensions of LT -2, and t has dimensions of T. Therefore, the equation, s = kamtn has dimensions of L = (LT- ) (T) 2 m n 1 0 or m n-2m L T =L T The powers of L and T must be the same on each side of the equation. Therefore, L1 = Lm and m = 1 Likewise, equating terms in T, we see that n – 2m must equal 0. Thus, n = 2m = 2 The value of k, a dimensionless constant, cannot be obtained by dimensional analysis . 1.14 2π g 1.15 (a) This is incorrect since the units of [ax] are m2/s2, while the units of [v] are m/s. (b) This is correct since the units of [y] are m, and cos(kx) is dimensionless if [k] is in m-1. 1.16 l = L L/T 2 = T2 = T Inserting the proper units for everything except G, kg m = G[kg]2 s2 [m]2 Multiply both sides by [m]2 and divide by [kg]2; the units of G are m3 kg · s2 1.17 One month is 1 mo = (30 day)(24 hr/day)(3600 s/hr) = 2.592 × 106 s Applying units to the equation, V = (1.50 Mft3/mo)t + (0.00800 Mft3/mo2)t2 Since 1 Mft3 = 106 ft3, V = (1.50 × 106 ft3/mo)t + (0.00800 × 106 ft3/mo2)t2 © 2000 by Harcourt College Publishers. All rights reserved. 5 Chapter 1 Solutions 6 Converting months to seconds, V= 1.50 × 106 ft3/mo 0.00800 × 106 ft3/mo2 2 t + t 2.592 × 106 s/mo (2.592 × 106 s/mo)2 Thus, V[ft3] = (0.579 ft3/s)t + (1.19 × 10 -9 ft3/s 2)t2 *1.18 Apply the following conversion factors: 1 in = 2.54 cm, 1 d = 86400 s, 100 cm = 1 m, and 109 nm = 1 m -2 9 1 in/day (2.54 cm/in)(10 m/cm)(10 nm/m) = 9.19 nm/s 86400 s/day 32 This means the proteins are assembled at a rate of many layers of atoms each second! 1.19 Area A = (100 ft)(150 ft) = 1.50 × 104 ft2, so A = (1.50 × 104 ft2)(9.29 × 10-2 m2/ft2) = 1.39 × 103 m2 Goal Solution A rectangular building lot is 100 ft by 150 ft. Determine the area of this lot in m2. G: We must calculate the area and convert units. Since a meter is about 3 feet, we should expect the area to be about A ≈ (30 m)(50 m) = 1 500 m2. O: Area = Length × Width. Use the conversion: 1 m = 3.281 ft. A: A = L × W = (100 ft) 1m 1m (150 ft ) = 1 390 m2 3.281 ft 3.281 ft L: Our calculated result agrees reasonably well with our initial estimate and has the proper units of m2. Unit conversion is a common technique that is applied to many problems. 1.20 (a) V = (40.0 m)(20.0 m)(12.0 m) = 9.60 × 103 m3 V = 9.60 × 103 m3 (3.28 ft/1 m)3 = 3.39 × 10 5 ft3 © 2000 by Harcourt College Publishers. All rights reserved. Chapter 1 Solutions (b) The mass of the air is m = ρairV = (1.20 kg/m3)(9.60 × 103 m3) = 1.15 × 104 kg The student must look up weight in the index to find Fg = mg = (1.15 × 104 kg)(9.80 m/s2) = 1.13 × 105 N Converting to pounds, Fg = (1.13 × 105 N)(1 lb/4.45 N) = 2.54 × 104 lb *1.21 (a) Seven minutes is 420 seconds, so the rate is r= 30.0 gal = 7.14 × 10-2 gal/s 420 s Converting gallons first to liters, then to m3, (b) r = 7.14 × 10 -2 gal 3.786 L 10-3 m3 s 1 gal 1 L r = 2.70 × 10-4 m3/s At that rate, to fill a 1-m3 tank would take (c) 1 m3 1 hr = 1.03 hr 4 3 2.70 × 10 m /s 3600 s t= 1.22 v = 5.00 furlongs 220 yd 0.9144 m 1 fortnight 1 day 1 hr fortnight 1 furlong 1 yd 14 days 24 hrs 3600 s = 8.32 × 10-4 m/s This speed is almost 1 mm/s; so we might guess the creature was a snail, or perhaps a sloth. 1.23 It is often useful to remember that the 1600-m race at track and field events is approximately 1 mile in length. To be precise, there are 1609 meters in a mile. Thus, 1 acre is equal in area to 1 mi2 1609 m 2 = 4.05 × 103 m2 640 acres mi (1 acre) © 2000 by Harcourt College Publishers. All rights reserved. 7 Chapter 1 Solutions 8 1.24 Volume of cube = L3 = 1 quart (Where L = length of one side of the cube.) Thus, 1 gallon 3.786 liters 1000 cm3 L3 = (1 quart) = 946 cm3, and 4 quarts 1 gallon 1 liter L = 9.82 cm 1.25 The mass and volume, in SI units, are m = (23.94 g) 1 kg = 0.02394 kg 1000 g V = (2.10 cm3)(10-2 m/cm)3 = 2.10 × 10-6 m3 Thus, the density is ρ= m 0.02394 kg = = 1.14 × 104 kg/m3 V 2.10 × 10-6 m 3 Goal Solution A solid piece of lead has a mass of 23.94 g and a volume of 2.10 cm3. From these data, calculate the density of lead in SI units (kg/m3). G: From Table 1.5, the density of lead is 1.13 × 104 kg/m3, so we should expect our calculated value to be close to this number. This density value tells us that lead is about 11 times denser than water, which agrees with our experience that lead sinks. m O: Density is defined as mass per volume, in ρ = . We must convert to SI units in the calculation. V A: ρ = 23.94 g 1 kg 100 cm 3 = 1.14 × 104 kg/m3 2.10 cm3 1000 g 1 m L: At one step in the calculation, we note that one million cubic centimeters make one cubic meter. Our result is indeed close to the expected value. Since the last reported significant digit is not certain, the difference in the two values is probably due to measurement uncertainty and should not be a concern. One important common-sense check on density values is that objects which sink in water must have a density greater than 1 g/cm3, and objects that float must be less dense than water. © 2000 by Harcourt College Publishers. All rights reserved. Chapter 1 Solutions 1.26 (a) We take information from Table 1.1: 1 LY = (9.46 × 1015 m) 1 AU = 6.31 × 104 AU 1.50 × 1011 m (b) The distance to the Andromeda galaxy is 2 × 1022 m = (2 × 1022 m) 1 AU = 1.33 × 1011 AU 11 1.50 × 10 m mSun 1.99 × 1030 kg = = 1.19 × 1057 atoms matom 1.67 × 10-27 kg 1.27 Natoms = 1.28 1 mi = 1609 m = 1.609 km; thus, to go from mph to km/h, multiply by 1.609. 1.29 (a) 1 mi/h = 1.609 km/h (b) 55 mi/h = 88.5 km/h (c) 65 mi/h = 104.6 km/h. Thus, ∆v = 16.1 km/h (a) 6 × 10 12 $ 1 hr 1 day 1 yr = 190 years 1000 $/s 3600 s 24 hr 365 days (b) The circumference of the Earth at the equator is 2π (6378 × 103 m) = 4.01 × 107 m. The length of one dollar bill is 0.155 m so that the length of 6 trillion bills is 9.30 × 1011 m. Thus, the 6 trillion dollars would encircle the Earth 9.30 × 1011 m = 2.32 × 104 times 4.01 × 107 m Goal Solution At the time of this book’s printing, the U.S. national debt is about $6 trillion. (a) If payments were made at the rate of $1 000 per second, how many years would it take to pay off a $6-trillion debt, assuming no interest were charged? (b) A dollar bill is about 15.5 cm long. If six trillion dollar bills were laid end to end around the Earth’s equator, how many times would they encircle the Earth? Take the radius of the Earth at the equator to be 6 378 km. (Note: Before doing any of these calculations, try to guess at the answers. You may be very surprised.) (a) G: $6 trillion is certainly a large amount of money, so even at a rate of $1000/second, we might guess that it will take a lifetime (~ 100 years) to pay off the debt. O: Time to repay the debt will be calculated by dividing the total debt by the rate at which it is repaid. © 2000 by Harcourt College Publishers. All rights reserved. 9 Chapter 1 Solutions 10 A: T = $6 trillion $6 × 10 12 = = 190 yr $1000/s ($1000/s)(3.16 × 107 s/yr) L: OK, so our estimate was a bit low. $6 trillion really is a lot of money! (b) G: We might guess that 6 trillion bills would encircle the Earth at least a few hundred times, maybe more since our first estimate was low. O: The number of bills can be found from the total length of the bills placed end to end divided by the circumference of the Earth. A: N = L (6 × 1012)(15.5 cm)(1 m/100 cm) = = 2.32 × 104 times C 2π 6.37 × 106 m L: OK, so again our estimate was low. Knowing that the bills could encircle the earth more than 20 000 times, it might be reasonable to think that 6 trillion bills could cover the entire surface of the earth, but the calculated result is a surprisingly small fraction of the earth’s surface area! 1.30 (a) (3600 s/hr)(24 hr/day)(365.25 days/yr) = 3.16 × 107 s/yr (b) Vmm = 4 3 4 πr = π (5.00 × 10-7 m)3 = 5.24 × 10-19 m3 3 3 Vcube 1 m3 = = 1.91 × 1018 micrometeorites Vmm 5.24 × 10-19 m 3 This would take 1.31 V = At, so 1.32 V= t= 1.91 × 1018 micrometeorites = 6.05 × 1010 yr 3.16 × 107 micrometeorites/yr V 3.78 × 10-3 m3 = = 1.51 × 10-4 m (or 151 µm) A 25.0 m2 1 [(13.0 acres)(43560 ft2/acre)] Bh = (481 ft) 3 3 = 9.08 × 107 ft3, or 2.83 × 10-2 m3 V = (9.08 × 107 ft3) 1 ft 3 6 = 2.57 × 10 m h B 3 © 2000 by Harcourt College Publishers. All rights reserved. Chapter 1 Solutions 6 1.33 Fg = (2.50 tons/block)(2.00 × 10 blocks)(2000 lb/ton) = 1.00 × 1010 lbs 1.34 The area covered by water is 2 Aw = 0.700 AEarth = (0.700)(4π REarth ) = (0.700)(4π)(6.37 × 106 m)2 = 3.57 × 1014 m2 The average depth of the water is d = (2.30 miles)(1609 m/l mile) = 3.70 × 103 m The volume of the water is V = Awd = (3.57 × 1014 m2)(3.70 × 103 m) = 1.32 × 1018 m3 and the mass is m = ρV = (1000 kg/m3)(1.32 × 1018 m3) = 1.32 × 1021 kg *1.35 SI units of volume are in m3: V = (25.0 acre-ft) *1.36 (a) 43560 ft 2 1 acre 3 0.3048 m = 3.08 × 104 m3 1 ft dnucleus, scale = dnucleus, real datom, scale datom, real 300 ft 1.06 × 10-10 m = (2.40 × 10-15 m) = 6.79 × 10-3 ft, or dnucleus, scale = (6.79 × 10-3 ft)(304.8 mm/1 ft) = 2.07 mm (b) 3 V atom ratom 4 π ratom/3 = = Vnucleus 4 π r3 r nucleus nucleus/3 1.06 × 10-10 m 2.40 × 10-15 m = 1.37 3 3 datom d nucleus = 3 = 8.62 × 10 13 times as large The scale factor used in the "dinner plate" model is S= 0.25 m = 2.5 × 10-6 m/lightyears 1.0 × 10 5 lightyears The distance to Andromeda in the scale model will be Dscale = DactualS = (2.0 × 106 lightyears)(2.5 × 10-6 m/lightyears) = 5.0 m © 2000 by Harcourt College Publishers. All rights reserved. 11 12 Chapter 1 Solutions (a) AEarth 4π rEarth rEarth 2 (6.37 × 106 m)(100 cm/m) 2 = = = = 13.4 2 A Moon 4πrMoon rMoon 1.74 × 108 cm (b) VEarth 4πrEarth /3 rEarth 3 (6.37 × 106 m)(100 cm/m) 3 = = = = 49.1 V Moon 4πrMoon3/3 rMoon 1.74 × 108 cm 2 1.38 3 1.39 To balance, mFe = mAl or ρFeVFe = ρAlVAl 4 4 3 3 ρFe π rFe = ρAl π rAl 3 3 ρFe 1/3 ρAl rAl = rFe 7.86 1/3 = 2.86 cm 2.70 rAl = (2.00 cm) 1.40 The mass of each sphere is 4πρAlrAl mA1 = ρAlVAl = 3 3 4π ρFerFe = 3 3 and mFe = ρFeVFe Setting these masses equal, 3 3 4πρFerFe 4πρFerFe = and rAl = rFe 3 3 1.41 3 ρFe/ρAl The volume of the room is 4 × 4 × 3 = 48 m3 , while the volume of one ball is 4π 0.038 m 3 = 2.87 × 10-5 m3. 3 2 Therefore, one can fit about 48 2.87 × 10-5 ∼ 106 ping-pong balls in the room. As an aside, the actual number is smaller than this because there will be a lot of space in the room that cannot be covered by balls. In fact, even in the best arrangement, the so-called "best π 2 = 0.74 so that at least 26% of the space will be empty. Therefore, the 6 above estimate reduces to 1.67 × 106 × 0.740 ∼ 106. packing fraction" is © 2000 by Harcourt College Publishers. All rights reserved. Chapter 1 Solutions 13 Goal Solution Estimate the number of Ping-Pong balls that would fit into an average-size room (without being crushed). In your solution state the quantities you measure or estimate and the values you take for them. G: Since the volume of a typical room is much larger than a Ping-Pong ball, we should expect that a very large number of balls (maybe a million) could fit in a room. O: Since we are only asked to find an estimate, we do not need to be too concerned about how the balls are arranged. Therefore, to find the number of balls we can simply divide the volume of an average-size room by the volume of an individual Ping-Pong ball. A: A typical room (like a living room) might have dimensions 15 ft × 20 ft × 8 ft. Using the approximate conversion 1 ft = 30 cm, we find 30 cm 3 = 7 × 107 cm3 1 ft Vroom ≈ 15 ft × 20 ft × 8 ft = 2400 ft3 A Ping-Pong ball has a diameter of about 3 cm, so we can estimate its volume as a cube: Vball ≈ (3 × 3 × 3) cm3 = 30 cm3 The number of Ping-Pong balls that can fill the room is N≈ V room 7 × 107 cm3 = = 2 × 106 balls ~ 106 balls V ball 30 cm3 L: So a typical room can hold about a million Ping-Pong balls. This problem gives us a sense of how big a million really is. *1.42 It might be reasonable to guess that, on average, McDonalds sells a 3 cm × 8 cm × 10 cm = 240 cm3 medium-sized box of fries, and that it is packed 3/4 full with fries that have a cross section of 1/2 cm × 1/2 cm. Thus, the typical box of fries would contain fries that stretched a total of 3 V 3 240 cm3 = = 720 cm = 7.2 m 4 A 4 (0.5 cm)2 L= 250 million boxes would stretch a total distance of (250 × 106 box)(7.2 m/box) = 1.8 × 109 m. But we require an order of magnitude, so our answer is 109 m = 1 million kilometers . *1.43 A reasonable guess for the diameter of a tire might be 2.5 ft, with a circumference of about 8 ft. Thus, the tire would make (50 000 mi)(5280 ft/mi)(1 rev/8 ft) = 3 × 107 rev ∼ 107 rev © 2000 by Harcourt College Publishers. All rights reserved. 14 1.44 Chapter 1 Solutions A typical raindrop is spherical and might have a radius of about 0.1 inch. Its volume is then approximately 4 × 10-3 in3. Since 1 acre = 43,560 ft2, the volume of water required to cover it to a depth of 1 inch is 43,560 ft2 144 in2 ≈ 6.3 × 106 in3. 1 acre 1 ft 2 (1 acre)(1 inch) = (1 acre · in) The number of raindrops required is n= *1.45 volume of water required volume of a single drop ≈ 6.3 × 106 in3 = 1.6 × 109 ∼ 109 3 3 4 × 10 in In order to reasonably carry on photosynthesis, we might expect a blade of grass to require at least 1/16 in2 = 43 × 10-5 ft2. Since 1 acre = 43,560 ft2, the number of blades of grass to be expected on a quarter-acre plot of land is about n= total area (0.25 acre)(43,560 ft2/acre) = area per blade 43 × 10-5 ft2/blade = 2.5 × 107 blades ∼ 107 blades 1.46 Since you have only 16 hours (57,600 s) available per day, you can count only $57,600 per day. Thus, the time required to count $1 billion dollars is t= 10 9 dollars 1 year = 47.6 years 4 5.76 × 10 dollars/day 365 days Since you are at least 18 years old, you would be beyond age 65 before you finished counting the money. It would provide a nice retirement, but a very boring life until then. We would not advise it. 1.47 Assume the tub measure 1.3 m by 0.5 m by 0.3 m. One-half of its volume is then V = (0.5)(1.3 m)(0.5 m)(0.3 m) = 0.10 m3 The mass of this volume of water is mwater = ρwaterV= (1000 kg/m3)(0.10 m3) = 100 kg ~102 kg Pennies are now mostly zinc, but consider copper pennies filling 50% of the volume of the tub. The mass of copper required is mcopper = ρcopperV = (8930 kg/m )(0.10 m ) = 893 kg 3 3 3 ~10 kg © 2000 by Harcourt College Publishers. All rights reserved. Chapter 1 Solutions *1.48 The typical person probably drinks 2 to 3 soft drinks daily. Perhaps half of these were in aluminum cans. Thus, we will estimate 1 aluminum can disposal per person per day. In the U.S. there are ∼250 million people, and 365 days in a year, so (250 × 106 cans/day)(365 days/year) ≈ 1010 cans are thrown away or recycled each year. Guessing that each can weighs around 1/10 of an ounce, we estimate this represents (1010 cans)(0.1 oz/can)(1 lb/16 oz)(1 ton/2000 lb) ≈ 3.1 × 105 tons/year. ∼105 tons 1.49 Assume: Total population = 107; one out of every 100 people has a piano; one tuner can serve about 1,000 pianos (about 4 per day for 250 weekdays, assuming each piano is tuned once per year). Therefore, # tuners ~ 1 tuner 1 piano (107 people) = 100 1000 pianos 100 people 1.50 (a) 1.51 (a) 2 (b) 4 (c) 3 (d) 2 πr2 = π (10.5 m ± 0.2 m)2 = π [ (10.5 m)2 ± 2(10.5 m)(0.2 m) + (0.2 m)2] = 346 m2 ± 13 m2 (b) 1.52 15 2πr = 2π (10.5 m ± 0.2 m) = 66.0 m ± 1.3 m ( a ) 756.?? 37.2? 0.83 + 2.5? 796./ 5/3 = 797 (b) 0.0032 (2 s.f.) × 356.3 (4 s.f.) = 1.14016 = (2 s.f.) 1.1 (c) 5.620 (4 s.f.) × π (> 4 s.f.) = 17.656 = (4 s.f.) 17.66 © 2000 by Harcourt College Publishers. All rights reserved. 16 1.53 Chapter 1 Solutions r = (6.50 ± 0.20) cm = (6.50 ± 0.20) × 10-2 m m = (1.85 ± 0.02) kg m ρ= also, 4 π r3 3 δρ δm 3δr = + ρ m r In other words, the percentages of uncertainty are cumulative. Therefore, ρ= δρ 0.02 3(0.20) = + = 0.103 ρ 1.85 6.50 1.85 4 π (6.5 × 10 -2 m) 3 3 = 1.61 × 103 kg/m3 and ρ ± δρ = (1.61 ± 0.17) × 103 kg/m3 1.54 (a) 3 1.55 The distance around is 38.44 m + 19.5 m + 38.44 m + 19.5 m = 115.88 m, but this answer must be rounded to 115.9 m because the distance 19.5 m carries information to only one place past the decimal. (b) 4 (c) 3 (d) 2 115.9 m 19.0 m 1.56 V = 2V1 + 2V2 = 2(V1 + V2) V1 = (17.0 m + 1.0 m + 1.0 m)(1.0 m)(0.09 m) = 1.70 m3 36.0 10.0cm m 3 V2 = (10.0 m)(1.0 m)(0.090 m) = 0.900 m V = 2(1.70 m3) + 2(0.900 m3) = 5.2 m3 δV δw 0.01 m = = 0.010 = 0.006 + 0.010 + 0.011 = 0.027 = 2.7% w 1.0 m V δt 0.1 cm = = 0.011 t 9.0 cm δ l1 0.12 m l1 = 19.0 m = 0.0063 1 1 1 1 © 2000 by Harcourt College Publishers. All rights reserved. Chapter 1 Solutions *1.57 It is desired to find the distance x such that 17 x = 100 m 1000 m (i.e., such that x is the same multiple of 100 m as the multiple that 1000 m is of x) . x 2 5 2 Thus, it is seen that x = (100 m)(1000 m) = 1.00 × 10 m , and therefore x = 5 1.00 × 10 m 2 = 316 m . 1.58 9.00 × 10-7 kg = 9.80 × 10–10 m3. If the diameter of a molecule is d, 918 kg/m3 then that same volume must equal d(πr2) = (thickness of slick)(area of oil slick) where r = 0.418 m. Thus, The volume of oil equals V = d= 1.59 9.80 × 10-10 m3 -9 2 = 1.79 × 10 m π (0.418 m) Vtotal Vtotal (Adrop) = 3 (4πr2) V drop 4πr /3 Atotal = (N)(Adrop) = 3Vtotal 30.0 × 10-6 m 3 2 = 3 = 4.50 m 5 r 2.00 × 10 m = 1.60 1.61 α' (deg) 15.0 20.0 25.0 24.0 24.4 24.5 24.6 24.7 2πr = 15.0 m α (rad) 0.262 0.349 0.436 0.419 0.426 0.428 0.429 0.431 tan( α ) 0.268 0.364 0.466 0.445 0.454 0.456 0.458 0.460 sin( α ) 0.259 0.342 0.423 0.407 0.413 0.415 0.416 0.418 difference 3.47% 6.43% 10.2% 9.34% 9.81% 9.87% 9.98% 10.1% r = 2.39 m h = tan55.0° r h = (2.39 m)tan(55.0°) = 3.41 m © 2000 by Harcourt College Publishers. All rights reserved. 24.6° 18 Chapter 1 Solutions h 55° r © 2000 by Harcourt College Publishers. All rights reserved. Chapter 1 Solutions *1.62 (a) [V] = L3, [A] = L2, [h] = L [V] = [A][h] L3 = L3L = L3. Thus, the equation is dimensionally correct. V cylinder = πR 2h = (πR 2)h = Ah, where A = π R 2 (b) Vrectangular object = 1.63 l wh = ( l w)h = Ah, where A = l w The actual number of seconds in a year is (86,400 s/day)(365.25 day/yr) = 31,557,600 s/yr The percentage error in the approximation is thus (π × 107 s/yr) – (31,557,600 s/yr) 31,557,600 s/yr *1.64 × 100% = 0.449% From the figure, we may see that the spacing between diagonal planes is half the distance between diagonally adjacent atoms on a flat plane. This diagonal distance may be obtained from the Pythagorean theorem, Ldiag = distance L2 + L2 . Thus, since the atoms are separated by a L = 0.200 nm, the diagonal planes are separated *1.65 (a) 1.67 t= (Vol rate of flow) 16.5 cm3/s = = 0.529 cm/s 2 (Area: π D /4) π (6.30 cm)2/4 Likewise, at a 1.35 cm diameter, v= *1.66 L2 + L2 = 0.141 nm The speed of flow may be found from v= (b) 1 2 16.5 cm3/s = 11.5 cm/s π (1.35 cm)2/4 6 V V 4(12.0 cm3 ) 1 m 10 µm = 289 µm = = = 0.0289 cm A π D2/4 π (23.0 cm)2 100 cm 1 m V20 mpg = (108 cars)(104 mi/yr) = 5.0 × 1010 gal/yr 20 mi/gal V25 mpg = (108 cars)(104 mi/yr) 10 = 4.0 × 10 gal/yr 25 mi/gal Fuel saved = V25 mpg – V20 mpg = 1.0 × 1010 gal/yr © 2000 by Harcourt College Publishers. All rights reserved. 19 20 Chapter 1 Solutions © 2000 by Harcourt College Publishers. All rights reserved. Chapter 1 Solutions 1.68 (a) 1 cubic meter of water has a mass m = ρV = (1.00 × 10-3 kg/cm3)(1.00 m3)(102 cm/m)3 = 1000 kg (b) As a rough calculation, we treat each item as if it were 100% water. cell: m = ρV = ρ Error! πR3 ) = ρ Error! π D3 ) 1 = ( 1000 kg/m3) π (1.0 × 10-6 m)3 = 5.2 × 10-16 kg 6 kidney: m = ρV = ρ Error! π R3 ) = (1.00 × 10-3 kg/cm3 )Error! 3 = Error! fly: m=ρ π D 2 h 4 π = (1 × 10-3 kg/cm3) (2.0 mm) 2(4.0 mm)(10-1 cm/mm)3 4 = 1.3 × 10-5 kg 1.69 The volume of the galaxy is πr2t = π (1021 m)2 1019 m ~ 1061 m3 If the distance between stars is 4 × 1016 m, then there is one star in a volume on the order of (4 × 1016 m)3 ~ 1050 m3. The number of stars is about 1061 m3 ~ 10 11 stars 3 10 m /star 50 © 2000 by Harcourt College Publishers. All rights reserved. 21 22 1.70 Al: Chapter 1 Solutions The density of each material is ρ = ρ= Cu: ρ = m m 4m = = V π r2h π D 2h 4(51.5 g) g = 2.75 π (2.52 cm)2(3.75 cm) cm3 The tabulated value 2.70 g is 2% cm3 smaller. 4(56.3 g) π (1.23 cm)2(5.06 cm) The tabulated value 8.92 g is 5% cm3 smaller. The tabulated value 7.86 g is 0.3% smaller. cm3 Brass: ρ = = 9.36 g cm3 4(94.4 g) g = 8.91 2 π (1.54 cm) (5.69 cm) cm3 Sn: ρ= 4(69.1 g) g = 7.68 2 π (1.75 cm) (3.74 cm) cm3 Fe: ρ= 4(216.1 g) g = 7.88 3 π (1.89 cm)2(9.77 cm) cm © 2000 by Harcourt College Publishers. All rights reserved. Chapter 2 Even Answers 2. 4. 6. 8. 10. 12. 14. 16. 18. 20. 22. 24. 26. 28. 30. 32. 34. 36. 38. 40. 42. 44. 46. 48. 50. 52. 54. 56. 58. 60. 62. 64. 66. 68. 70. 72. (a) 180 km (b) 63.4 km/h (a) 50.0 m/s (b) 41.0 m/s (a) 2v1v2/(v1 + v2) (b) 0 (a) 27.0 m (b) xf = 27.0 m + (18.0 m/s)∆t + (3.00 m/s2)(∆t)2 (c) 18.0 m/s (b) vt = 5.0 s = 23 m/s, vt = 4.0 s = 18 m/s, vt = 3.0 s = 14 m/s, vt = 2.0 s = 9.0 m/s (c) 4.6 m/s2 (d) 0 –4.00 m/s2, sign indicates that acceleration is in negative x direction (a) 20.0 m/s, 5.00 m/s (b) 262 m (c) –4 m/s2 (d) 34 m (e) 28 m (a) 13.0 m/s (b) 10.0 m/s, 16.0 m/s (c) 6.00 m/s2 (d) 6.00 m/s2 (f) The spacing of the successive positions would change with less regularity. (a) 5.25 m/s2 (b) 168 m (c) 52.5 m/s 160 ft (a) 1.87 km (b) 1.46 km (c) a1 = 3.3 m/s2 (0 ≤ t ≤ 15 s), a2 = 0 (15 s ≤ t ≤ 40 s), a3 = –5.0 m/s2 (40 s ≤ t ≤ 50 s) (d) (i) x1 = (1.67 m/s2)t2, (ii) x2 = (50 m/s)t – 375 m, (iii) x3 = (250 m/s)t – (2.5 m/s2)t2 – 4375 m (e) 37.5 m/s (a) 12.7 m/s (b) -2.30 m/s (a) x = (30.0t – t2) m, v = (30.0 – 2.00t) m/s (b) 225 m 3.10 m/s (a) –4.90 × 105 m/s2 (b) 3.57 × 10–4 s (c) 18.0 cm 200 m (a) 4.98 × 10–9 s (b) 1.20 × 1015 m/s2 11.4 s, 212 m $99.4/h 1.79 s gh (a) 96.0 ft/s downward (b) 3.07 × 103 ft/s2 upward (c) 3.13 × 10–2 s (a) 98.0 m/s (b) 490 m 7.96 s (a) a = –(10.0 × 107 m/s3)t + 3.00 × 105 m/s2; x = –(1.67 × 107 m/s3)t3 + (1.50 × 105 m/s2)t2 (b) 3.00 × 10–3 s (c) 450 m/s (d) 0.900 m (a) 0.111 s (b) 5.53 m/s 48.0 mm (a) 15.0 s (b) 30.0 m/s (c) 225 m 155 s, 129 s ~ 103 m/s2 (a) 26.4 m (b) 6.82% 1.38 × 103 m 2 vboy (c) , 0 (d) vboy, 0 h (b) a = 1.63 m/s2 downward © 2000 by Harcourt College Publishers. All rights reserved. 2 Chapter 2 Even Answers © 2000 by Harcourt College Publishers. All rights reserved. Chapter 2 Solutions *2.1 2.2 (a) – v = 2.30 m/s (b) ∆x 57.5 m – 9.20 m – v = = = 16.1 m/s ∆t 3.00 s (c) ∆x 57.5 m – 0 m – v = = = 11.5 m/s ∆t 5.00 s (a) Displacement = (8.50 × 104 m/h) 35.0 h + 130 × 103 m 60.0 x = (49.6 + 130) × 103 m = 180 km 2.3 2.4 displacement 180 km = = 63.4 km/h time (35.0 + 15.0) 60.0 + 2.00 h (b) Average velocity = (a) vav = ∆x 10 m = = 5 m/s ∆t 2s (b) vav = 5m = 1.2 m/s 4s (c) vav = x2 – x1 5 m – 10 m = = –2.5 m/s t2 – t1 4s–2s (d) vav = x2 – x1 –5 m – 5 m = = –3.3 m/s t2 – t1 7s–4s (e) vav = x2 – x1 0–0 = = 0 m/s t2 – t1 8–0 x = 10t2 For t(s) = 2.0 2.1 3.0 x(m) = 40 44.1 90 (a) ∆x 50 m – v = = = 50.0 m/s ∆t 1.0 s (b) ∆x 4.1 m – v = = = 41.0 m/s ∆t 0.1 s © 2000 by Harcourt College Publishers. All rights reserved. 2 2.5 Chapter 2 Solutions (a) Let d represent the distance between A and B. Let t1 be the time for which the walker d has the higher speed in 5.00 m/s = . Let t2 represent the longer time for the return trip t1 d d d in –3.00 m/s = – . Then the times are t1 = and t2 = . The average t2 (5.00 m/s) (3.00 m/s) speed is: Total distance – v = = Total time d+d 2d = (8.00 m/s)d d d (5.00 m/s) + (3.00 m/s) (15.0 m2/s2) 2(15.0 m2/s2) – v = = 3.75 m/s 8.00 m/s (b) She starts and finishes at the same point A. With total displacement = 0, average velocity = 0 2.6 (a) Total distance – v = Total time Let d be the distance from A to B. Then the time required is d d + . v1 v2 – And the average speed is v = (b) 2.7 2v 1 v 2 2d = d d v1 + v2 + v1 v2 With total displacement zero, her average velocity is 0 . (a) x (m) 5 0 t (s) 2 4 6 —5 (b) v = slope = 5.00 m – (–3.00 m) 8.00 m = = 1.60 m/s (6.00 s – 1.00 s) 5.00 s © 2000 by Harcourt College Publishers. All rights reserved. Chapter 2 Solutions 2.8 (a) At any time, t, the displacement is given by x = (3.00 m/s2)t2. Thus, at ti = 3.00 s: (b) 3 xi = (3.00 m/s2)(3.00 s)2 = 27.0 m At tf = 3.00 s + ∆t : xf = (3.00 m/s2)(3.00 s + ∆t)2, or xf = 27.0 m + (18.0 m/s)∆t + (3.00 m/s2)(∆t)2 (c) The instantaneous velocity at t = 3.00 s is: xf – xi = lim [(18.0 m/s) + (3.00 m/s2)∆t], or ∆t → o ∆ t ∆t → 0 v = lim v = 18.0 m/s 2.9 (a) x (m) at ti = 1.5 s, xi = 8.0 m (Point A) 12 at tf = 4.0 s, xf = 2.0 m (Point B) 10 C 8 (2.0 – 8.0) m 6.0 m – xf – xi v = = =– = –2.4 m/s tf – ti (4 – 1.5) s 2.5 s A 6 4 B 2 (b) The slope of the tangent line is found from points C and D. D 0 0 1 2 t (s) 3 4 5 6 (tC = 1.0 s, xC = 9.5 m) and (tD = 3.5 s, xD = 0), v ≅ –3.8 m/s 2.10 (c) The velocity is zero when x is a minimum. This is at t ≈ 4 s . (b) At t = 5.0 s, the slope is v ≅ 58 m ≅ 23 m/s 2.5 s 60 At t = 4.0 s, the slope is v ≅ 54 m ≅ 18 m/s 3s 40 At t = 3.0 s, the slope is v ≅ 49 m ≅ 14 m/s 3.4 s 20 At t = 2.0 s, the slope is v ≅ 36 m ≅ 9.0 m/s 4.0 s 0 (c) x (m) t (s) 0 ∆v 23 m/s – a = ≅ ≅ 4.6 m/s2 ∆t 5.0 s 2 4 v (m/s) 20 (d) Initial velocity of the car was zero . 0 t (s) 0 © 2000 by Harcourt College Publishers. All rights reserved. 2 4 4 2.11 Chapter 2 Solutions (a) v= (5 – 0) m = 5 m/s (1 – 0) s (5 – 10) m = –2.5 m/s (4 – 2) s (b) v= (c) (5 m – 5 m) v= = 0 (5 s – 4 s) x (m) 10 8 6 4 2 0 −2 (d) 2.12 0 – (–5 m) v= = +5 m/s (8 s – 7 s) 1 2 3 4 5 6 7 8 t (s) −4 −6 vf – vi 0 – 60.0 m/s – a = = = – 4.00 m/s2 tf – ti 15.0 s – 0 The negative sign in the result shows that the acceleration is in the negative x direction. *2.13 Choose the positive direction to be the outward perpendicular to the wall. v = vi + at a= 2.14 (a) ∆v 22.0 m/s – (–25.0 m/s) = = 1.34 × 104 m/s2 ∆t 3.50 × 10–3 s Acceleration is constant over the first ten seconds, so at the end v = vi + at = 0 + (2.00 m/s2)(10.0 s) = 20.0 m/s Then a = 0 so v is constant from t = 10.0 s to t = 15.0 s. And over the last five seconds the velocity changes to v = vi + at = 20.0 m/s – (3.00 m/s2)(5.00 s) = 5.00 m/s (b) In the first ten seconds x = xi + vit + 1 2 1 at = 0 + 0 + (2.00 m/s2)(10.0 s) 2 = 100 m 2 2 Over the next five seconds the position changes to x = xi + vit + 1 2 at = 100 m + 20.0 m/s (5.00 s) + 0 = 200 m 2 And at t = 20.0 s x = xi + vit + 1 2 1 at = 200 m + 20.0 m/s (5.00 s) + (–3.00 m/s2)(5.00 s) 2 = 262 m 2 2 © 2000 by Harcourt College Publishers. All rights reserved. Chapter 2 Solutions *2.15 (a) Acceleration is the slope of the graph of v vs t. a (m/s2) For 0 < t < 5.00 s, a = 0 2.0 For 15.0 s < t < 20.0 s, a = 0 For 5.0 s < t < 15.0 s, a= a= vf – vi tf – ti 8.00 – (–8.00) = 1.60 m/s2 15.0 – 5.00 1.6 1.0 0.0 t (s) 0 5 We can plot a(t) as shown. (b) a= (i) vf – vi tf – ti For 5.00 s < t < 15.0 s, ti = 5.00 s, vi = –8.00 m/s tf = 15.0 s, vf = 8.00 m/s; a= (ii) vf – vi 8.00 – (–8.00) = = 1.60 m/s2 tf – ti 15.0 – 5.00 ti = 0, vi = –8.00 m/s, tf = 20.0 s, vf = 8.00 m/s a= vf – vi 8.00 – (–8.00) = = 0.800 m/s2 tf – ti 20.0 – 0 © 2000 by Harcourt College Publishers. All rights reserved. 10 15 20 5 6 2.16 Chapter 2 Solutions (a) See the Graphs at the right. Choose x = 0 at t = 0 40 1 At t = 3 s, x = (8 m/s)(3 s) = 12 m 2 20 At t = 5 s, x = 12 m + (8 m/s)(2 s) = 28 m At t = 7 s, x = 28 m + (b) x (m) 1 (8 m/s)(2 s) = 36 m 2 0 t (s) 0 5 10 5 10 v (m/s) 10 For 0 < t < 3 s, a = (8 m/s)/3 s = 2.67 m/s2 For 3 < t < 5 s, a = 0 0 (c) For 5 s < t < 9 s, a = –(16 m/s)/4 s = – 4 m/s2 (d) At t = 6 s, x = 28 m + (6 m/s)(1 s) = 34 m (e) At t = 9 s, x = 36 m + 1 (– 8 m/s) 2 s = 28 m 2 −10 a (m/s2) 5 0 −5 2.17 x = 2.00 + 3.00t – t2, v = dx dv = 3.00 – 2.00t, a = = dt dt –2.00 At t = 3.00 s: 2.18 (a) x = (2.00 + 9.00 – 9.00) m = 2.00 m (b) v = (3.00 – 6.00) m/s = –3.00 m/s (c) a = –2.00 m/s2 (a) At t = 2.00 s, x = [3.00(2.00)2 –2.00(2.00) + 3.00] m = 11.0 m At t = 3.00 s, x = [3.00(9.00)2 –2.00(3.00) + 3.00] m = 24.0 m so t (s) ∆x 24.0 m – 11.0 m – v = = = 13.0 m/s ∆t 3.00 s – 2.00 s © 2000 by Harcourt College Publishers. All rights reserved. 5 10 t (s) Chapter 2 Solutions (b) At all times the instantaneous velocity is v= d (3.00t2 – 2.00t + 3.00) = (6.00t – 2.00) m/s dt At t = 2.00 s, v = [6.00(2.00) – 2.00] m/s = 10.0 m/s At t = 3.00 s, v = [6.00(3.00) – 2.00] m/s = 16.0 m/s (c) ∆v 16.0 m/s – 10.0 m/s – a = = = 6.00 m/s2 ∆t 3.00 s – 2.00 s (d) At all times a= d (6.00 – 2.00) = 6.00 m/s2 dt (This includes both t= 2.00 s and t= 3.00 s). 2.19 ∆v 8.00 m/s 4 = = m/s2 ∆t 6.00 s 3 (a) a= (b) Maximum positive acceleration is at t = 3 s, and is approximately 2 m/s2 (c) a = 0, at t = 6 s , and also for t > 10 s (d) Maximum negative acceleration is at t = 8 s, and is approximately –1.5 m/s2 © 2000 by Harcourt College Publishers. All rights reserved. 7 8 Chapter 2 Solutions *2.20 = reading order = velocity = acceleration a b c d e f *2.21 One way of phrasing the answer: The spacing of the successive positions would change with less regularity. Another way: The object would move with some combination of the kinds of motion shown in (a) through (e). Within one drawing, the acceleration vectors would vary in magnitude and direction. 2 From v f = vi2 + 2ax, we have (10.97 × 103 m/s)2 = 0 + 2a(220 m), so that a = 2.74 × 105 m/s2 2.22 (a) Assuming a constant acceleration: a= (b) vf – vi 42.0 m/s = = 5.25 m/s2 t 8.00 s Taking the origin at the original position of the car, x= (c) which is 2.79 × 104 times g 1 1 (v + vf) t = (42.0 m/s)(8.00 s) = 168 m 2 i 2 From vf = vi + at, the velocity 10.0 s after the car starts from rest is: vf = 0 + (5.25 m/s2)(10.0 s) = 52.5 m/s © 2000 by Harcourt College Publishers. All rights reserved. Chapter 2 Solutions *2.23 (a) x – xi = 1 1 (v + v) t becomes 40 m = (vi + 2.80 m/s)(8.50 s) 2 i 2 which yields (b) 2.24 a= 9 vi = 6.61 m/s v – vi 2.80 m/s – 6.61 m/s = = – 0.448 m/s2 t 8.50 s Suppose the unknown acceleration is constant as a car moving at vi = 35.0 mi/h comes to a v = 0 stop in x – xi = 40.0 ft. We find its acceleration from 2 v 2 = v i + 2a(x – xi ) 2 (v 2 – v i ) a= 2(x – x i ) = 0 – (35.0 mi/h)2 5280 ft 2 1 h 2 = – 32.9 ft/s2 2(40.0 ft) 1 mi 3600 s Now consider a car moving at vi = 70.0 mi/h and stopping to v = 0 with a = – 32.9 ft/s2. From the same equation its stopping distance is 2 x – xi = = v 2 – vi 2a 0 – (70.0 mi/h)2 5280 ft 2 1 h 2(–32.9 ft/s2) 1 mi 3600 s 2 = 160 ft 2.25 Given vi = 12.0 cm/s when xi = 3.00 cm (t = 0), and at t = 2.00 s, x = –5.00 cm ∆x = vit + 1 2 at ; 2 ⇒ x – xi = vit + 1 2 at ; 2 – 5 . 0 0 – 3.00 = 12.0(2.00) + 1 a (2.00)2 ; 2 ⇒ – 8 . 0 0 = 24.0 + 2a a=– 32.0 = –16.0 cm/s2 2 © 2000 by Harcourt College Publishers. All rights reserved. 10 Chapter 2 Solutions Goal Solution A body moving with uniform acceleration has a velocity of 12.0 cm/s in the positive x direction when its x coordinate is 3.00 cm. If its x coordinate 2.00s later is –5.00 cm, what is the magnitude of its acceleration? G: Since the object must slow down as it moves to the right and then speeds up to the left, the acceleration must be negative and should have units of cm/s2. O: First we should sketch the problem to see what is happening: x (cm) —5 0 initial 5 final —5 0 5 Here we can see that the object travels along the x-axis, first to the right, slowing down, and then speeding up as it travels to the left in the negative x direction. We can show the position as a function of time with the notation: x(t) x(0) = 3.00 cm, x(2.00) = –5.00 cm, and v(0) = 12.0 cm/s A: Use the kinematic equation x – xi = vit + 1 2 at , and solve for a. 2 a= 2(x – x i – v it) t2 a= 2[–5.00 cm – 3.00 cm – (12.0 cm/s)(2.00 s)] (2.00 s)2 a = –16.0 cm/s2 L: The acceleration is negative as expected and it has the correct units of cm/s2. It also makes sense that the magnitude of the acceleration must be greater than 12 cm/s2 since this is the acceleration that would cause the object to stop after 1 second and just return the object to its starting point after 2 seconds. © 2000 by Harcourt College Publishers. All rights reserved. Chapter 2 Solutions 2.26 (a) Total displacement = area under the (v, t) curve from t = 0 to 50 s. 1 1 ∆x = (50 m/s)(15 s) + (50 m/s)(40 – 15)s + (50 m/s)(10 s) = 1875 m 2 2 (b) From t = 10 s to t = 40 s, displacement (area under the curve) is 1 ∆x = (50 m/s + 33 m/s)(5 s) + (50 m/s)(25 s) = 1457 m 2 a (m/s2) (c) 5 ∆v (50 – 0) m/s a1 = = = 3.3 m/s2 ∆t 15 s – 0 0 ≤ t ≤ 15 s: 0 15 s < t < 40 s: a 2 = 0 40 s ≤ t ≤ 50 s: (d) (i) x1 = 0 + (ii) x2 = a3 = 11 t (s) 10 ∆v (0 – 50) m/s = = –5.0 m/s2 ∆t 50 s – 40 s 1 1 a t2 = (3.3 m/s2) t2, 2 1 2 or 20 30 40 50 —5 x1 = (1.67 m/s2)t2 1 (15 s) [50 m/s – 0] + (50 m/s)(t – 15 s), or 2 x2 = (50 m/s)t – 375 m area under v vs t 1 ( i i i ) For 40 s ≤ t ≤ 50 s, x3 = from t = 0 to 40 s + a3(t – 40 s)2 + (50 m/s)(t – 40 s) 2 or x3 = 375 m + 1250 m + 1 (–5.0 m/s2)(t – 40 s) 2 + (50 m/s)(t – 40 s) which reduces to 2 x3 = (250 m/s)t – (2.5 m/s2)t2 – 4375 m (e) *2.27 (a) total displacement 1875 m – v = = = 37.5 m/s total elapsed time 50 s Compare the position equation x = 2.00 + 3.00t – 4.00t2 to the general form 1 x = xi + vit + at2 to recognize that: 2 xi = 2.00 m, vi = 3.00 m/s, and a = –8.00 m/s2 The velocity equation, v = vi + at, is then v = 3.00 m/s – (8.00 m/s2)t. 3 The particle changes direction when v = 0, which occurs at t = s. 8 The position at this time is: 2 x = 2.00 m + (3.00 m/s)Error! s) – (4.00 m/s2)Error! s) = Error! © 2000 by Harcourt College Publishers. All rights reserved. 12 Chapter 2 Solutions (b) From x = xi + vit + t=– 2vi . a t= 1 2 at , observe that when x = xi, the time is given by 2 Thus, when the particle returns to its initial position, the time is –2(3.00 m/s) 3 = s and the velocity is – 8.00 m/s2 4 v = 3.00 m/s – (8.00 m/s2) 3 s = –3.00 m/s 4 2.28 2.29 vi = 5.20 m/s (a) v(t = 2.50 s) = vi + at = 5.20 m/s + (3.00 m/s2)(2.50 s) = 12.7 m/s (b) v(t = 2.50 s) = vi + at = 5.20 m/s + (–3.00 m/s2)(2.50 s) = –2.30 m/s (a) x= 1 2 at 2 (Eq 2.11) 1 400 m = (10.0 m/s2) t2 2 t = 8.94 s (b) v = at (Eq 2.8) v = (10.0 m/s2)(8.94 s) = 89.4 m/s 2.30 (a) Take ti = 0 at the bottom of the hill where xi = 0, vi = 30.0 m/s, and a = –2.00 m/s2. Use these values in the general equation x = xi + vi t + to find 1 2 at 2 x = 0 + 30.0t m/s + when t is in seconds 1 (–2.00 m/s2) t2 2 x = (30.0t – t2)m To find an equation for the velocity, use v = vi + at = 30.0 m/s + (–2.00 m/s2)t v = (30.0 – 2.00t) m/s © 2000 by Harcourt College Publishers. All rights reserved. Chapter 2 Solutions (b) The distance of travel x becomes a maximum, xmax, when v = 0 (turning point in the motion). Use the expressions found in part (a) for v to find the value of t when x has its maximum value: From v = (30.0 – 2.00t) m/s , v=0 when t = 15.0 s Then xmax = (30.0t – t2) m = (30.0)(15.0) – (15.0)2 = 225 m 2.31 (a) vi = 100 m/s, a = –5.00 m/s2 2 v 2 = v i + 2ax 0 = (100)2 – 2(5.00)x x = 1000 m (b) *2.32 and t = 20.0 s No, at this acceleration the plane would overshoot the runway. In the simultaneous equations v = v + a t x – x = 1 (v + v t ) 2 x xi i x xi x we have v = v – (5.60 m/s )(4.20 s) 1 62.4 m = 2 (v + v )4.20 s x 2 xi xi x So substituting for vxi gives 62.4 m = 1 [v + (5.60 m/s2)(4.20 s) + vx]4.20 s 2 x 1 14.9 m/s = vx + (5.60 m/s2)(4.20 s) 2 vx = 3.10 m/s © 2000 by Harcourt College Publishers. All rights reserved. 13 14 Chapter 2 Solutions *2.33 Take any two of the standard four equations, such as v x = v xi + a x t 1 x – xi = 2 (v xi + v x )t solve one for vxi, and substitute into the other: v xi = v x – a x t x – xi = 1 (v – a x t + v x ) t 2 x Thus x – x i = v x t – 1 a t2 2 x Back in problem 32, 1 62.4 m = vx(4.20 s) – (–5.60 m/s2)(4.20 s) 2 vx = 2.34 2 62.4 m – 49.4 m = 3.10 m/s 4.20 s We assume the bullet is a cylinder. It slows down just as its front end pushes apart wood fibers. 2 (a) a= v2 – v i (280 m/s)2 – (420 m/s)2 = = –4.90 × 105 m/s2 2x 2(0.100 m) (b) t= 0.100 0.020 + = 3.57 × 10-4 s 350 280 (c) vi = 420 m/s, 2 v = 0; a = – 4.90 × 105 m/s2; v2 = vi2 + 2ax 2 vi v 2 – vi (420 m/s)2 x= = =– 2a 2a (–2 × 4.90 × 105 m/s2) x = 0.180 m *2.35 (a) The time it takes the truck to reach 20.0 m/s is found from v = vi + at, solving for t yields t = v – vi 20.0 m/s – 0 m/s = = 10.0 s a 2.00 m/s2 The total time is thus 10.0 s + 20.0 + 5.00 s = 35.0 s © 2000 by Harcourt College Publishers. All rights reserved. Chapter 2 Solutions (b) The average velocity is the total distance traveled divided by the total time taken. The distance traveled during the first 10.0 s is – x1 = v t = 0 + 20.0 (10.0) = 100 m 2 The distance traveled during the next 20.0 s is x2 = vit + 1 2 at = (20.0)(20.0) + 0 = 400 m, a being 0 for this interval. 2 The distance traveled in the last 5.00 s is – x3 = v t = 20.0 + 0 (5.00) = 50.0 m 2 The total distance x = x1 + x2 + x3 = 100 + 400 + 50.0 = 550 m, and the average velocity is given by x 550 – v = = = 15.7 m/s t 35.0 *2.36 1 2 at yields x = 20.0(40.0) – 1.00(40.0)2/2 = 0, which is obviously 2 wrong. The error occurs because the equation used is for uniformly accelerated motion, which this is not. The acceleration is –1.00 m/s2 for the first 20.0 s and 0 for the last 20.0 s. The distance traveled in the first 20.0 s is: Using the equation x = vit + x = vit + 1 2 at = (20.0)(20.0) – 1.00(20.02)/2 = 200 m 2 During the last 20.0 s, the train is at rest. Thus, the total distance traveled in the 40.0 s interval is 200 m . 2.37 2.38 v – vi 632(5280/3600) = = – 662 ft/s2 = –202 m/s2 t 1.40 (a) a= (b) x = vit + 1 2 1 at = (632)(5280/3600)(1.40) – 662(1.40)2 = 649 ft = 198 m 2 2 We have vi = 2.00 × 104 m/s, v = 6.00 × 106 m/s, x – xi = 1.50 × 10–2 m (a) 15 1 x – xi = (vi + v) t 2 t= 2(x – xi) 2(1.50 × 10–2 m) = = 4.98 × 10–9 s vi + v 2.00 × 104 m/s + 6.00 × 106 m/s © 2000 by Harcourt College Publishers. All rights reserved. 16 Chapter 2 Solutions (b) 2 v 2 = v i + 2a(x – xi ) 2 v2 – v i (6.00 × 106 m/s)2 – (2.00 × 104 m/s)2 a= = = 1.20 × 1015 m/s2 2(x – x i) 2(1.50 × 10–2 m) 2.39 (a) Take initial and final points at top and bottom of the incline. If the ball starts from rest, vi = 0, a = 0.500 m/s2, x – xi = 9.00 m 2 Then v2 = v i + 2a(x – xi) = 02 + 2(0.5 00 m/s2) 9.00 m v = 3.00 m/s (b) x – xi = vit + 1 2 at 2 1 9.00 m = 0 + (0.500 m/s2) t2 2 t = 6.00 s (c) Take initial and final points at the bottom of the planes and the top of the second plane, respectively. vi = 3.00 m/s v=0 x – xi = 15.00 m 2 v 2 = v i + 2a(x – xi) gives 2 a= (v 2 – v i ) [0 – (3.00 m/s)2] = 2(x – x i ) 2(15.0 m) = – 0.300 m/s2 (d) Take initial point at the bottom of the planes and final point 8.00 m along the second: vi = 3.00 m/s x – xi = 8.00 m a = – 0.300 m/s2 2 v 2 = v i + 2a(x – xi) = (3.00 m/s)2 + 2(– 0.300 m/s2)(8.00 m) = 4.20 m2/s2 v = 2.05 m/s © 2000 by Harcourt College Publishers. All rights reserved. Chapter 2 Solutions 2.40 17 Take the original point to be when Sue notices the van. Choose the origin of the x-axis at Sue's car. For her we have xis = 0 as = –2.00 m/s2 vis = 30.0 m/s so her position is given by xs(t) = xis + vis t + 1 a t2 2 s 1 = (30.0 m/s)t + (–2.00 m/s2) t2 2 For the van, xiv = 155 m xv(t) = xiv + vivt + viv = 5.00 m/s av = 0 and 1 a t2 = 155 m + (5.00 m/s)t + 0 2 v To test for a collision, we look for an instant tc when both are at the same place: 2 30.0tc – tc = 155 + 5.00tc 2 0 = tc – 25.0tc + 155 From the quadratic formula tc = 25.0 ± (25.0)2 – 4(155) = 13.6 s or 11.4 s 2 The smaller value is the collision time. (The larger value tells when the van would pull ahead again if the vehicles could move through each other). The wreck happens at position 155 m + (5.00 m/s)(11.4 s) = 212 m . 2.41 Choose the origin (y = 0, t = 0) at the starting point of the ball and take upward as positive. Then, yi = 0, vi = 0, and a = –g = –9.80 m/s2. The position and the velocity at time t become: y – yi = vit + 1 2 1 1 at ⇒ y = – gt2 = – (9.80 m/s2) t2 2 2 2 and v = vi + at ⇒ v = – gt = –(9.80 m/s2)t (a) 1 at t = 1.00 s: y = – (9.80 m/s2)(1.00 s) 2 = –4.90 m 2 1 at t = 2.00 s: y = – (9.80 m/s2)(2.00 s) 2 = –19.6 m 2 1 at t = 3.00 s: y = – (9.80 m/s2)(3.00 s) 2 = –44.1 m 2 © 2000 by Harcourt College Publishers. All rights reserved. 18 Chapter 2 Solutions (b) at t = 1.00 s: v = –(9.80 m/s2)(1.00 s) = –9.80 m/s at t = 2.00 s: v = –(9.80 m/s2)(2.00 s) = –19.6 m/s at t = 3.00 s: v = –(9.80 m/s2)(3.00 s) = –29.4 m/s *2.42 Assume that air resistance may be neglected. Then, the acceleration at all times during the flight is that due to gravity, a = –g = –9.80 m/s2. During the flight, Goff went 1 mile (1609 m) up and then 1 mile back down. Determine his speed just after launch by considering his upward flight: 2 2 v 2 = v i + 2a(y – yi) ⇒ 0 = v i – 2(9.80 m/s2)(1609 m) ⇒ vi = 178 m/s His time in the air may be found by considering his motion from just after launch to just before impact: y – yi = vit + 1 2 1 at ⇒ 0 = (178 m/s)t – (–9.80 m/s2) t2 2 2 The root t = 0 describes launch; the other root, t = 36.2 s, describes his flight time. His rate of pay may then be found from pay rate = 2.43 (a) $1.00 $ 3600 s = 0.0276 = $99.4/h 36.2 s s 1 h 1 y = vit + 2 at 2 4.00 = (1.50)vi – (4.90)(1.50)2 and vi = 10.0 m/s upward (b) v = vi + at = 10.0 – (9.80)(1.50) = – 4.68 m/s v = 4.68 m/s downward 2.44 We have y=– 1 2 gt + vit + yi 2 0 = – (4.90 m/s2)t2 – (8.00 m/s)t + 30.0 m Solving for t, t= 8.00 ± 64.0 + 588 – 9.80 Using only the positive value for t, we find t = 1.79 s © 2000 by Harcourt College Publishers. All rights reserved. Chapter 2 Solutions *2.45 19 The bill starts from rest vi = 0 and falls with a downward acceleration of 9.80 m/s2 (due to gravity). Thus, in 0.20 s it will fall a distance of ∆y = vit – 1 2 gt = 0 – (4.90 m/s2)(0.20 s)2 = –0.20 m 2 This distance is about twice the distance between the center of the bill and its top edge (≅ 8 cm). Thus, David will be unsuccessful . Goal Solution Emily challenges her friend David to catch a dollar bill as follows. She holds the bill vertically, as in Figure P2.45, with the center of the bill between David’s index finger and thumb. David must catch the bill after Emily releases it without moving his hand downward. If his reaction time is 0.2 s, will he succeed? Explain your reasoning. G: David will be successful if his reaction time is short enough that he can catch the bill before it falls half of its length (about 8 cm). Anyone who has tried this challenge knows that this is a difficult task unless the catcher “cheats” by anticipating the moment the bill is released. Since David’s reaction time of 0.2 s is typical of most people, we should suspect that he will not succeed in meeting Emily’s challenge. O: Since the bill is released from rest and experiences free fall, we can use the equation y = 1 2 gt 2 to find the distance y the bill falls in t = 0.2 s 1 A: y = (9.80 m/s2)(0.2 s) 2 = 0.196 m > 0.08 m 2 Since the bill falls below David’s fingers before he reacts, he will not catch it. L: It appears that even if David held his fingers at the bottom of the bill (about 16 cm below the top edge), he still would not catch the bill unless he reduced his reaction time by tensing his arm muscles or anticipating the drop. *2.46 At any time t, the position of the ball released from rest is given by y1 = h – position of the ball thrown vertically upward is described by y2 = vit – 1 2 gt . At time t, the 2 1 2 gt . 2 The time at which the first ball has a position of y1 = h/2 is found from the first equation as 1 h/2 = h – gt2, which yields t = h/g . To require that the second ball have a position of 2 1 y2 = h/2 at this time, use the second equation to obtain h/2 = vi h/g – g(h/g). This gives the 2 required initial upward velocity of the second ball as v i = gh . © 2000 by Harcourt College Publishers. All rights reserved. Chapter 2 Solutions 20 2.47 (a) v = vi – gt (Eq. 2.8) v = 0 when t = 3.00 s, g = 9.80 m/s2, ∴ vi = gt = (9.80 m/s2)(3.00 s) = 29.4 m/s (b) 1 1 y = 2 (v + vi) t = 2 (29.4 m/s)(3.00 s) = 44.1 m Goal Solution A baseball is hit such that it travels straight upward after being struck by the bat. A fan observes that it requires 3.00 s for the ball to reach its maximum height. Find (a) its initial velocity and (b) the maximum height it reaches. G: We can expect the initial speed of the ball to be somewhat greater than the speed of the pitch, which might be about 60 mph (~30 m/s), so an initial upward velocity off the bat of somewhere between 20 and 100 m/s would be reasonable. We also know that the length of a ball field is about 300 ft. (~100m), and a pop-fly usually does not go higher than this distance, so a maximum height of 10 to 100 m would be reasonable for the situation described in this problem. O: Since the ball’s motion is entirely vertical, we can use the equation for free fall to find the initial velocity and maximum height from the elapsed time. A: Choose the “initial” point when the ball has just left contact with the bat. Choose the “final” point at the top of its flight. In between, the ball is in free fall for t = 3.00 s and has constant acceleration, a = -g = -9.80 m/s2. Solve the equation vyf = vyi – gt for vyi when vyf = 0 (when the ball reaches its maximum height). ( a ) vyi = vyf + gt = 0 + (9.80 m/s2)(3.00 s) = 29.4 m/s (upward) (b) The maximum height in the vertical direction is 1 1 yf = vyi t + at2 = (29.4 m/s)(3.00 s) + (–9.80 m/s2)(3.00 s) 2 = 44.1 m 2 2 L: The calculated answers seem reasonable since they lie within our expected ranges, and they have the correct units and direction. We must remember that it is possible to solve a problem like this correctly, yet the answers may not seem reasonable simply because the conditions stated in the problem may not be physically possible (e.g. a time of 10 seconds for a pop fly would not be realistic). © 2000 by Harcourt College Publishers. All rights reserved. Chapter 2 Solutions 2.48 Take downward as the positive (a) y direction. While the woman was in free fall, ∆y = 144 ft, vi = 0, and a = g = 32.0 ft/s2 Thus, ∆y = vit + 1 2 at → 144 ft = 0 + (16.0 ft/s2)t2 2 giving tfall = 3.00 s. Her velocity just before impact is: v = vi + gt = 0 + (32.0 ft/s2)(3.00 s) = 96.0 ft/s . (b) While crushing the box, vi = 96.0 ft/s, v = 0, and ∆y = 18.0 in = 1.50 ft. 2 Therefore, a = or (c) a = 3.07 × 103 ft/s2 upward Time to crush box: ∆t = or 2.49 v2 – v i 0 – (96.0 ft/s)2 = = –3.07 × 103 ft/s2, 2(∆y) 2(1.50 ft) ∆y ∆y 2(1.50 ft) = = – (v + v i)/2 0 + 96.0 ft/s v ∆t = 3.13 × 10 –2 s Time to fall 3.00 m is found from Eq. 2.11 with vi = 0, 1 3.00 m = (9.80 m/s2) t2; 2 (a) t = 0.782 s With the horse galloping at 10.0 m/s, the horizontal distance is vt = 7.82 m (b) t = 0.782 s © 2000 by Harcourt College Publishers. All rights reserved. 21 22 2.50 Chapter 2 Solutions Time to top = 10.0 s. v = vi – gt (a) At the top, v = 0. Then, t = (b) h = v it – vi = 10.0 s g 1 2 gt 2 At t = 10.0 s, h = (98.0)(10.0) – 2.51 vi = 98.0 m/s 1 (9.80)(10.0) 2 = 490 m 2 vi = 15.0 m/s (a) v = vi – gt = 0 t= vi 15.0 m/s = = 1.53 s g 9.80 m/s2 2 vi 1 225 gt2 = = m = 11.5 m 2 2g 19.6 (b) h = v it – (c) At t = 2.00 s v = vi – gt = 15.0 – 19.6 = – 4.60 m/s a = –g = –9.80 m/s2 2.52 y = 3.00t3 At t = 2.00 s, y = 3.00(2.00)3 = 24.0 m, and vy = dy = 9.00t2 = 36.0 m/s ↑ dt If the helicopter releases a small mailbag at this time, the equation of motion of the mailbag is yb = ybi + vit – Setting yb = 0, 1 1 gt2 = 24.0 + 36.0t – (9.80) t2 2 2 0 = 24.0 + 36.0t – 4.90t2 Solving for t, (only positive values of t count), t = 7.96 s © 2000 by Harcourt College Publishers. All rights reserved. Chapter 2 Solutions 2.53 (a) J= da = constant dt a = J ∫ dt = Jt + c2 da = Jdt but a = ai when t = 0 so c1 = ai, Therefore, a = Jt + a i a= dv ; dv = adt dt v = ∫ adt = ∫ (Jt + ai)dt = 1 2 Jt + ait + c2 2 but v = vi when t = 0, so c2 = vi v= and v = 1 2 Jt + a it + v i 2 dx ; dx = vdt dt 1 x = ∫ vdt = ∫ Jt 2 + a it + v i dt 2 x= 1 3 1 2 Jt + ait + vit + c3 6 2 x = xi when t = 0, so c3 = xi Therefore, x = (b) 1 3 1 2 Jt + a it + v it + x i 6 2 2 a2 = (Jt + ai)2 = J2t2 + a i + 2Jait 2 a 2 = a i + (J2t2 + 2Jait) 1 2 a 2 = a i + 2J Jt 2 + a it 2 Recall the expression for v: v = So (v – vi) = 1 2 Jt + ait + vi 2 1 2 Jt + ait 2 2 Therefore, a 2 = a i + 2J(v – v i) © 2000 by Harcourt College Publishers. All rights reserved. 23 24 2.54 Chapter 2 Solutions (a) a= dv d = [– 5.00 × 107 t2 + 3.00 × 105 t] dt dt Error! t + 3.00 × 105 m/s2 ) Take xi = 0 at t = 0. t Then v = dx dt t ⌠ (–5.00 × 107 t2 + 3.00 × 105t) dt x–0=⌠ ⌡ vdt = ⌡ 0 0 t3 t2 + 3.00 × 105 3 2 x = – 5.00 × 107 x = – (1.67 × 107 m/s3)t3 + (1.50 × 105 m/s2)t2 The bullet escapes when a = 0, at – (10.0 × 107 m/s3)t + 3.00 × 105 m/s2 = 0 (b) t= 3.00 × 105 s = 3.00 × 10-3 s 10.0 × 107 New v = (– 5.00 × 107)(3.00 × 10-3)2 + (3.00 × 105)(3.00 × 10-3) (c) v = – 450 m/s + 900 m/s = 450 m/s x = – (1.67 × 107)(3.00 × 10-3)3 + (1.50 × 105)(3.00 × 10-3)2 (d) x = – 0.450 m + 1.35 m = 0.900 m 2.55 a= dv = –3.00v2, vi = 1.50 m/s dt Solving for v, dv = –3.00v2 dt v 0 ⌠ ⌡ v–2dv = –3.00 ⌠ ⌡ dt v = vi – t=0 1 1 + = –3.00t v vi When v = or 3.00t = 1 1 – v vi vi 1 , t= = 0.222 s 2 3.00 v i © 2000 by Harcourt College Publishers. All rights reserved. Chapter 2 Solutions 2.56 (a) 25 The minimum distance required for the motorist to stop, from an initial speed of 18.0 m/s, is 2 v 2 – v i 0 – (18.0 m/s)2 ∆x = = = 36.0 m 2a 2(–4.50 m/s2) Thus, the motorist can travel at most (38.0 m – 36.0 m) = 2.00 m before putting on the brakes if he is to avoid hitting the deer. The maximum acceptable reaction time is then tmax = (b) 2.00 m 2.00 m = = 0.111 s vi 18.0 m/s In 0.300 s, the distance traveled at 18.0 m/s is x = vit1 = (18.0 m/s)(0.300) = 5.40 m ∴ The displacement for an acceleration – 4.50 m/s2 is 38.0 – 5.40 = 32.6 m. 2 v 2 = v i + 2ax = (18.0 m/s)2 – 2(4.50 m/s2)(32.6 m) = 30.6 m2/s2 v= 2.57 30.6 = 5.53 m/s The total time to reach the ground is given by y – yi = vit + 1 2 at 2 1 0 – 25.0 m = 0 + (–9.80 m/s2) t2 2 t= 2(25.0 m) = 2.26 s 9.80 m/s2 The time to fall the first fifteen meters is found similarly: 1 2 –15.0 m = 0 – (9.80 m/s2) t 1 2 t1 = 1.75 s So t – t1 = 2.26 s – 1.75 s = 0.509 s suffices for the last ten meters. © 2000 by Harcourt College Publishers. All rights reserved. 26 Chapter 2 Solutions *2.58 The rate of hair growth is a velocity and the rate of its increase is an acceleration. Then mm/d vi = 1.04 mm/d and a = 0.132 . The increase in the length of the hair (i.e., w displacement) during a time of t = 5.00 w = 35.0 d is ∆x = vit + 1 2 at 2 1 ∆x = (1.04 mm/d)(35.0 d) + (0.132 mm/d ⋅ w)(35.0 d)(5.00 w) 2 or 2.59 ∆x = 48.0 mm Let path (#1) correspond to the motion of the rocket accelerating under its own power. Path (#2) is the motion of the rocket under the influence of gravity with the rocket still rising. Path (#3) is the motion of the rocket under the influence of gravity, but with the rocket falling. The data in the table is found for each phase of the rocket's motion. 2 1 (#1): v2 – (80.0)2 = 2(4.00)(1000); therefore v = 120 m/s 120 = 80.0 + (4.00)t giving t = 10.0 s 0 (#2): 0 – (120)2 = 2(–9.80)∆x giving ∆x = 735 m 0 – 120 = –9.80t giving t = 12.2 s This is the time of maximum height of the rocket. (#3): v2 – 0 = 2(–9.80)(–1735) v = –184 = (–9.80)t giving t = 18.8 s (a) ttotal = 10 + 12.2 + 18.8 = 41.0 s (b) ∆xtotal = 1.73 km (c) vfinal = –184 m/s 0 #1 #2 #3 Launch End Thrust Rise Upwards Fall to Earth t 0 10.0 22.2 41.0 x 0 1000 1735 0 v 80 120 0 –184 © 2000 by Harcourt College Publishers. All rights reserved. a +4.00 +4.00 –9.80 –9.80 3 Chapter 2 Solutions 2.60 Distance traveled by motorist = (15.0 m/s)t Distance traveled by policeman = *2.61 27 1 (2.00 m/s2) t2 2 (a) intercept occurs when 15.0t =t2 (b) v (officer) = (2.00 m/s2)t = 30.0 m/s (c) x (officer) = t = 15.0 s 1 (2.00 m/s2) t2 = 225 m 2 Area A1 is a rectangle. Thus, A1 = hw = vit. Area A 2 is triangular. Therefore A2 = 1 1 bh = t(v – v i). 2 2 The total area under the curve is A = A1 + A2 = vit + (v – vi)t/2 v v A2 and since v – vi = at vi A = v it + 1 2 at 2 A1 t The displacement given by the equation is: x = vit + 2.62 1 2 at , the same result as above for the total area. 2 a1 = 0.100 m/s2, a2 = –0.500 m/s2 x = 1000 m = 1 1 2 2 a t + v1t2 + a 2 t2 2 11 2 t = t1 + t2 and v1 = a1t1 = –a2t2 1000 = a 1t1 1 a 1t1 2 1 2 a 1 t 1 + a 1t1 – + a2 2 a2 2 a2 © 2000 by Harcourt College Publishers. All rights reserved. 0 t 28 Chapter 2 Solutions 1000 = a1 2 1 a1 1 – t1 2 a2 t1 = 20,000 = 129 s 1.20 t2 = a 1 t 1 12.9 = ≈ 26 s –a 2 0.500 Total time = t = 155 s 2.63 (a) Let x be the distance traveled at acceleration a until maximum speed v is reached. If this achieved in time t1 we can use the following three equations: x= (v + v i) t1, 2 100 – x = v(10.2 – t1) and v = vi + at1 The first two give 100 = 10.2 – 2 t 1 v = 10.2 – 2 t 1 at 1 1 a= 200 . (20.4 – t 1 )t 1 For Maggie a = For Judy a = (b) 1 200 = 5.43 m/s2 (18.4)(2.00) 200 = 3.83 m/s2 (17.4)(3.00) v = at 1 Maggie: v = (5.43)(2.00) = 10.9 m/s Judy: v = (3.83)(3.00) = 11.5 m/s (c) 1 2 At the six-second mark x = 2 at1 + v(6.00 – t1) 1 Maggie: x = 2 (5.43)(2.00) 2 + (10.9)(4.00) = 54.3 m 1 Judy: x = 2 (3.83)(3.00) 2 + (11.5)(3.00) = 51.7 m Maggie is ahead by 2.62 m . © 2000 by Harcourt College Publishers. All rights reserved. Chapter 2 Solutions *2.64 Let the ball fall 1.50 m. It strikes at speed given by: 2 2 v x = vxi + 2a(x – xi ) 2 v x = 0 + 2(–9.80 m/s2)(–1.50 m) vx = –5.42 m/s and its stopping is described by 2 2 v x = vxi + 2ax(x – xi) 0 = (–5.42 m/s)2 + 2ax(–10–2 m) ax = –29.4 m2/s2 = +1.47 × 103 m/s2 –2.00 × 10–2 m Its maximum acceleration will be larger than the average acceleration we estimate by imagining constant acceleration, but will still be of order of magnitude ~ 103 m/s2 . 2.65 Acceleration a = 3.00 m/s2 (a) Deceleration a' = – 4.50 m/s2 Keeping track of speed and time for each phase of motion, v0 = 0, v1 = 12.0 m/s v1 = 12.0 m/s ∆t01 = 4.00 s t1 = 5.00 s v1 = 12.0 m/s, v2 = 0 ∆t12 = 2.67 s v2 = 0 m/s, v3 = 18.0 m/s ∆t23 = 6.00 s v3 = 18.0 m/s t3 = 20 .0 s v3 = 18.0 m/s, v4 = 6.00 m/s v4 = 6.00 m/s ∆t34 = 2.67 s t4 = 4 .00 s v4 = 6.00 m/s, v5 = 0 ∆t45 = 1.33 s Σt = 45.7 s (b) – x = Σ vi ti = 6.00(4.00) + 12.0(5.00) + 6.00(2.67) + 9.00(6.00) + 18.0(20.0) + 12.0(2.67) + 6.00(4.00) + 3.00(1.33) = 574 m (c) ∆x 574 m – v = = = 12.6 m/s ∆t 45.7 s © 2000 by Harcourt College Publishers. All rights reserved. 29 30 Chapter 2 Solutions (d) tWALK = 2∆x 2(574 m) = = 765 s vWALK (1.50 m/s) (about 13 minutes, and better exercise!) 2.66 (a) 1 2 d = 2 (9.80) t1 d = 336t2 t1 + t2 = 2.40 336t2 = 4.90(2.40 – t2)2 2 4.90t2 – 359.5t2 + 28.22 = 0 t2 = t2 = 359.5 ± (359.5)2 – 4(4.90)(28.22) 9.80 359.5 ± 358.75 = 0.0765 s 9.80 ∴d = 336t2 = 26.4 m 2.67 1 (9.80)(2.40) 2 = 28.2 m, an error of 6.82% . 2 (b) Ignoring the sound travel time, d = (a) y = vi1t + 2 at2 = 50.0 = 2.00t + 2 (9.80) t2 1 t = 2.99 s (b) 1 after the first stone is thrown. 1 y = vi2t + 2 at2 and t = 2.99 – 1.00 = 1.99 s 1 substitute 50.0 = vi2(1.99) + 2 (9.80)(1.99) vi2 = 15.4 m/s (c) 2 downward v1 = vi1 + at = –2.00 + (–9.80)(2.99) = –31.3 m/s v2 = vi2 + at = –15.3 + (–9.80)(1.99) = –34.9 m/s 2.68 The time required for the car to come to rest and the time required to regain its original speed of |∆v| 25.0 m/s 25.0 m/s are both given by ∆t = = . The total distance the car travels in these |a| 2.50 m/s2 two intervals is xcar = ∆x1 + ∆x2 = (25.0 m/s + 0) (0 + 25.0 m/s) (10.0 s) + (10.0 s) = 250 m 2 2 © 2000 by Harcourt College Publishers. All rights reserved. Chapter 2 Solutions 31 The total elapsed time when the car regains its original speed is ∆ttotal = 10.0 s + 45.0 s + 10.0 s = 65.0 s The distance the train has traveled in this time is xtrain = (25.0 m/s)(65.0 s) = 1.63 × 103 m Thus, the train is 1.63 × 103 m – 250 m = 1.38 × 103 m 2.69 (a) ahead of the car. We require xs = xk when ts = tk + 1.00 1 1 xs = (3.50 m/s2)(tk + 1.00) 2 = (4.90 m/s2)(tk) 2 = xk 2 2 tk + 1.00 = 1.183tk tk = 5.46 s (b) 1 xk = (4.90 m/s2)(5.46 s) 2 (c) vk = (4.90 m/s2)(5.46 s) = 26.7 m/s 2 = 73.0 m vs = (3.50 m/s2)(6.46 s) = 22.6 m/s 2.70 (a) In walking a distance ∆x, in a time ∆t, the length of rope l is only increased by ∆x sin θ. ∆x ∴ The pack lifts at a rate sin θ. ∆t ∆x x v= sin θ = vboy = v boy ∆t l (b) a= va h m x x2 + h2 vboy dx dv d 1 = + vboyx dt d t l l dt a = vboy 2 vboy l – vboyx dl dl , but =v dt l2 d t 2 2 vboy vboy h 2 h 2 v boy x2 ∴ a = l 1 – 2 = l 2 = l (x 2 + h 2 ) 3 / 2 l 2 (c) v boy ,0 h (d) vboy, 0 l © 2000 by Harcourt College Publishers. All rights reserved. x vboy 32 2.71 Chapter 2 Solutions h = 6.00 m, vboy = 2.00 m/s v= v boyx ∆x x sin θ = vboy = 2 ∆t l (x + h 2 ) 1 / 2 However x = vboyt 2 ∴v= v boyt 2 (v boy t 2 + h 2 ) 1/2 = (4t2 4t + 36) 1/2 (a) t(s) 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 (b) v(m/s) 0 0.32 0.63 0.89 1.11 1.28 1.41 1.52 1.60 1.66 1.71 v (m/s) 1.8 1.5 1.2 0.9 0.6 0.3 0.0 t (s) 0 1 2 3 4 5 From problem 2.70 above, 2 2 h 2 v boy h 2 v boy 144 a= 2 = = (x + h 2 ) 3 / 2 (v 2 t 2 + h 2 ) 3/2 (4t2 + 36) 3/2 boy t(s) 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 a(m/s2) 0.67 0.64 0.57 0.48 0.38 0.30 0.24 0.18 0.14 0.11 0.09 a (m/s2) 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 t (s) 0 1 2 3 © 2000 by Harcourt College Publishers. All rights reserved. 4 5 Chapter 2 Solutions 33 2.72 Time t(s) Height h(m) 0.00 5.00 0.25 5.75 0.50 6.40 0.75 6.94 1.00 7.38 1.25 1.50 1.75 ∆h (m) ∆t (s) – v (m/s) midpt time t (s) 0.75 0.25 3.00 0.13 0.65 0.25 2.60 0.38 0.54 0.25 2.16 0.63 0.44 0.25 1.76 0.88 0.34 0.25 1.36 1.13 0.24 0.25 0.96 1.38 7.72 7.96 0.14 0.25 0.56 1.63 0.03 0.25 0.12 1.88 8.13 2.25 8.07 –0.06 0.25 –0.24 2.13 2.50 7.90 –0.17 0.25 –0.68 2.38 2.75 7.62 –0.28 0.25 –1.12 2.63 3.00 7.25 –0.37 0.25 –1.48 2.88 3.25 6.77 –0.48 0.25 –1.92 3.13 3.50 6.20 –0.57 0.25 –2.28 3.38 3.75 5.52 –0.68 0.25 –2.72 3.63 4.00 4.73 –0.79 0.25 –3.16 3.88 4.25 3.85 –0.88 0.25 –3.52 4.13 4.50 2.86 –0.99 0.25 –3.96 4.38 4.75 1.77 –1.09 0.25 –4.36 4.63 5.00 0.58 –1.19 0.25 –4.76 4.88 4.00 2.00 0.00 −2.00 1 −4.00 −6.00 8.10 2.00 − v (m/s) acceleration = slope of line is constant. –a = –1.63 m/s2 = 1.63 m/s2 downward © 2000 by Harcourt College Publishers. All rights reserved. 2 3 4 5 t (s) Chapter 2 Solutions 34 2.73 The distance x and y are always related by x2 + y2 = L2. Differentiating this equation with respect to time, we have 2x Now dx dy + 2y =0 dt dt y dy dx is vB, the unknown velocity of B; and = –v. dt dt From the equation resulting from differentiation, we have B x L y v α dy x dx x = – = – (–v) dt y d t y But y = tan α x O A x so vB = 1 v tan α When α = 60.0°, vB = v v 3 = = 0.577v tan 60.0° 3 Goal Solution Two objects, A and B, are connected by a rigid rod that has a length L. The objects slide along perpendicular guide rails, as shown in Figure P2.73. If A slides to the left with a constant speed v, find the velocity of B when α = 60.0°. G: The solution to this problem may not seem obvious, but if we consider the range of motion of the two objects, we realize that B will have the same speed as A when α = 45, and when α = 90, then vB = 0. Therefore when α = 60, we should expect vB to be between 0 and v. O: Since we know a distance relationship and we are looking for a velocity, we might try differentiating with respect to time to go from what we know to what we want. We can express the fact that the distance between A and B is always L, with the relation: x2+ y2 = L2. By differentiating this equation with respect to time, we can find vB = dy/dt in terms of dx/dt = vA = –v. dx dy + 2y =0 dt dt dy x dx x Substituting and solving for the speed of B: vB = = – = – (–v) dt y d t y y v Now from the geometry of the figure, we notice that = tan α, so vB = x tan α v v When α = 60.0°, vB = = = 0.577v (B is moving up) tan 60 3 A: Differentiating x2 + y2 = L2 gives us 2x L: Our answer seems reasonable since we have specified both a magnitude and direction for the velocity of B, and the speed is between 0 and v in agreement with our earlier prediction. In this and many other physics problems, we can find it helpful to examine the limiting cases that define boundaries for the answer. © 2000 by Harcourt College Publishers. All rights reserved. Chapter 2 Solutions © 2000 by Harcourt College Publishers. All rights reserved. 35 Chapter 3 Even Answers 2. 4. 6. 8. 10. 12. 14. 16. 18. 20. 22. 24. 26. 28. 30. 32. 34. 36. 38. 40. 42. 44. 46. 48. 50. 52. 54. 56. 58. 60. (a) 8.60 m (b) 4.47 m, -63.4˚, 4.24 m, 135˚ (a) (2.17, 1.25) m and (–1.90, 3.29) m (b) 4.55 m (a) r, 180˚ – θ (b) 2r, 180˚ + θ (c) 3r, –θ 14 km, 65˚ N of E 310 km at 57˚ S of W 9.54 N, 57.0˚ above the x-axis 7.92 m at 4.34˚ N of W (a) ~105 m upward (b) ~103 m upward 5.24 km at 25.9˚ N of W 86.6 m, - 50.0 m 358 m at 2.00˚ S of E |B| = 7.81, α = 59.2˚, β = 39.8˚, γ = 67.4˚ 788 miles at 48.0˚ NE of Dallas (b) 5.00i + 4.00j, 6.40 at 38.7˚, –1.00i + 8.00j, 8.06 at 97.2˚ Cx = 7.30 cm, Cy = –7.20 cm 6.22 blocks at 110˚ counterclockwise from east (a) 4.47 m at θ = 63.4˚ (b) 8.49 m at θ = 135˚ 42.7 yards 4.64 m at 78.6˚ N of E 1.43 × 104 m at 32.2˚ above the horizontal 106˚ - 220i + 57.6j, 227 paces at 165˚ (a) (3.12i + 5.02j – 2.20k) km (b) 6.31 km (a) (15.1i + 7.72j) cm (b) (–7.72i + 15.1j) cm (c) (+7.72i + 15.1j) cm (a) 74.6˚ N of E (b) 470 km a = 5.00, b = 7.00 2 tan–1(1/n) (3.60i + 7.00j) N, 7.87 N at 97.8˚counterclockwise from horizontal –2.00 m/s j, it is the velocity vector (a) (10.0 m, 16.0 m) © 2000 by Harcourt College Publishers. All rights reserved. 2 Chapter 3 Even Answers © 2000 by Harcourt College Publishers. All rights reserved. Chapter 3 Solutions *3.1 x = r cos θ = (5.50 m) cos 240° = (5.50 m)(–0.5) = –2.75 m y = r sin θ = (5.50 m) sin 240° = (5.50 m)(–0.866) = – 4 .76 m 3.2 (a) (b) d= (x 2 – x 1 ) 2 + (y 2 – y 1 ) 2 = d= 25.0 + 49.0 = 8.60 m r1 = (2.00)2 + (–4.00)2 = θ1 = tan–1 – r2 = (2.00 – [–3.00]2) + (–4.00 – 3.00)2 20.0 = 4.47 m 4.00 = –63.4° 2.00 (–3.00)2 + (3.00)2 = 18.0 = 4.24 m θ2 = 135° measured from + x axis. 3.3 We have 2.00 = r cos 30.0° r= 2.00 = 2.31 cos 30.0° and y = r sin 30.0° = 2.31 sin 30.0° = 1.15 3.4 (a) x = r cos θ and y = r sin θ, therefore x1 = (2.50 m) cos 30.0°, y1 = (2.50 m) sin 30.0°, and (x1, y1) = (2.17, 1.25) m x2 = (3.80 m) cos 120°, y2 = (3.80 m) sin 120°, and (x2, y2) = (–1.90, 3.29) m (b) d= (∆x)2 + (∆y)2 = 16.6 + 4.16 = 4.55 m © 2000 by Harcourt College Publishers. All rights reserved. 2 3.5 Chapter 3 Solutions The x distance out to the fly is 2.00 m and the y distance up to the fly is 1.00 m. (a) We can use the Pythagorean theorem to find the distance from the origin to the fly, distance = (b) 3.6 (2.00 m)2 + (1.00 m)2 = 5.00 m2 = 2.24 m 1 θ = Arctan = 26.6°; r = 2.24 m, 26.6° 2 We have r = (a) x2 + y2 = x2 + y2 and θ = Arctan y x The radius for this new point is (–x)2 + y2 = x2 + y2 = r and its angle is Arctan y = 180° – θ (–x) (–2x)2 + (–2y)2 = 2r (b) This point is in the third quadrant if (x, y) is in the first quadrant or in the fourth quadrant if (x, y) is in the second quadrant. It is at angle 180° + θ . (3x)2 + (–3y)2 = 3r (c) This point is in the fourth quadrant if (x, y) is in the first quadrant or in the third quadrant if (x, y) is in the second quadrant. It is at angle –θ . 3.7 C (a) The distance d from A to C is d 300 km d= x2 + y2 φ 30° where x = (200) + (300 cos 30.0°) = 460 km B and y = 0 + (300 sin 30.0°) = 150 km ∴ d = (460)2 + (150)2 = 484 km (b) tan φ = y 150 = = 0.326 x 460 R φ = tan-1(0.326) = 18.1° N of W 3.8 R ≅ 14 km θ = 65° N of E θ 6 km © 2000 by Harcourt College Publishers. All rights reserved. 13 km 200 km A Chapter 3 Solutions 3.9 tan 35.0° = 3 x 100 m x = (100 m)(tan 35.0°) = 70.0 m x 35.0° 100 m 3.10 –R = 310 km at 57° S of W B −R A E base 0 3.11 100 km 200 km (a) Using graphical methods, place the tail of vector B at the head of vector A. The new vector A + B has a magnitude of 6.1 a t 112° from the x-axis. (b) The vector difference A – B is found by placing the negative of vector B at the head of vector A. The resultant vector A – B has magnitude 14.8 units at an angle of 22° from the + x-axis. y —B B A A+B A—B x O © 2000 by Harcourt College Publishers. All rights reserved. 4 3.12 Chapter 3 Solutions Find the resultant F 1 + F 2 graphically by placing the tail of F 2 at the head of F 1. The resultant force vector F1 + F2 is of magnitude 9.5 N and at an angle of 57° above the x-axis . y F2 F1 + F2 F1 x 0 3.13 (a) 1 2 3 N d = –10.0i = 10.0 m since the displacement is a straight line from point A to point B. C (b) The actual distance walked is not equal to the straight-line displacement. The distance follows the curved path of the semi-circle (ACB). 5.00 m 1 s = (2π r) = 5π = 15.7 m 2 B (c) d A If the circle is complete, d begins and ends at point A. Hence, d = 0 . 3.14 Your sketch should be drawn to scale, and should look somewhat like that pictured below. The angle from the westward direction, θ, can be measured to be 4° N of W , and the distance R from the sketch can be converted according to the scale to be 7.9 m . N W 15.0 meters θ R 3.50 meters E 8.20 meters 30.0° S © 2000 by Harcourt College Publishers. All rights reserved. Chapter 3 Solutions 3.15 5 To find these vector expressions graphically, we draw each set of vectors. Measurements of the results are taken using a ruler and protractor. (a) A + B = 5.2 m at 60° (b) A – B = 3.0 m at 330° B A A+B −B A−B A b a 0 (c) 2m 0 4m B – A = 3.0 m at 150° (d) 2m 4m A – 2B = 5.2 m at 300° A B−A − 2B B A − 2B −A c 0 *3.16 (a) 2m d 4m 0 2m 4m The large majority of people are standing or sitting at this hour. Their instantaneous foot-to-head vectors have upward vertical components on the order of 1 m and randomly oriented horizontal components. The citywide sum will be ~105 m upward . (b) Most people are lying in bed early Saturday morning. We suppose their beds are oriented north, south, east, west quite at random. Then the horizontal component of their total vector height is very nearly zero. If their compressed pillows give their height vectors vertical components averaging 3 cm, and if one-tenth of one percent of the population are on-duty nurses or police officers, we estimate the total vector height as ~ 105(0.03 m) + 102(1 m) ~103 m upward . © 2000 by Harcourt College Publishers. All rights reserved. 6 3.17 Chapter 3 Solutions The scale drawing for the graphical solution should be similar to the figure at the right. The magnitude and direction of the final displacement from the starting point are obtained by measuring d and θ on the drawing and applying the scale factor used in making the drawing. The results should be y 135 ft 40.0° 135 ft 30.0° 200 ft x d d ≈ 420 ft and θ ≈ –3° θ N 3.18 x 0 km 1.41 –4.00 –2.12 –4.71 R= 4.00 km y 3.00 km 1.41 0 –2.12 2.29 45.0° 2.00 km 45.0° 3.00 km R |x|2 + |y|2 = 5.24 km 3.00 km φ θt E θ = tan–1 3.19 y = 154° x or φ = 25.9° N of W Call his first direction the x direction. R = 10.0 m i + 5.00 m(–j) + 7.00 m(–i) = 3.00 m i – 5.00 m j = (3.00)2 + (5.00)2 m at Arctan 5 to the right 3 R = 5.83 m at 59.0° to the right from his original motion 3.20 Coordinates of super-hero are: x = (100 m) cos (–30.0°) = 86.6 m y = (100 m) sin (–30.0°) = –50.0 m © 2000 by Harcourt College Publishers. All rights reserved. Chapter 3 Solutions y 30.0° x 100 m © 2000 by Harcourt College Publishers. All rights reserved. 7 8 3.21 Chapter 3 Solutions The person would have to walk 3.10 sin(25.0°) = 1.31 km north , and 3.10 cos(25.0°) = 2.81 km east . 3.22 + x East, + y North Σx = 250 + 125 cos 30° = 358 m Σy = 75 + 125 sin 30° – 150 = –12.5 m d= (Σx)2 + (Σy)2 = tan θ = (358)2 + (–12.5)2 = 358 m (Σy) 12.5 =– = –0.0349 θ = –2.00° (Σx) 358 d = 358 m at 2.00° S of E *3.23 Let the positive x-direction be eastward, positive y-direction be vertically upward, and the positive z-direction be southward. The total displacement is then d = (4.80 cm i + 4.80 cm j) + (3.70 cm j – 3.70 cm k) 3.24 or d = 4.80 cm i + 8.50 cm j – 3.70 cm k (a) The magnitude is d = (b) Its angle with the y-axis follows from cos θ = (4.80)2 + (8.50)2 + (–3.70)2 cm = 10.4 cm 8.50 , giving θ = 35.5° . 10.4 B = Bxi + Byj + B2k B = 4.00i + 6.00j + 3.00k |B| = (4.00)2 + (6.00)2 + (3.00)2 = 7.81 α = cos–1 4.00 = 59.2° 7.81 β = cos–1 6.00 = 39.8° 7.81 3.00 γ = cos–1 7.81 = 67.4° © 2000 by Harcourt College Publishers. All rights reserved. Chapter 3 Solutions 9 y 3.25 Ax = –25.0 Ay = 40.0 A= 2 2 Ax + Ay = A (–25.0)2 + (40.0)2 = 47.2 units From the triangle, we find that φ 40.0 40.0 φφ = 58.0°, so that θ = 122° −25.0 −25.0 θt x Goal Solution A vector has an x component of –25.0 units and a y component of 40.0 units. Find the magnitude and direction of this vector. y 40 r 30 20 10 x —30—25—20—15—10 —5 0 5 10 G: First we should visualize the vector either in our mind or with a sketch. Since the hypotenuse of the right triangle must be greater than either the x or y components that form the legs, we can estimate the magnitude of the vector to be about 50 units. The direction of the vector appears to be about 120° from the +x axis. O: The graphical analysis and visual estimates above may be sufficient for some situations, but we can use trigonometry to obtain a more precise result. A: The magnitude can be found by the Pythagorean theorem: r = r= x2 + y2 (–25.0 units)2 + (40 units)2 = 47.2 units y (if we consider x and y to both be positive) . x y 40.0 φ = tan–1 = tan–1 = tan–1 (1.60) = 58.0° x 25.0 We observe that tan φ = The angle from the +x axis can be found by subtracting from 180. = 180 – 58 = 122° L: Our calculated results agree with our graphical estimates. We should always remember to check that our answers are reasonable and make sense, especially for problems like this where it is easy to mistakenly calculate the wrong angle by confusing coordinates or overlooking a minus sign. Quite often the direction angle of a vector can be specified in more than one way, and we must choose a notation that makes the most sense for the given problem. If compass directions were stated in this question, we could have reported the vector angle to be 32.0° west of north or a compass heading of 328°. © 2000 by Harcourt College Publishers. All rights reserved. 10 Chapter 3 Solutions *3.26 The east and north components of the displacement from Dallas (D) to Chicago (C) are the sums of the east and north components of the displacements from Dallas to Atlanta (A) and from Atlanta to Chicago. In equation form: dDCeast = dDAeast + dACeast = 730 cos 5.00° – 560 sin 21.0° = 527 miles. dDCnorth = dDAnorth + dACnorth = 730 sin 5.00° + 560 cos 21.0° = 586 miles. (dDCeast)2 + (dDCnorth)2 = 788 mi By the Pythagorean theorem, d = Then tan θ = dDCnorth = 1.11 and θ = 48.0°. dDCeast Thus, Chicago is 788 miles at 48.0° north east of Dallas . 3.27 x = d cos θ = (50.0 m)cos(120) = –25.0 m y = d sin θ = (50.0 m)sin(120) = 43.3 m d = (–25.0 m)i + (43.3 m)j 3.28 (a) −B B A−B A A+B B (b) C = A + B = 2.00i + 6.00j + 3.00i – 2.00j = 5.00i + 4.00j C= 25.0 + 16.0 at Arctan 4 5 C = 6.40 at 38.7° D = A – B = 2.00i + 6.00j – 3.00i + 2.00j = –1.00i + 8.00j D= (–1.00)2 + (8.00)2 at Arctan 8.00 (–1.00) © 2000 by Harcourt College Publishers. All rights reserved. Chapter 3 Solutions D = 8.06 at (180° – 82.9°) = 8.06 at 97.2° © 2000 by Harcourt College Publishers. All rights reserved. 11 12 3.29 Chapter 3 Solutions d= (x 1 + x 2 + x 3 ) 2 + (y 1 + y 2 + y 3 ) 2 = (3.00 – 5.00 + 6.00)2 + (2.00 + 3.00 + 1.00)2 = 52.0 = 7.21 m θ = tan-1 6.00 = 56.3° 4.00 3.30 A = –8.70i + 15.0j B = 13.2i – 6.60j A – B + 3C = 0 3C = B – A = 21.9i – 21.6j C = 7.30i – 7.20j or Cx = 7.30 cm Cy = –7.20 cm 3.31 (a) (A + B) = (3i – 2j) + (–i – 4j) = 2i – 6j (b) (A – B) = (3i – 2j) – (–i – 4j) = 4i + 2j (c) A + B = 22 + 62 = 6.32 (d) A – B = 42 + 22 = 4.47 (e) 6 θA + B = tan–1 – = –71.6° = 288° 2 2 θA – B = tan–1 = 26.6° 4 3.32 Let i = east and j = north. R = 3.00b j + 4.00b cos 45° i + 4.00b sin 45° j – 5.00b i R = – 2.17b i + 5.83b j R= 2.17 2 + 5.832 b at Arctan 5.83 N of W 2.17 = 6.22 blocks at 110° counterclockwise from east © 2000 by Harcourt College Publishers. All rights reserved. Chapter 3 Solutions 3.33 3.34 x = r cos θ and y = r sin θ, therefore: (a) x = 12.8 cos 150°, y = 12.8 sin 150°, and (x, y) = (–11.1i + 6.40j) m (b) x = 3.30 cos 60.0°, y = 3.30 sin 60.0°, and (x, y) = (1.65i + 2.86j) cm (c) x = 22.0 cos 215°, y = 22.0 sin 215°, and (x, y) = (–18.0i – 12.6j) in (a) D = A + B + C = 2i + 4j D = (b) 22 + 42 = 4.47 m at θ = 63.4° E = –A – B + C = –6i + 6j E = 62 + 62 = 8.49 m at θ = 135° 3.35 d1 = (–3.50j) m d2 = 8.20 cos 45.0°i + 8.20 sin 45.0°j = (5.80i + 5.80j) m d3 = (–15.0i) m R = d1 + d2 + d3 = (–15.0 + 5.80)i + (5.80 – 3.50)j = (–9.20i + 2.30j) m (or 9.20 m west and 2.30 m north) The magnitude of the resultant displacement is |R| = 2 2 Rx + R y = (–9.20)2 + (2.30)2 = 9.48 m The direction is θ = Arctan 2.30 = 166° –9.20 3.36 Refer to the sketch R = A + B + C = –10.0i – 15.0j + 50.0i = 40.0i – 15.0j R = [(40.0)2 + (–15.0)2]1/2 = 42.7 yards |A| = 10.0 R |B| = 15.0 |C| = 50.0 © 2000 by Harcourt College Publishers. All rights reserved. 13 Chapter 3 Solutions 14 3.37 (a) F = F1 + F2 F = 120 cos (60.0˚)i + 120 sin (60.0˚)j – 80.0 cos (75.0˚)i + 80.0 sin (75.0˚)j F = 60.0i + 104j – 20.7i + 77.3j = (39.3i + 181j) N F (b) = (39.3)2 + (181)2 = 185 N ; θ = tan–1 181 = 77.8° 39.3 F3 = –F = (–39.3i – 181j) N Goal Solution The helicopter view in Figure P3.37 shows two people pulling on a stubborn mule. Find (a) the single force that is equivalent to the two forces shown and (b) the force that a third person would have to exert on the mule to make the resultant force equal to zero. The forces are measured in units of newtons. G: The resultant force will be larger than either of the two individual forces, and since the two people are not pulling in exactly the same direction, the magnitude of the resultant should be less than the sum of the magnitudes of the two forces. Therefore, we should expect 120 N < R < 200 N. The angle of the resultant force appears to be straight ahead and perhaps slightly to the right. If the stubborn mule remains at rest, the ground must be exerting on the animal a force equal to the resultant R but in the opposite direction. 120 N 80 N 75° 60° O: We can find R by adding the components of the two force vectors. A: F1 = (120 cos 60)i N + (120 sin 60)j N = 60.0i N + 103.9j N F2 = –(80 cos 75)i N + (80 sin 75)j N = –20.7i N + 77.3j N R = F1 + F2 = 39.3i N + 181.2j N R = |R| = (39.3)2 + (181.2)2 = 185 N The angle can be found from the arctan of the resultant components. y 181.2 θ = tan–1 = tan–1 = tan–1 (4.61) = 77.8° counterclockwise from the +x axis x 39.3 The opposing force that the either the ground or a third person must exert on the mule, in order for the overall resultant to be zero, is 185 N at 258° counterclockwise from +x. L: The resulting force is indeed between 120 N and 200 N as we expected, and the angle seems reasonable as well. The process applied to solve this problem can be used for other “statics” problems encountered in physics and engineering. If another force is added to act on a system that is already in equilibrium (sum of the forces is equal to zero), then the system may accelerate. Such a system is now a “dynamic” one and will be the topic of Chapter 5. © 2000 by Harcourt College Publishers. All rights reserved. Chapter 3 Solutions 15 3.38 East x 0m 1.41 –0.500 +0.914 R = 3.39 North y 4.00 m 1.41 –0.866 4.55 2 x 2 + y = 4.64 m at 78.6° N of E A = 3.00 m, θA = 30.0°, B = 3.00 m, θB = 90.0° Ax = A cos θA = 3.00 cos 30.0° = 2.60 m, Ay = A sin θA = 3.00 sin 30.0° = 1.50 m so, A = Axi + Ayj = (2.60i + 1.50j) m Bx = 0, By = 3.00 m so B = 3.00j m A + B = (2.60i + 1.50j) + 3.00j = (2.60i + 4.50j) m *3.40 The y coordinate of the airplane is constant and equal to 7.60 × 103 m whereas the x coordinate is given by x = vit where vi is the constant speed in the horizontal direction. At t = 30.0 s we have x = 8.04 × 103, so vi = 268 m/s. The position vector as a function of time is P = (268 m/s)t i + (7.60 × 103 m)j. At t = 45.0 s, P = [1.21 × 104 i + 7.60 × 103 j] m. The magnitude is P= (1.21 × 104)2 + (7.60 × 103)2 m = 1.43 × 104 m and the direction is 7.60 × 103 = 1.21 × 104 θ = Arctan y 32.2° above the horizontal 3.41 We have B = R – A B θ Ax = 150 cos 120° = –75.0 cm Ay = 150 sin 120° = 130 cm A R =A+B 120.0° Rx = 140 cos 35.0° = 115 cm Ry = 140 sin 35.0° = 80.3 cm Therefore, B = [115 – (–75)]i + [80.3 – 130]j = (190i – 49.7j) cm © 2000 by Harcourt College Publishers. All rights reserved. 35.0° x 16 Chapter 3 Solutions B = [1902 + (49.7)2]1/2 = 196 cm , θ = tan–1 – 49.7 = –14.7° 190 © 2000 by Harcourt College Publishers. All rights reserved. Chapter 3 Solutions *3.42 Since A + B = 6.00j, we have (Ax + Bx)i + (Ay + By)j = 0i + 6.00 j giving Ax + Bx = 0, or Ax = –Bx y (1) and Ay + By = 6.00 (2) Since both vectors have a magnitude of 5.00, we also have: 2 2 2 A+B 2 A x + Ay = Bx + By = (5.00)2 B 2 2 2 2 θt From Ax = –Bx, it is seen that Ax = Bx . Therefore Ax + Ay = 2 2 2 φ = 2t 2θ θ t A 2 B x + By gives Ay = By . Then Ay = By, and Equation (2) gives Ay = By = 3.00. Defining θ as the angle between either A or B and the y axis, it is seen that Ay By 3.00 = = = 0.600 A B 5.00 and θ = 53.1° cos θ = The angle between A and B is then φ = 2θ = 106° . 3.43 3.44 (a) A = 8.00i + 12.0j – 4.00 k (b) B = A/4 = 2.00i + 3.00j – 1.00k (c) C = –3A = –24.0i – 36.0j + 12.0k R = 75.0 cos 240°i + 75.0 sin 240°j + 125 cos 135°i + 125 sin 135°j + 100 cos 160°i + 100 sin 160°j R = –37.5i – 65.0j – 88.4i + 88.4j – 94.0i + 34.2j R = –220i + 57.6j R= (–220)2 + 57.62 at Arctan 57.6 above the –x-axis 220 R = 227 paces at 165° 3.45 17 (a) C = A + B = (5.00i – 1.00j – 3.00k) m |C| = (5.00)2 + (1.00)2 + (3.00)2 m = 5.92 m (b) D = 2A – B = (4.00i – 11.0j + 15.0k) m © 2000 by Harcourt College Publishers. All rights reserved. x 18 Chapter 3 Solutions |D| = (4.00)2 + (11.0)2 + (15.0)2 m = 19.0 m © 2000 by Harcourt College Publishers. All rights reserved. Chapter 3 Solutions *3.46 19 The displacement from radar station to ship is S = (17.3 sin 136˚i + 17.3 cos 136˚j) km = (12.0i – 12.4j) km From station to plane, the displacement is P = (19.6 sin 153°i + 19.6 cos 153˚j + 2.20k) km, or P = (8.90i – 17.5j + 2.20k) km. (a) From plane to ship the displacement is D = S – P = (3.12i + 5.02j – 2.20k) km (b) The distance the plane must travel is D = |D| = (3.12)2 + (5.02)2 + (2.20)2 km = 6.31 km 3.47 The hurricane's first displacement is 41.0 km (3.00 h) at 60.0° N of W, and its second h displacement is 25.0 km (1.50 h) due North. With i representing east and j representing h north, its total displacement is: 41.0 km cos 60.0° (3.00 h)(–i) + 41.0 km sin 60.0° (3.00 h) j h h + 25.0 with magnitude *3.48 (a) km (1.50 h) j = 61.5 km (–i) + 144 km j h (61.5 km)2 + (144 km)2 = 157 km E = (17.0 cm) cos 27.0˚i + (17.0 cm) sin 27.0˚j y E = (15.1i + 7.72j) cm 27.0° (b) F = –(17.0 cm) sin 27.0˚i + (17.0 cm) cos 27.0˚j F 27.0° G E F = (–7.72i + 15.1j) cm 27.0° (c) G = +(17.0 cm) sin 27.0˚i + (17.0 cm) cos 27.0˚j G = (+7.72i + 15.1j) cm © 2000 by Harcourt College Publishers. All rights reserved. x 20 3.49 Chapter 3 Solutions Ax = –3.00, Ay = 2.00 (a) A = Axi + Ayj = –3.00i + 2.00j (b) |A| = tanθ = 2 2 Ax + Ay = (–3.00)2 + (2.00)2 = 3.61 Ay 2.00 = = –0.667, tan–1(–0.667) = –33.7° Ax (–3.00) θ is in the 2nd quadrant, so θ = 180° + (–33.7°) = 146° (c) Rx = 0, Ry = –4.00, R = A + B thus B = R – A and Bx = Rx – Ax = 0 – (–3.00) = 3.00, By = Ry – Ay = – 4.00 – 2.00 = – 6.00 Therefore, B = 3.00i – 6.00j 3.50 Let +x = East, +y = North, x 300 –175 0 125 3.51 y 0 303 150 453 (a) θ = tan–1 (b) R = y = 74.6° N of E x x2 + y2 = 470 km Refer to Figure P3.51 in the textbook. (a) Rx = 40.0 cos 45.0° + 30.0 cos 45.0° = 49.5 Ry = 40.0 sin 45.0° – 30.0 sin 45.0° + 20.0 = 27.1 R = 49.5i + 27.1j (b) R = (49.4)2 + (27.1)2 = 56.4 θ = tan–1 27.1 = 28.7° 49.5 © 2000 by Harcourt College Publishers. All rights reserved. Chapter 3 Solutions 3.52 21 Taking components along i and j , we get two equations: 6.00a – 8.00b + 26.0 = 0 –8.00a + 3.00b + 19.0 = 0 Solving simultaneously, a = 5.00, b = 7.00 Therefore, 5.00A + 7.00B + C = 0 *3.53 Let θ represent the angle between the directions of A and B. Since A and B have the same magnitudes, A, B, and R = A + B form an isosceles triangle in which the angles are 180˚ – θ , θ /2, and θ /2. The magnitude of R is then R = 2A cos(θ /2). [Hint: apply the law of cosines to the isosceles triangle and use the fact that B = A.] Again, A, –B, and D = A – B form an isosceles triangle with apex angle θ . Applying the law of cosines and the identity (1 – cosθ ) = 2 sin2(θ /2) gives the magnitude of D as D = 2A sin(θ /2). The problem requires that R = 100D. Thus, 2A cos(θ /2) = 200A sin(θ /2). This gives tan(θ /2) = 0.010 and θ = 1.15° . R /2 θt/2 θ t A B D A *3.54 θt −B Let θ represent the angle between the directions of A and B. Since A and B have the same magnitudes, A, B, and R = A + B form an isosceles triangle in which the angles are 180˚ – θ , θ /2, and θ /2. The magnitude of R is then R = 2A cos(θ /2). [Hint: apply the law of cosines to the isosceles triangle and use the fact that B = A.] Again, A, –B, and D = A – B form an isosceles triangle with apex angle θ . Applying the law of cosines and the identity (1 – cosθ ) = 2 sin2(θ /2) gives the magnitude of D as D = 2A sin(θ /2). The problem requires that R = nD, or cos(θ /2) = nsin(θ /2), giving θ = 2 tan–1 (1/n) . R /2 θt/2 θ t A A B D θ t −B © 2000 by Harcourt College Publishers. All rights reserved. 22 3.55 Chapter 3 Solutions (a) Rx = 2.00 , Ry = 1.00 , Rz = 3.00 (b) R = (c) cos θx = cos θy = cos θz = *3.56 2 2 2 Rx + R y + R Z Rx R Ry R Rz R = 4.00 + 1.00 + 9.00 ⇒ θx = cos–1 Rx ⇒ θy = cos–1 Ry ⇒ θz = cos–1 Rz = 14.0 = 3.74 = 57.7° from + x R = 74.5° from + y R R = 36.7° from + z Choose the +x-axis in the direction of the first force. The total force, in newtons, is then 12.0i + 31.0j – 8.40i – 24.0j = (3.60i) + (7.00j) N The magnitude of the total force is (3.60)2 + (7.00)2 N = 7.87 N and the angle it makes with our +x-axis is given by tanθ = (7.00) , θ = 62.8˚. Thus, its angle (3.60) counterclockwise from the horizontal is 35.0˚ + 62.8˚ = 97.8˚ y x 31 N R 12 N 35.0° 8.4 N horizontal 24 N © 2000 by Harcourt College Publishers. All rights reserved. Chapter 3 Solutions 3.57 d1 = 100i d2 = –300j 23 y d3 = –150 cos (30.0˚)i – 150 sin (30.0˚)j = –130i – 75.0j θt φ d4 = –200 cos (60.0˚)i + 200 sin (60.0˚)j = –100i + 173j x R = d1 + d2 + d3 + d4 = –130i – 202j R = [(–130)2 + (–202)2]1/2 = 240 m R φ = tan–1 202 = 57.2° 130 θ = 180 + φ = 237° *3.58 dP/dt = d(4i + 3j – 2t j)/dt = 0 + 0 – 2j = – (2.00 m/s)j The position vector at t = 0 is 4i + 3j. At t = 1 s, the position is 4i + 1j, and so on. The object is moving straight downward at 2 m/s, so dP/dt represents its velocity vector . 3.59 v = vxi + vy j = (300 + 100 cos 30.0°)i + (100 sin 30.0°)j v = (387i + 50.0j) mi/h v 3.60 (a) = 390 mi/h at 7.37° N of E You start at r1 = rA = 30.0 m i – 20.0 m j. The displacement to B is rB – rA = 60.0i + 80.0j – 30.0i + 20.0j = 30.0i + 100j You cover one–half of this, 15.0i + 50.0j, to move to r2 = 30.0i – 20.0j + 15.0i + 50.0j = 45.0i + 30.0j Now the displacement from your current position to C is rC – r2 = –10.0i – 10.0j – 45.0i – 30.0j = –55.0i – 40.0j You cover one–third, moving to 1 r3 = r2 + ∆r23 = 45.0i + 30.0j + (–55.0i –40.0j) = 26.7i + 16.7j 3 © 2000 by Harcourt College Publishers. All rights reserved. 24 Chapter 3 Solutions The displacement from where you are to D is rD – r3 = 40.0i – 30.0j – 26.7i – 16.7j = 13.3i – 46.7j You traverse one–quarter of it, moving to 1 1 r4 = r3 + (rD – r3) = 26.7i + 16.7j + (13.3i – 46.7j) = 30.0i + 5.00j 4 4 The displacement from your new location to E is rE – r4 = –70.0i + 60.0j – 30.0i – 5.00j = –100i + 55.0j of which you cover one–fifth, –20.0i + 11.0j, moving to r4 + ∆r45 = 30.0i + 5.00j – 20.0i + 11.0j = 10.0i + 16.0j. The treasure is at (10.0 m, 16.0 m) (b) Following the directions brings you to the average position of the trees. The steps we took numerically in part (a) bring you to rA + rB 1 rA + (rB – rA) = 2 2 then to (rA + rB) rC –(rA + rB)/2 (rA + rB + rC) + = 2 3 3 then to rA + rB + rC rD – (rA + rB + rC) /3 (rA + rB + rC + rD) + = 3 4 4 and at last to rA + rB + rC + rD rE – (rA + rB + rC + rD)/4 + 4 5 = rA + rB + rC + rD + rE 5 This center of mass of the tree distribution is in the same location whatever order we take the trees in. © 2000 by Harcourt College Publishers. All rights reserved. Chapter 3 Solutions 3.61 (a) From the picture R1 = ai + bj and R1 (b) R2 = ai + bj + ck. Its magnitude is 25 = a2 + b2 R1 2 + c2 = a2 + b2 + c2 z a b O R2 c R1 x y 3.62 (a) r1 + d = r2 defines the displacement d, so d = r2 – r1. (b) r2 d r1 3.63 The displacement of point P is invariant under rotation of the coordinates. Therefore, r = r' and r2 = (r')2 or, x2 + y2 = (x')2 + (y')2 y Also, from the figure, β = θ – α y′ P y' y ∴ tan–1 x' = tan–1 x – α y' = x' r y – tanα x α β αt O y 1 + tan α x Which we simplify by multiplying top and bottom by x cos α . Then, x' = x cosα + y sinα , y' = – x sinα + y cosα © 2000 by Harcourt College Publishers. All rights reserved. x′ θ β x 26 Chapter 3 Solutions © 2000 by Harcourt College Publishers. All rights reserved. Chapter 4 Even Answers 2. 4. 6. 8. 10. 12. 14. 16. 18. 20. 22. 24. 26. 28. 30. 32. 34. 36. 38. 40. ( a ) (1.00i + 0.750j) m/s (b) (1.00i + 0.500j) m/s, 1.12 m/s ( a ) (–5.00ωi + 0j) m/s, (0i + 5.00ω2j)m/s2 (b) r = (4.00 m)i + (5.00 m)(–i sin ωt – j cos ωt), v = (5.00 m)ω(–i cos ωt + j sin ωt), a = (5.00 m)ω2 (i sin ωt + j cos ωt) (c) path is a circle of 5.00 m radius and centered at (0, 4.00) m ( a ) –12.0t j m/s, –12.0 j m/s2 (b) (3.00i – 6.00j) m, –12.0j m/s 1 ( a ) r = [5.00ti + (3.00t2) j] m, v = [5.00i + (3.00t)j] m/s 2 (b) (10.0 m, 6.00 m), 7.81 m/s g 2h (a) v = d horizontally (b) θ = tan–1 below the horizontal 2h d 48.6 m/s 0.600 m/s2 ( a ) 22.6 m (b) 52.3 m (c) 1.18 s 2 2 v i sin θi cos θi v i sin 2θi xh = ,R= g g 18.7 m 9.91 m/s (a) 1.02 × 103 m/s (b) 2.72 × 10–3 m/s2 0.0337 m/s2 directed toward center of the Earth 0.186 s–1 7.58 × 103 m/s, 5.80 × 103 s (96.7 min) ( a ) 0.600 m/s2 (b) 0.800 m/s2 (c) 1.00 m/s2 (d) 53.1˚ inward from path ( a ) 30.8 m/s2 down (b) 70.4 m/s2 upward ( a ) 26.9 m/s (b) 67.3 m (c) (2.00i – 5.00j) m/s2 18.0 s 2L/c 2L/c tAlan = ,t = , Beth returns first. 1 – v2/c2 Beth 1 – v2/c2 58. ( a ) 10.1 m/s2 at 14.3˚ south of vertical (b) 9.80 m/s2 vertically downward R 1 13gR 13R 13R (a) 2 (b) 3gR (c) gR/3 (d) (e) 33.7° (f) (g) 3g 2 12 24 12 54.4 m/s2 g (b) A = – 2 (c) 14.5 m/s 2v i ( a ) 1.69 km/s (b) 1.80 h 10.7 m/s ( a ) 26.6˚ (b) 0.949 R 2 7.50 m/s in direction ball was thrown 60. 62. (a) v i > gR (b) ( 2 – 1) R (18.8, –17.3) m 42. 44. 46. 48. 50. 52. 54. 56. © 2000 by Harcourt College Publishers. All rights reserved. 2 64. 66. 68. 70. Chapter 4 Even Answers 0.139 m/s (a) 22.9 m/s (b) 360 m from base of cliff (c) (114i – 44.3j)m/s (a) 5.14 s (b) (–1.30i + 4.68j) m/s where + i is eastward and + j is northward (c) 19.4 m Safe distances are less than 270 m or greater than 3.48 × 103 m from the eastern shore. © 2000 by Harcourt College Publishers. All rights reserved. Chapter 4 Solutions *4.1 x(m) 0 – 3000 – 1270 – 4270 m (a) y(m) – 3600 0 1270 – 2330 m Net displacement = x2 + y2 = 4.87 km at 28.6° S of W (b) Average speed = (20.0 m/s)(180 s) + (25.0 m/s)(120 s) + (30.0 m/s)(60.0 s) (180 s + 120 s + 60.0 s) = 23.3 m/s (c) Average velocity = 4.87 × 103 m = 13.5 m/s along R 360 s N E R 3600 m 1800 m 3000 m 4.2 (a) For the average velocity, we have – x(4.00) – x(2.00) y(4.00) – y(2.00) v = i + j 4.00 s – 2.00 s 4.00 s – 2.00 s = 5.00 m – 3.00 m 3.00 m – 1.50 m i + j 2.00 s 2.00 s – v = (1.00i + 0.750j) m/s © 2000 by Harcourt College Publishers. All rights reserved. 2 Chapter 4 Solutions (b) For the velocity components, we have vx = dx = a = 1.00 m/s dt vy = dy = 2ct = (0.250 m/s2)t dt Therefore, v = vxi + vyj = (1.00 m/s)i + (0.250 m/s2) tj v(2.00) = (1.00 m/s)i + (0.500 m/s)j and the speed is v 4.3 4.4 t = 2.00 s = (1.00 m/s)2 + (0.500 m/s)2 = 1.12 m/s (a) r = 18.0t i + (4.00t – 4.90t2)j (b) v = (18.0 m/s)i + [4.00 m/s – (9.80 m/s2)t]j (c) a = (–9.80 m/s2)j (d) r(3.00 s) = (54.0 m)i – (32.1 m)j (e) v(3.00 s) = (18.0 m/s)i – (25.4 m/s)j (f) a(3.00 s) = (– 9.80 m/s2)j (a) From x = –5.00 sinωt, the x-component of velocity is vx = and a x = dx d = (–5.00ω sinωt) = –5.00ω cosωt dt d t dv x = +5.00ω 2 sin ω t dt similarly, v y = d (4.00 – 5.00 cosωt) = 0 + 5.00ω sinωt d t and ay = d (5.00ω sinωt) = 5.00ω 2cosωt d t At t = 0, v = –5.00ω cos 0 i + 5.00ω sin 0 j = (–5.00ω i + 0 j) m/s and a = 5.00ω 2 sin 0 i + 5.00ω 2 cos 0 j = (0 i + 5.00ω 2 j) m/s2 © 2000 by Harcourt College Publishers. All rights reserved. Chapter 4 Solutions (b) r = xi + yj = (4.00 m)j +(5.00 m)(–sinω t i – cosω t j) v = (5.00 m)ω [ –cosω t i + sinω t j] a = (5.00 m)ω 2[sinω t i + cosω t j] 4.5 (c) The object moves in a circle of radius 5.00 m centered at (0, 4.00 m) . (a) v = vi + at a= (b) (v – v i) [(9.00i + 7.00j) – (3.00i – 2.00j)] = = (2.00i + 3.00j) m/s2 t 3.00 r = ri + vit 1 2 1 2 at = (3.00i – 2.00j)t + (2.00i + 3.00j) t ; 2 2 x = (3.00t + t2) m 4.6 (a) (b) 4.7 and y = (1.50t2 – 2.00t) m v= dr d = (3.00i – 6.00t2j) = – 12.0tj m/s d t d t a= dv d = (– 12.0tj) = – 12.0j m/s2 dt d t r = (3.00i – 6.00j) m; v = – 12.0j m/s vi = (4.00i + 1.00j) m/s and v(20.0) = (20.0i – 5.00j) m/s. (a) ax = ∆v x 20.0 – 4.00 = m/s2 = 0.800 m/s2 ∆t 20.0 ay = ∆v y – 5.00 – 1.00 = m/s2 = – 0.300 m/s2 ∆t 20.0 (b) θ = tan–1 – 0.300 = – 20.6° = 339° from +x axis 0.800 (c) At t = 25.0 s, x = xi + vxit + 1 1 a t2 = 10.0 + 4.00(25.0) + (0.800)(25.0) 2 = 360 m 2 x 2 y = yi + vyit + 1 1 a t2 = – 4.00 + 1.00(25.0) + (– 0.300)(25.0) 2 = –72.7 m 2 y 2 θ = tan–1 v y – 6.50 = tan–1 = –15.2° v x 24.0 © 2000 by Harcourt College Publishers. All rights reserved. 3 4 Chapter 4 Solutions *4.8 a = 3.00j m/s2; vi = 5.00i m/s; ri = 0i + 0j (a) r = ri + vit + 1 at2 = 2 5.00ti + 1 3.00t2j m 2 v = vi + at = (5.00i + 3.00tj) m/s (b) t = 2.00 s, r = (5.00)(2.00)i + 1 (3.00)(2.00) 2j = (10.0i + 6.00j) m 2 so x = 10.0 m , y = 6.00 m v = 5.00i + (3.00)(2.00)j = (5.00i + 6.00j) m/s v= v 4.9 (a) = 2 2 v x + vy = (5.00)2 + (6.00)2 = 7.81 m/s The mug leaves the counter horizontally with a velocity vxi (say). If time t elapses before it hits the ground, then since there is no horizontal acceleration, x = vxit. i.e., x (1.40 m) t= = . In the same time it falls a distance 0.860 m with acceleration vxi vxi downward of 9.80 m/s2. Then using y = yi + vyit + 1 a t2 2 y we have 1 1.40 m 2 0 = 0.860 m – (9.80 m/s2) 2 vxi i.e., v xi = (b) (4.90 m/s2)(1.96 m2) = 3.34 m/s 0.860 m The vertical velocity component with which it hits the floor is vy = vyi + ayt = –(9.80 m/s2) 1.40 m = – 4.11 m/s 3.34 m/s Hence, the angle θ at which the mug strikes the floor is given by θ = tan–1 v y – 4.11 = tan–1 = –50.9° v x 3.34 © 2000 by Harcourt College Publishers. All rights reserved. Chapter 4 Solutions Goal Solution G: Based on our everyday experiences and the description of the problem, a reasonable speed of the mug would be a few m/s and it will hit the floor at some angle between 0 and 90°, probably about 45°. O: We are looking for two different velocities, but we are only given two distances. Our approach will be to separate the vertical and horizontal motions. By using the height that the mug falls, we can find the time of the fall. Once we know the time, we can find the horizontal and vertical components of the velocity. For convenience, we will set the origin to be the point where the mug leaves the counter. A: Vertical motion: y = –0.860 m, vyi = 0, vy = unknown, ay = –9.80 m/s2 Horizontal motion: x = 1.40 m, vx = constant (unknown), ax = 0 ( a ) To find the time of fall, we use the free fall equation: y = vyit + 1 a t2 2 y 1 Solving: –0.860 m = 0 + (–9.80 m/s2) t2 so that t = 0.419 s 2 Then vx = x 1.40 m = = 3.34 m/s t 0.419 s (b) As the mug hits the floor, vy = vyi + ayt = 0 – (9.8 m/s2)(0.419 s) = –4.11 m/s The impact angle is θ = tan–1 v y 4.11 m/s = tan–1 = 50.9° below the horizontal v x 3.34 m/ L: This was a multi-step problem that required several physics equations to solve; our answers do agree with our initial expectations. Since the problem did not ask for the time, we could have eliminated this variable by substitution, but then we would have had to substitute the algebraic expression t = 2y/g into two other equations. So in this case it was easier to find a numerical value for the time as an intermediate step. Sometimes the most efficient method is not realized until each alternative solution is attempted. © 2000 by Harcourt College Publishers. All rights reserved. 5 6 Chapter 4 Solutions *4.10 The mug is a projectile from just after leaving the counter until just before it reaches the floor. Taking the origin at the point where the mug leaves the bar, the coordinates of the mug at any time are x = vxit + 1 1 1 a t2 = vxit + 0 and y = vyit + ayt2 = 0 – gt2 2 x 2 2 When the mug reaches the floor, y = –h, so –h = – 1 2 gt which gives the time of impact as 2 t= 2h . g (a) Since x = d when the mug reaches the floor, x = vxit becomes d = vxi 2h g giving the initial velocity as v xi = d (b) g . 2h Just before impact, the x-component of velocity is still vxf = vxi while the y-component is 2h . Then the direction of motion just before impact is below the g |vyf| horizontal at an angle of θ = tan–1 , or vxf vyf = vyi + ayt = 0 – g θ = tan–1 g 4.11 2h g /d g 2h = tan –1 2h d (a) The time of flight of the first snowball is the nonzero root of y = yi + vyit1 + 1 2 at 2 y 1 1 2 0 = 0 + (25.0 m/s) sin 70.0°t1 – (9.80 m/s2) t1 2 t1 = 2(25.0 m/s) sin 70.0° = 4.79 s 9.80 m/s2 The distance to your target is x – xi = vxit1 = (25.0 m/s) cos 70.0° (4.79 s) = 41.0 m © 2000 by Harcourt College Publishers. All rights reserved. Chapter 4 Solutions 7 Now the second snowball we describe by y = yi + vyit2 + 1 2 a t 2 y 2 2 0 = (25.0 m/s) sin θ2t2 – (4.90 m/s2)t 2 t2 = (5.10 s) sin θ2 x – xi = vxit2 41.0 m = (25.0 m/s) cos θ2 (5.10 s) sin θ2 = (128 m) sin θ2 cos θ2 0.321 = sin θ2 cos θ2 Using sin 2θ = 2 sin θ cos θ we can solve 0.321 = (b) 1 sin 2θ2 2 2θ2 = Arcsin 0.643 θ2 = 20.0° The second snowball is in the air for time t2 = (5.10 s) sin θ2 = (5.10 s) sin 20.0° = 1.75 s, so you throw it after the first by t1 – t2 = 4.79 s – 1.75 s = 3.05 s . *4.12 1 y = vi (sin 3.00°)t – 2 gt2, vy = vi sin 3.00° – gt When y = 0.330 m, vy = 0 and vi sin 3.00° = gt y = vi (sin 3.00°) vi sin 3.00° 1 vi sin 3.00° 2 – g g 2 g 2 v i sin2 3.00° y= = 0.330 m 2g ∴ vi = 48.6 m/s The 12.6 m is unnecessary information. *4.13 x = vxit = vi cos θit x = (300 m/s)(cos 55.0°)(42.0 s) x = 7.23 × 103 m y = vyit – 1 1 gt2 = vi sin θit – gt2 2 2 © 2000 by Harcourt College Publishers. All rights reserved. 8 Chapter 4 Solutions 1 y = (300 m/s)(sin 55.0°)(42.0 s) – (9.80 m/s2)(42.0 s) 2 = 1.68 × 103 m 2 © 2000 by Harcourt College Publishers. All rights reserved. Chapter 4 Solutions *4.14 From Equation 4.14, R = 15.0 m, vi = 3.00 m/s, θmax = 45.0° 2 vi 9.00 = = 0.600 m/s2 ∴g= 15.0 R 2 4.15 h= 2 v i sin2 θi v i (sin 2θ i) ;R= ; 3h = R, 2g g 2 2 so 3v i sin2 θi v i (sin 2θ i) = 2g g or 2 sin2 θi tan θi = = 3 sin 2θi 2 thus θi = tan–1 4 = 53.1° 3 4.16 (a) x = vxit = (8.00 cos 20.0°)(3.00) = 22.6 m (b) Taking y positive downwards, y = vyit + 1 gt2 2 = 8.00(cos 20.0°)3.00 + (c) 10.0 = 8.00 cos 20.0° t + 1 (9.80)(3.00) 2 = 52.3 m 2 1 (9.80) t2 2 4.90t2 + 2.74t – 10.0 = 0 t= –2.74 ± (2.74)2 + 196 = 1.18 s 9.80 © 2000 by Harcourt College Publishers. All rights reserved. 9 10 4.17 Chapter 4 Solutions x = vxit 2000 m = (1000 m/s) cos θit t= 2.00 s cos θi y = vyit + 1 a t2 2 y 1 800 m = (1000 m/s) sin θit – (9.80 m/s2) t2 2 800 m = (1000 m/s) sin θi 2 2.00 s – 1 (9.80 m/s2) 2.00 s cos θi 2 cos θi 800 m cos2 θi = (2000 m) sin θi cos θi –19.6 m 19.6 m + 800 m cos2 θi = 2000 m 1 – cos2 θi cos θi 384 m2 + 31 360 m2 cos2 θi + 640 000 m2 cos4 θi = 4 000 000 m2 cos2 θi – 4 000 000 m2 cos4 θi 4 640 000 cos4 θi – 3 968 640 cos2 θi + 384 = 0 cos2 θi = 3 968 640 ± (3 968 640)2 – 4(4 640 000)(384) 9 280 cos θi = 0.925 or 0.00984 θi = 22.4° *4.18 or 89.4° Both solutions are valid. The equation y = (tan θi)x – g 2 2v i x2 describes the trajectory of the projectile. When y is cos2 θi a maximum (at x = xh), the slope is zero ie., dy = 0 at x = x h . This gives dx g d y 2xh = 0, so the x-coordinate at which the maximum height = tan θi – 2 dx x = xh 2v i cos2 θi 2 occurs is x h = v i sin θi cos θi . The maximum-height point is halfway through the entire g 2 symmetrical trajectory. Thus, the horizontal range is R = 2xh = 2 v i 2 sin θi cos θi v i sin 2θi = . g g © 2000 by Harcourt College Publishers. All rights reserved. Chapter 4 Solutions 4.19 11 (a) We use Equation 4.12: y = x tan θi – gx2 2 2v i cos2 θi With x = 36.0 m, vi = 20.0 m/s, and θ = 53.0°, we find y = (36.0 m)(tan 53.0°) – (9.80 m/s2)(36.0 m)2 = 3.94 m (2)(20.0 m/s)2 cos2 53.0° The ball clears the bar by (3.94 – 3.05) m = 0.889 m . (b) The time the ball takes to reach the maximum height is t1 = vi sin θi g = (20.0 m/s)(sin 53.0°) = 1.63 s 9.80 m/s2 The time to travel 36.0 m horizontally is t2 = t2 = x vix 36.0 m = 2.99 s (20.0 m/s)(cos 53.0°) Since t2 > t1 the ball clears the goal on its way down . 4.20 (40.0 m/s)(cos 30.0°)t = 50.0 m. (Eq. 4.10) The stream of water takes t = 1.44 s to reach the building, which it strikes at a height y = v yit – 1 gt2 2 = (40.0 sin 30.0°)t – 4.21 1 1 (9.80) t2 = (40.0) (1.44) – (4.90)(1.44)2 = 18.7 m 2 2 From Equation 4.10, x = vxit = (vi cos θi)t. Therefore, the time required to reach the building a d distance d away is t = . At this time, the altitude of the water is vi cos θi y = vyit + 1 d –g d 2 ayt2 = (vi sin θi) 2 vi cos θi 2 vi cos θi Therefore the water strikes the building at a height of y = d tan θ i – gd 2 2 2v i level. © 2000 by Harcourt College Publishers. All rights reserved. cos2 θi above ground 12 4.22 Chapter 4 Solutions The horizontal kick gives zero vertical velocity to the ball. Then its time of flight follows from y = yi + vyit + 1 a t2 2 y 1 – 40.0 m = 0 + 0 + (–9.80 m/s2) t2 2 t = 2.86 s The extra time 3.00 s – 2.86 s = 0.143 s is the time required for the sound she hears to travel straight back to the player. It covers distance (343 m/s)0.143 s = 49.0 m = x2 + (40.0 m)2 where x represents the horizontal distance the ball travels. x = 28.3 m = vxit + 0t2 ∴ vxi = *4.23 28.3 m = 9.91 m/s 2.86 s From the instant he leaves the floor until just before he lands, the basketball star is a projectile. His vertical velocity and vertical displacement are related by the equation 2 2 v yf = vyi + 2ay(yf – yi). Applying this to the upward part of his flight gives 2 0 = vyi + 2(–9.80 m/s2)(1.85 – 1.02) m. From this, vyi = 4.03 m/s. [Note that this is the answer to part (c) of this problem.] For the downward part of the flight, the equation becomes 2 v yf = 0 + 2(–9.80 m/s2)(0.900 – 1.85) m, giving vyf = –4.32 m/s as the vertical velocity just before he lands. (a) His hang time may then be found from vyf = vyi + ayt as follows: –4.32 m/s = 4.03 m/s + (–9.80 m/s2)t or (b) t = 0.852 s Looking at the total horizontal displacement during the leap, x = vxit becomes 2.80 m = vxi(0.852 s), which yields vxi = 3.29 m/s . (c) vyi = 4.03 m/s See above for proof. © 2000 by Harcourt College Publishers. All rights reserved. Chapter 4 Solutions (d) The takeoff angle is: θ = tan–1 v y i 4.03 m/s = tan–1 = 50.8° . v x i 3.29 m/s (e) Similarly for the deer, the upward part of the flight gives 2 2 2 v yf = vyi + 2ay (yf – yi): 0 = vyi + 2(–9.80 m/s2)(2.50 – 1.20) m so vyi = 5.04 m/s 2 2 For the downward part, vyf = vyi + 2ay(yf – yi) yields 2 v yf = 0 + 2(–9.80 m/s2)(0.700 – 2.50) m and vyf = –5.94 m/s The hang time is then found as vyf = vyi + ayt: –5.94 m/s = 5.04 m/s + (–9.80 m/s2)t and t = 1.12 s 4.24 ∆x 2π (3.84 × 108 m) = = 1.02 × 103 m/s ∆t [(27.3 d)(24 h/d)(3600 s/h)] (a) v= (b) Since v is constant and only direction changes, a= 4.25 ar = v2 (1.02 × 103)2 = = 2.72 × 10–3 m/s2 r (3.84 × 10 8) v2 (20.0 m/s)2 = = 377 m/s2 r (1.06 m) The mass is unnecessary information. 4.26 4.27 T = (24 h) 3600 s = 86400 s h a= v2 R v= 2πR 2π(6.37 × 106 m) = = 463 m/s T 86400 s a= (463 m/s)2 = 0.0337 m/s2 6.37 × 106 m r = 0.500 m; vt = a= (directed toward the center of the Earth) 2πr 2π(0.500 m) = = 10.47 m/s T (60.0 s/200 rev) 10.5 m/s v2 (10.47)2 = = 219 m/s2 (inward) r 0.5 © 2000 by Harcourt College Publishers. All rights reserved. 13 14 Chapter 4 Solutions *4.28 The centripetal acceleration is ar = v= a rr = v2 , so the required speed is r 1.40(9.80 m/s2)(10.0 m) = 11.7 m/s The period (time for one rotation) is given by T = 2πr/v and the rotation rate is the frequency: f= 4.29. 1 v 11.7 m/s = = = 0.186 s–1 T 2πr 2π(10.0 m) ( a ) v = rω At 8.00 rev/s, v = (0.600 m)(8.00 rev/s)(2π rad/rev) = 30.2 m/s = 9.60π m/s At 6.00 rev/s, v = (0.900 m)(6.00 rev/s)(2π rad/rev) = 33.9 m/s = 10.8π m/s 6.00 rev/s (b) Acceleration = gives the larger linear speed. v2 (9.60π m/s)2 = = 1.52 × 103 m/s2 r 0.600 m (c) At 6.00 rev/s, acceleration = *4.30. (10.8π m/s)2 = 1.28 × 103 m/s2 0.900 m The satellite is in free fall. Its acceleration is due to the acceleration of gravity and is by effect a centripetal acceleration: ar = g v2 =g r v= v= rg = (6400 + 600)(103 m)(8.21 m/s2) = 7.58 × 103 m/s 2π r 2π r 2π(7000 × 103 m) and T = = = 5.80 × 103 s T v (7.58 × 103 m/s) T = (5.80 × 103 s) 1 min = 96.7 min 60 s © 2000 by Harcourt College Publishers. All rights reserved. Chapter 4 Solutions 4.31 We assume the train is still slowing down at the instant in question. v2 ar = = 1.29 m/s2 r at = a= ∆v (– 40.0 km/h)(103 m/km)(1 h/3600 s) = = – 0.741 m/s2 ∆t 15.0 s 2 2 ar + at = (1.29 m/s2)2 + (– 0.741 m/s2)2 = 1.48 m/s2 inward and 29.9° backward Goal Solution G: If the train is taking this turn at a safe speed, then its acceleration should be significantly less than g, perhaps a few m/s2 (otherwise it might jump the tracks!), and it should be directed toward the center of the curve and backward since the train is slowing. O: Since the train is changing both its speed and direction, the acceleration vector will be the vector sum of the tangential and radial acceleration components. The tangential acceleration can be found from the changing speed and elapsed time, while the radial acceleration can be found from the radius of curvature and the train’s speed. A: First, let’s convert the speeds to units from km/h to m/s: vi = 90.0 km/h = (90.0 km/h) 103 m 1 h = 25.0 m/s 1 km 3600 s vf = 50.0 km/h = (50.0 km/h) 103 m 1 h = 13.9 m/s 1 km 3600 s Tangential accel.: at = ∆v 13.9 m/s – 25.0 m/s = = –0.741 m/s2 (backward) ∆t 15.0 s Radial acceleration: a r = a= 2 v 2 (13.9 m/s)2 = = 1.29 m/s2 (inward) r 150 m 2 a t + a r = (–0.741 m/s2)2 + (1.29 m/s2)2 = 1.48 m/s2 at 0.741 m/s2 θ = tan–1 = tan–1 = 29.9° (backwards from a radial line) a r 1.29 m/s2 at ar a L: The acceleration is clearly less than g, and it appears that most of the acceleration comes from the radial component, so it makes sense that the acceleration vector should point mostly toward the center of the curve and slightly backwards due to the negative tangential acceleration. © 2000 by Harcourt College Publishers. All rights reserved. 15 16 Chapter 4 Solutions *4.32 (a) at = 0.600 m/s2 (b) ar = (c) a= v2 (4.00 m/s)2 = = 0.800 m/s2 r 20.0 m 2 2 a t + ar θ = tan–1 4.33 ar = 53.1° inward from path at r = 2.50 m, a = 15.0 m/s2 (a) ar = a cos 30.0° = (15.0 m/s2) cos 30.0° = 13.0 m/s2 (b) ar = 2.50 m v2 r so 30° (c) 4.35 v a v2 = rar = (2.50 m)(13.0 m/s2) = 32.5 m2/s2 v= 4.34 = 1.00 m/s2 2 32.5 m/s = 5.70 m/s 2 a 2 = a t + ar so 2 at = a 2 – a r = (15.0 m/s2)2 – (13 .0 m/s2)2 = 7.50 m/s2 (a) atop = v2 (4.30 m/s)2 = = 30.8 m/s2 down r 0.600 m (b) abottom = v2 (6.50 m/s)2 = = 70.4 m/s2 upward r 0.600 m (a) 36.9° 20.2 m/s2 22.5 m/s2 (b) The components of the 20.2 and the 22.5 m/s2 along the rope together constitute the radial acceleration: ar = (22.5 m/s2) cos (90.0° – 36.9°) + (20.2 m/s2) cos 36.9° © 2000 by Harcourt College Publishers. All rights reserved. Chapter 4 Solutions ar = 29.7 m/s2 © 2000 by Harcourt College Publishers. All rights reserved. 17 18 Chapter 4 Solutions (c) ar = v= v2 r arr = 29.7 m/s2 (1.50 m) = 6.67 m/s tangent to circle v = 6.67 m/s at 36.9° above the horizontal 4.36 (a) vH = 0 + aHt = (3.00i – 2.00j) m/s2 (5.00 s) vH = (15.0i – 10.0j) m/s vJ = 0 + ajt = (1.00i + 3.00j) m/s2 (5.00 s) vJ = (5.00i + 15.0j) m/s vHJ = vH – vJ = (15.0i – 10.0j – 5.00i – 15.0j) m/s vHJ = (10.0i – 25.0j) m/s vHJ = (b) (10.0)2 + (25.0)2 m/s = 26.9 m/s rH = 0 + 0 + 1 1 a t2 = (3.00i – 2.00j) m/s2 (5.00 s)2 2 H 2 rH = (37.5i – 25.0j) m 1 rJ = (1.00i + 3.00j) m/s2 (5.00 s)2 = (12.5i – 37.5j) m 2 rHJ = rH – rJ = (37.5i – 25.0j – 12.5i – 37.5j) m rHJ = (25.0i – 62.5j) m rHJ= (c) (25.0)2 + (62.5)2 m = 67.3 m aHJ = aH – aJ = (3.00i – 2.00j – 1.00i – 3.00j) m/s2 aHJ = (2.00i – 5.00j) m/s2 © 2000 by Harcourt College Publishers. All rights reserved. Chapter 4 Solutions 4.37 Total time in still water t = d 2000 = = 1.67 × 103 s v 1.20 Total time = time upstream plus time downstream tup = 1000 = 1.43 × 103 s (1.20 – 0.500) tdown = 1000 = 588 s (1.20 + 0.500) ttotal = 1.43 × 103 + 588 = 2.02 × 103 s Goal Solution G: If we think about the time for the trip as a function of the stream’s speed, we realize that if the stream is flowing at the same rate or faster than the student can swim, he will never reach the 1.00 km mark even after an infinite amount of time. Since the student can swim 1.20 km in 1000 s, we should expect that the trip will definitely take longer than in still water, maybe about 2000 s (~30 minutes). O: The total time in the river is the longer time upstream (against the current) plus the shorter time downstream (with the current). For each part, we will use the basic equation t = d/v, where v is the speed of the student relative to the shore. A: tup = tdn = d 1000 m = = 1429 s vstudent – vstream 1.20 m/s – 0.500 m/s d 1000 m = = 588 s vstudent + vstream 1.20 m/s + 0.500 m/s Total time in river, triver = tup + tdn = 2.02 × 103 s In still water, tstill = d 2000 m = = 1.67 × 103 s therefore, tR = 1.21tstill v 1.20 m/s L: As we predicted, it does take the student longer to swim up and back in the moving stream than in still water (21% longer in this case), and the amount of time agrees with our estimation. 4.38 The bumpers are initially 100 m = 0.100 km apart. After time t the bumper of the leading car travels 40.0t, while the bumper of the chasing car travels 60.0t. Since the cars are bumper-to-bumper at time t, we have 0.100 + 40.0t = 60.0t, yielding t = 5.00 × 10–3 h = 18.0 s © 2000 by Harcourt College Publishers. All rights reserved. 19 20 4.39 Chapter 4 Solutions v = (1502 + 30.02)1/2 = 153 km/h θ = tan–1 30.0 = 11.3° 150 4.40 north of west For Alan, his speed downstream is c + v, while his speed upstream is c – v. Therefore, the total time for Alan is t1 = L L 2L/c + = c+v c–v 1 – v2/c2 For Beth, her cross-stream speed (both ways) is c2 – v2 Thus, the total time for Beth is t2 = Since 1 – 4.41 2L c2 – v2 = 2L/c 1 – v2/c2 v2 < 1, t1 > t2, or Beth, who swims cross-stream, returns first. c2 α = Heading with respect to the shore β = Angle of boat with respect to the shore ( a ) The boat should always steer for the child at heading α = tan–1 (b) 0.600 = 36.9° 0.800 y 2.50 km/h vx = 20.0 cos α – 2.50 = 13.5 km/h Rescue boat 0.600 km vy = 20.0 sin α = 12.0 km/h β β = tan–1 (c) t= 12.0 km/h = 41.6° 13.5 km/h dy 0.600 km = = 5.00 × 10–2 h = 3.00 min v y 12.0 km/h © 2000 by Harcourt College Publishers. All rights reserved. α x 0.800 km Chapter 4 Solutions 4.42 (a) 21 To an observer at rest in the train car, the bolt accelerates downward and toward the rear of the train. (2.50 m/s)2 + (9.80 m/s)2 = 10.1 m/s2 a= tan θ = 2.50 m/s2 = 0.255 9.80 m/s2 θ = 14.3° to the south from the vertical (b) 4.43 a = 9.80 m/s2 vertically downward Identify the student as the S' observer and the professor as the S observer. For the initial motion in S', we have v y' v x' y′ S′ S u = tan 60.0° = 3 Then, because there is no x-motion in S, we can write v = v ' + u = 0 so that v' = –u = –10.0 m/s. Hence the x y y x ball is thrown backwards in S'. Then, O′ x′ x O a v y = v'y = 3 v x' = 10.0 3 m/s 2 Using vy = 2gh (from Eq. 4.13), we find h= (10.0 3 m/s)2 = 15.3 m 2(9.80 m/s2) The motion of the ball as seen by the student in S' is shown in diagram (b). The view of the professor in S is shown in diagram (c). y S′ u S 60.0° x′ O′ b x O c © 2000 by Harcourt College Publishers. All rights reserved. 22 Chapter 4 Solutions 2 4.44 v i sin2 θi Equation 4.13: h = 2g 2 2 v i sin 2θi 2v i sin θi cos θi Equation 4.14: R = = g g If h = R/6, Equation 4.13 yields [vi sin θi = gR/3 ] (1) Substituting the result given in Equation (1) above into Equation 4.14 gives R= 2( gR/3)vi cos θi g which reduces to [vi cos θi = (a) 1 3gR ] (2) 2 From vyf = vyi + ayt, the time to reach the peak of the path (where vyf = 0) is found to be vi sin θi R tpeak = . Using Equation (1), this gives tpeak = . The total time of the ball’s g 3g flight is then tflight = 2tpeak = 2 (b) R 3g At the peak of the path, the ball moves horizontally with speed vpeak = vxi = vi cos θi Using Equation (1), this becomes vpeak = (c) 3gR . The initial vertical component of velocity is vyi = vi sin θi and, from Equation (1), this is vyi = (d) 1 2 gR/3 Squaring Equations (1) and (2) and adding the results gives 2 v i (sin2 θi + cos2 θi) = gR 3gR 13gR + = 3 4 12 Thus, the initial speed is vi = 13gR 12 . © 2000 by Harcourt College Publishers. All rights reserved. Chapter 4 Solutions (e) 23 Dividing Equation (1) by (2) yields tan θi = vi sin θi ( gR/3) 2 = vi cos θi 1 =3 3gR 2 Therefore, θi = tan–1 2 = 33.7° . 3 (f) For a given initial speed, the projection angle yielding maximum peak height is θi = 90.0°. With the speed found in (d), Equation 4.13 then yields hmax = (g) For a given initial speed, the projection angle yielding maximum range is θi = 45.0°. With the speed found in (d), Equation 4.14 then gives Rmax = 4.45 (13 gR/12) sin2 90.0° 13R = 2g 24 (13gR/12) sin 90.0° 13R = g 12 At any time t, the two drops have identical y-coordinates. The distance between the two drops is then just twice the magnitude of the horizontal displacement either drop has undergone. Therefore, d = 2|x(t)| = 2(vxit) = 2(vi cos θi)t = 2vit cos θi 4.46 After the string breaks the ball is a projectile, for time t in y = vyit + 1 a t2 2 y 1 –1.20 m = 0 + (–9.80 m/s2) t2 2 t = 0.495 s Its constant horizontal speed is vx = x 2.00 m = = 4.04 m/s t 0.495 s so before the string breaks ac = v2 (4.04 m/s)2 = = 54.4 m/s2 r 0.300 m © 2000 by Harcourt College Publishers. All rights reserved. 24 4.47 Chapter 4 Solutions (a) y = tan(θi) x – g 2 2v i cos2 ( θ i) Path of the projectile x2 vi θi Setting x = d cos(φ), and y = d sin(φ ), we have d sin(φ) = tan (θi)d cos(φ) – g 2 2v i cos2 (θi) d φ (d cos(φ))2 Solving for d yields, 2 d= 2v i cos (θi) [sin (θi) cos (φ) – sin (φ) cos (θi)] g cos2 (φ ) 2 or (b) d= Setting 2v i cos (θ i) sin (θ i – φ ) g cos2 (φ ) dd φ = 0 leads to θi = 45° + dθi 2 and 2 d max v i (1 – sin φ ) = g cos2 φ 4.48 (a)(b) Since the shot leaves the gun horizontally, the time it takes to reach the target is t = The vertical distance traveled in this time is y=– where (c) 1 g x 2 gt2 = – = Ax2 2 2 v i A=– g 2 2v i If x = 3.00 m, y = – 0.210 m, then A = vi = –g = 2A –0.210 = –2.33 × 10–2 9.00 –9.80 m/s = 14.5 m/s –4.66 × 10–2 © 2000 by Harcourt College Publishers. All rights reserved. x . vi Chapter 4 Solutions 4.49 25 Refer to the sketch: (a) & (b) ∆x = vxit ; substitution yields 130 = (vi cos 35.0°)t ∆y = vyit + 1 2 at substitution yields 2 20.0 = (vi sin 35.0°)t + 1 (–9.80) t2 2 vi Solving the above gives t = 3.81 s 21.0 m vi = 41.7 m/s (c) 35.0° 1.00 m 130 m vy = vi sin θi – gt vx = vi cos θi At t = 3.81 s, vy = 41.7 sin 35.0° – (9.80)(3.81) = –13.4 m/s vx = (41.7 cos 35.0°) = 34.1 m/s v= 4.50 (a) 2 2 v x + v y = 36.6 m/s The moon's gravitational acceleration is the bullet's centripetal acceleration: (For the moon's radius, see endpapers of text.) a= v2 r v2 1 9.80 m/s2 = 1.74 × 106 m 6 v= (b) 2.84 × 106 m2/s2 = 1.69 km/s v= 2π r T T= 2π r 2π(1.74 × 106 m) = = 6.47 × 103 s = 1.80 h v 1.69 × 103 m/s © 2000 by Harcourt College Publishers. All rights reserved. 26 4.51 Chapter 4 Solutions (a) ar = v2 (5.00 m/s)2 = = 25.0 m/s2 r 1.00 m aT = g = 9.80 m/s2 (b) 9.80 (c) a= 25.0 φt 25.0 φt a a 2 9.80 2 a r + a t = (25.0 m/s2)2 + (9.80 m/s)2 = 26.8 m/s2 at 9.80 m/s2 φ = tan–1 = tan–1 = 21.4° 25.0 m/s2 a r 4.52 x = vixt = vit cos 40.0° Thus, when x = 10.0 m, t= 10.0 m vi cos 40.0° At this time, y should be 3.05 m – 2.00 m = 1.05 m. Thus, 1.05 m = (vi sin 40.0°) 10.0 m 1 10.0 m 2 + (–9.80 m/s2) vi cos 40.0° 2 vi cos 40.0° From this, vi = 10.7 m/s *4.53 At t = 2.00 s, vx = 4.00 m/s vy = –8 .00 m/s v= 2 2 v x + v y = 8.94 m/s θ = tan–1 vy = – 63.4°, below horizontal vx © 2000 by Harcourt College Publishers. All rights reserved. Chapter 4 Solutions 4.54 27 The special conditions allowing use of Equation 4.14 apply. 2 vi 1 For the ball thrown at 45.0°, D = R45 = g vi 2 v i sin 2θ 2 sin 2θ For the bouncing ball, D = R1 + R2 = + where θ is the angle it makes with g g the ground when thrown and when bouncing. (a) 2 We require: 2 2 2 vi vi sin 2θ v i sin 2θ = + g g 4g sin 2θ = (b) 4 5 θ = 26.6° The time for any symmetric parabolic flight is given by 1 y = vyit – 2 gt2 1 0 = vi sin θit – 2 gt2 If t = 0 is the time the ball is thrown, then t = 2vi sin θi g is the time at landing. So, for the ball thrown at 45.0° t4 5 = 2vi sin 45.0° g For the bouncing ball, vi 2vi sin 26.6° 2 2 sin 26.6° 3vi sin 26.6° t = t1 + t2 = + = g g g The ratio of this time to that for no bounce is 3vi sin 26.6°/g 1.34 = = 0.949 2vi sin 45.0°/g 1.41 θ 45.0° θ D © 2000 by Harcourt College Publishers. All rights reserved. 28 4.55 Chapter 4 Solutions From Equation 4.13, the maximum height a ball can reach is 2 v i sin θi h= 2g 2 2 vi For a throw straight up, θi = 90° and h = . 2g 2 From Equation 4.14 the range a ball can be thrown is R = v i sin 2θ . g 2 For maximum range, θ = 45° and R = Therefore for the same vi , h = 4.56. vi . g R 40.0 m = = 20.0 m 2 2 Using the range equation (Equation 4.14) 2 v i sin (2θ i) R= g 2 the maximum range occurs when θi = 45°, and has a value R = Given R, this yields vi = vi . g gR If the boy uses the same speed to throw the ball vertically upward, then vy = gR – gt and y= gR t – gt2 2 at any time, t. At the maximum height, vy = 0, giving t = ymax = gR R g – g 2 R g R , and so the maximum height reached is g 2 =R–R = R 2 2 © 2000 by Harcourt College Publishers. All rights reserved. Chapter 4 Solutions 4.57 29 Choose upward as the positive y-direction and leftward as the positive x-direction. The vertical height of the stone when released from A or B is yi = (1.50 + 1.20 sin 30.0°) m = 2.10 m (a) vi The equations of motion after release at A are vy = vi sin 60.0° – gt = (1.30 – 9.80t) m/s B vx = vi cos 60.0° = 0.750 m/s y = (2.10 + 1.30t – 4.90t2) m 30° 1.20 m vi ∆xA = (0.750t) m When y = 0, t = –1.30 ± (1.30)2 + 41.2 = 0.800 s –9.80 Then, ∆xA = (0.750)(0.800) m = 0.600 m (b) The equations of motion after release at point B are vy = vi(–sin 60.0°) – gt = (–1.30 – 9.80t) m/s vx = vi cos 60.0° = 0.750 m/s yi = (2.10 – 1.30t – 4.90t2) m When y = 0, t = +1.30 ± (–1.30)2 + 41.2 = 0.536 s –9.80 Then, ∆xB = (0.750)(0.536) m = 0.402 m v 2 (1.50 m/s)2 = = 1.87 m/s2 toward the center r 1.20 m (c) ar = (d) After release, a = –gj = 9.80 m/s2 downward A © 2000 by Harcourt College Publishers. All rights reserved. 30° 30 4.58 Chapter 4 Solutions The football travels a horizontal distance 2 v i sin (2θ i) (20.0)2 sin (60.0°) R= = = 35.3 m g 9.80 30° ∆x 20 m Time of flight of ball is t= R 2vi sin θi 2(20.0) sin 30.0° = = 2.04 s g 9.80 The receiver is ∆x away from where the ball lands and ∆x = 35.3 – 20.0 = 15.3 m. To cover this distance in 2.04 s, he travels with a velocity v= 4.59 (a) 15.3 = 7.50 m/s in the direction the ball was thrown 2.04 ∆y = – 1 2 gt ; ∆x = vit. Combine the equations eliminating t: 2 ∆y = – 1 ∆x 2 –2∆y 2 g from this (∆x)2 = v 2 vi g i –2 ∆y = 275 g thus ∆x = vi (b) –2(–300) = 6.80 × 103 m = 6.80 km 9.80 The plane has the same velocity as the bomb in the x direction. Therefore, the plane will be 3000 m directly above the bomb (c) When θ is measured from the vertical, tan θ = θ = tan–1 when it hits the ground. ∆x ; therefore, ∆y ∆x 6800 = tan–1 = 66.2° ∆y 3000 © 2000 by Harcourt College Publishers. All rights reserved. Chapter 4 Solutions 4.60 31 Measure heights above the level ground. The elevation yb of the ball follows yb = R + 0 – 1 2 gt 2 2 yb = R – gx2/2vi with x = vit so (a) 2 The elevation yr of points on the rock is described by yr + x2 = R2. We will have yb = yr at x = 0, but for all other x we require the ball to be above the rock surface as in yb > yr. Then 2 y b + x2 > R2 2 2 R – gx 2 + x2 > R2 2v i R2 – g 2x 4 4 4v i gx2R 2 vi + + x2 > g2x4 4 4v i + x2 > R2 gx2R 2 vi We get the strictest requirement for x approaching zero. If the ball's parabolic trajectory has large enough radius of curvature at the start, the ball will clear the whole rock: gR 1> (b) 2 vi With v i = vi > gR gR and yb = 0, we have 0 = R – gx2 or x = 2 R 2gR The distance from the rock's base is x – R = ( 2 – 1)R 4.61 (a) From Part (C), the raptor dives for 6.34 – 2.00 = 4.34 s undergoing displacement 197 m downward and (10.0)(4.34) = 43.4 m forward. v= ∆d (197) 2 + (43.4) 2 = = 46.5 m/s ∆t 4.34 (b) α = tan–1 –197 = –77.6° 43.4 (c) 197 = 1 2 gt 2 t = 6.34 s © 2000 by Harcourt College Publishers. All rights reserved. 32 Chapter 4 Solutions Goal Solution G: We should first recognize that the hawk cannot instantaneously change from slow horizontal motion to rapid downward motion. The hawk cannot move with infinite acceleration. We assume that the time required for the hawk to accelerate is short compared to two seconds. Based on our everyday experiences, a reasonable diving speed for the hawk might be about 100 mph (~ 50 m/s) downwards and should last only a few seconds. O: We know the distance that the mouse and hawk fall, but to find the diving speed of the hawk, we must know the time of descent. If the hawk and mouse both maintain their original horizontal velocity of 10 m/s (as they should without air resistance), then the hawk only needs to think about diving straight down, but to a ground-based observer, the path will appear to be a straight line angled less than 90° below horizontal. A: The mouse falls a total vertical distance, y = 200 m – 3.00 m = 197 m The time of fall is found from y = vyit – t= 1 2 gt 2 2(197 m) = 6.34 s 9.80 m/s2 To find the diving speed of the hawk, we must first calculate the total distance covered from the vertical and horizontal components. We already know the vertical distance, y, so we just need the horizontal distance during the same time (minus the 2.00 s late start). x = vxi(t – 2.00 s) = (10.0 m/s)(6.34 s – 2.00 s) = 43.4 m The total distance is d = x2 + y2 = So the hawk’s diving speed is vhawk = (43.4 m)2 + (197 m)2 = 202 m d 202 m = = 46.5 m/s t – 2.00 s 4.34 s At an angle of θ = tan–1 y 197 m = tan–1 = 77.6° below the horizontal x 43.4 m L: The answers appear to be consistent with our predictions, even thought it is not possible for the hawk to reach its diving speed in zero time. Sometimes we must make simplifying assumptions to solve complex physics problems, and sometimes these assumptions are not physically possible. Once the idealized problem is understood, we can attempt to analyze the more complex, real-world problem. For this problem, if we considered the realistic effects of air resistance and the maximum diving acceleration attainable by the hawk, we might find that the hawk could not catch the mouse before it hit the ground. © 2000 by Harcourt College Publishers. All rights reserved. Chapter 4 Solutions 4.62 (1) Equation of bank (1) y2 = 16x (2) and (3) are the equations of motion 1 (3) y = – 2 gt2 (2) x = vit Substitute for t from (2) into (3) y = – 1 x2 g 2 v 2 i Equate y from the bank equation to y from the equations of motion: 16x = – 1 x2 2 g2x4 g 2x 3 g 2 ⇒ 4 – 16x = x 4 – 16 = 0 2 v 4v 4v i i 4 From this, x = 0 or x3 = Also, y = – 64vi g2 i and x = 4 104 1/3 = 18.8 m 9.802 1 x2 1 (9.80)(18.8)2 g = = –17.3 m 2 v 2 2 (10.0) 2 i vi = 10 m/s 4.63 Consider the rocket's trajectory in 3 parts as shown in the sketch. 2 3 1 53° Our initial conditions give: ay = 30.0 sin 53.0° = 24.0 m/s2; ax = 30.0 cos 53.0° = 18.1 m/s2 vyi = 100 sin 53.0° = 79.9 m/s; vxi = 100 cos 53.0° = 60.2 m/s © 2000 by Harcourt College Publishers. All rights reserved. 33 34 Chapter 4 Solutions The distances traveled during each phase of the motion are given in the table. Path #1 : vyf – 79.9 = (24.0)(3.00) or vyf = 152 m/s vxf – 60.2 = (18.1)(3.00) or vxf = 114 m/s Path #2 : ∆y = (79.9)(3.00) + 1 (24.0)(3.00) 2 = 347 m 2 ∆x = (60.2)(3.00) + 1 (18.1)(3.00) 2 = 262 m 2 ax = 0, vxf = vxi = 114 m/s Path Part 0 – 152 = –(9.80)t or t = 15.5 s ∆x = (114)(15.5) = 1.77 × ∆y = (152)(15.5) – 103 m; 1 (9.80)(15.5) 2 = 1.17 × 103 m Path #3 : #1 (vyf)2 – 0 = 2(–9.80)(–1.52 × 103) 2 #2 #3 ay 24.0 –9.80 –9.80 ax 18.1 0.0 0.00 vyf 152 vxf 114 0.0 114 v yi 79.9 152 v xi 60.2 114 –173 114 0.00 114 or vyf = –173 m/s ∆y 347 1.17 × 103 –1.52 × 103 vxf = vxi = 114 m/s since ax = 0 ∆x 262 1.77 × 103 2.02 × 103 –173 – 0 = –(9.80)t or t = 17.6 s t 3.00 15.5 17.6 ∆x = (114)(17.7) = 2.02 × 103 m 4.64 (a) ∆y(max) = 1.52 × 103 m (b) t(net) = 3.00 + 15.5 + 17.7 = 36.1 s (c) ∆x(net) = 262 + 1.77 × 103 + 2.02 × 103 = 4.05 × 103 m Let V = boat's speed in still water and v = river's speed and let d = distance traveled upstream in t1 = 60.0 min and t2 = time of return. Then, for the log, 1000 m = vt = v(t1 + t2), and for the boat, d = (V – v)t1; (d + 1000) = (V +v)t2; and t = t1 + t2 Combining the above gives 1000 d d + 1000 = + v (V – v) (V + v) Substituting for d = (V – v)(3600) gives v = 0.139 m/s © 2000 by Harcourt College Publishers. All rights reserved. Chapter 4 Solutions 4.65 (a) While on the incline: 2 (b) v 2 – v i = 2a ∆x v – v i = at v2 – 0 = 2(4.00)(50.0) 20.0 – 0 = 4.00t v = 20.0 m/s t = 5.00 s Initial free-flight conditions give us vxi = 20.0 cos 37.0° = 16.0 m/s; vyi = –20.0 sin 37.0° = –12.0 m/s vx = vxi since ax = 0; 2 vy = –(2ay ∆y + vyi )1/2 = –[2(–9.80)(–30.0) + (–12.0)2]1/2 = –27.1 m/s 2 2 v = (vx + vy )1/2 = [(16.0)2 + (–27.1)2]1/2 = 31.5 m/s at 59.4° below the horizontal (c) t1 = 5 s; t2 = (vy – vyi) (–27.1 + 12.0) = = 1.54 s ay –9.80 t = t1 + t2 = 6.54 s (d) ∆x = vxit1 = (16.0)(1.54) = 24.6 m 50 m 37° 30 m © 2000 by Harcourt College Publishers. All rights reserved. 35 36 4.66 Chapter 4 Solutions (a) Coyote: ∆x = 1 2 1 at ; 70.0 = (15.0) t2 2 2 Coyoté Stupidus Roadrunner: ∆x = vit; 70.0 = vit Solving the above, we get vi = 22.9 m/s (b) At the edge of the cliff vxi = at = (15.0)(3.06) = 45.8 m/s ∆y = 1 a t2 2 y Substituting we find –100 = ∆x = vxit + 1 (–9.80) t2 2 1 1 a t2 = (45.8)t + (15.0) t2 2 x 2 Solving the above gives ∆x = 360 m (c) and t = 3.06 s t = 4.52 s For the Coyote's motion through the air v xf = v xi + a xt vxf = 45.8 + 15(4.52) vxf = 114 m/s v yf = v yi + a yt = 0 – 9.80(4.52) vyf = –44.3 m/s 4.67 (a) ∆x = vxit, ∆y = vyit + 1 gt2, 2 d cos 50.0° = (10.0 cos 15.0°)t, and d sin 50.0° = (10.0 sin 15.0°)t + 1 (–9.80) t2 2 Solving the above gives d = 43.2 m t = 2.87 s © 2000 by Harcourt College Publishers. All rights reserved. Chicken Delightus EP BE BEE P Chapter 4 Solutions (b) 37 Since ax = 0, vxf = vxi = 10.0 cos 15.0° = 9.66 m/s vyf = vyi + ay t = (10.0 sin 15.0°) – (9.80)(2.87) = –25.5 m/s Air resistance would decrease the values of the range and maximum height. As an air foil he can get some lift and increase his distance. 4.68 Define i to be directed East, and j to be directed North. N According to the figure, set v je = velocity of Jane, relative to the earth vjm vje vme = velocity of Mary, relative to the earth vjm = velocity of Jane, relative to Mary, Such that vje = vjm + vme 60.0° E Solve for part (b) first. By the figure, vje = [5.40(cos 60.0°)i + 5.40(sin 60.0°)j] m/s = (2.70i + 4.68j) m/s and vme = 4.00i m/s So, (b) vjm = (–1.30i + 4.68j) m/s The distance between the two players increases at a rate of |vjm|: |vjm| = (a) (1.30)2 + (4.68)2 m/s = 4.86 m/s Therefore, t = d 25.0 m = = 5.14 s vjm 4.86 m/s (c) After 4 s, d = vjmt = (4.86 m/s)(4.00) = 19.4 m apart © 2000 by Harcourt College Publishers. All rights reserved. vme 38 Chapter 4 Solutions *4.69 Think of shaking down the mercury in an old fever thermometer. Swing your hand through a circular arc, quickly reversing direction at the bottom end. Suppose your hand moves through one-quarter of a circle of radius 60 cm in 0.1 s. Its speed is 1 (2π )(0.6 m) 4 ≅ 9 m/s 0.1 s and its centripetal acceleration is v2 (9 m/s)2 ≅ ~102 m/s2 r 0.6 m The tangential acceleration of stopping and reversing the motion will make the total acceleration somewhat larger, but will not affect its order of magnitude. 4.70 Find the highest elevation θ H that will clear the mountain peak; this will yield the range of the closest point of bombardment. Next find the lowest elevation θL that will clear the mountain peak; this will yield the maximum range under these conditions if both θH and θL are > 45°; x = 2500 m, y = 1800 m, vi = 250 m/s. y = vyit – 1 1 gt2 = vi (sin θ)t – gt2 2 2 x = vxit = vi (cos θ)t Thus t = x vi cos θ Substitute into the expression for y y = vi (sin θ) but x 1 – g Error! vi cos θ 2 1 gx2 = tan2 θ + 1 thus y = x tan θ – 2 (tan2 θ + 1) and 2 cos θ 2v i 0= gx2 2 2v i tan2 θ – x tan θ + gx2 2 2v i +y Substitute values, use the quadratic formula and find tan θ = 3.905 or 1.197 which gives θH = 75.6° and θL = 50.1° © 2000 by Harcourt College Publishers. All rights reserved. Chapter 4 Solutions 2 v i sin 2θH Range (at θH) = = 3.07 × 103 m from enemy ship g 3.07 × 103 – 2500 – 300 = 270 m from shore 2 v i sin 2θL Range (at θL) = = 6.28 × 103 m from enemy ship g 6.28 × 103 – 2500 – 300 = 3.48 × 103 m from shore Therefore, safe distance is < 270 m vi = 250 m/s vi or > 3.48 × 103 m from the shore. 1800 m θH θ L 2500 m 300 m © 2000 by Harcourt College Publishers. All rights reserved. 39 40 Chapter 4 Solutions © 2000 by Harcourt College Publishers. All rights reserved. Chapter 5 Even Answers 2. 4. 6. 8. 10. 12. 14. 16. 18. 20. 22. 24. 26. 28. 30. 32. 34. 36. 38. 40. 42. 44. 46. 48. 50. 52. 54. 56. 58. 60. 62. 64. 66. 68. (a) 5.00 m/s2 (b) 19.6 N (c) 10.0 m/s2 444 s (a) 1.44 m (b) (50.9i + 1.40j) N 4.45 N (a) –4.47 × 1015 m/s2 (b) +2.09 × 10–10 N (a) 534 N (b) 54.5 kg 2.55 N for a 88.7 kg person (16.3i + 14.6j) N 5.15 m/s2 at 14.0° S of E (a) 181˚ counterclockwise from x-axis (b) 11.2 kg (c) 37.5 m/s (d) (–37.5i – 0.893j) m/s 112 N T1 = 296 N, T2 = 163 N, T3 = 325 N (a) T = Fg/sin θ (b) 1.79 N (a) 5.10 × 103 N (b) 3.62 × 103 kg (a) a = g tan θ (b) 4.16 m/s2 (a) 2.54 m/s2 down the incline (b) 3.18 m/s (a) 3.57 m/s2 (b) 26.7 N (c) 7.14 m/s (a) 36.8 N (b) 2.45 m/s2 (c) 1.23 m m 1m 2 m 1m 2 m 2g m 2g (a) a1 = 2a2 (b) T1 = g, T2 = g (c) a1 = , a2 = 1 1 1 4m 1 + m2 2m 1 + 2 m 2 m1 + 4 m 2 2m 1 + 2 m 2 7.84 m/s2 independent of the mass 0.456 (a) 55.2˚ (b) 167 N µs = 0.727, µk = 0.577 221 m (a) 2.31 m/s2, down for 4.00 kg, left for 1.00 kg, up for 2.00 kg (b) Tleft = 30.0 N, Tright = 24.2 N (a) 0.931 m/s2 (b) 6.10 cm (a) 3.00 s (b) 20.1 m (c) (18.0i – 9.00j) m (a) 2.00 m/s2 (b) 4.00 N on m1, 6.00 on m2, 8.00 on m3 (c) 14.0 N between m1 and m2, 8.00 N between m2 and m3 g sin θ (a) M = 3m sin θ (b) T1 = 2 mg sin θ, T2 = 3 mg sin θ (c) a = 1 + 2 sin θ 1 + sin θ 1 + sin θ (d) T1 = 4mg , T2 = 6mg (e) Mmax = 3m(sin θ + µs cos θ) 1 + 2 sin θ 1 + 2 sin θ (f) Mmin = 3m(sin θ – µs cos θ) (g) T2,max – T2,min = (Mmax – Mmin)g = 6µs mg cos θ (a) (–45.0i + 15.0j) m/s (b) 162˚ from +x-axis (c) (–225i + 75.0j) m (d) (–227, 79.0) m (a) 4.90 m/s2 (b) 3.13 m/s (c) 1.35 m (d) 1.14 s (e) no The system does not start to move when released, f1 + f2 = 29.4 N a = 0.143 m/s2, approximately 4% high (b) T = 9.80 N, a = 0.580 m/s2 © 2000 by Harcourt College Publishers. All rights reserved. 2 70. 72. 74. 76. 78. Chapter 5 Even Answers (a) m 2g m 1M (b) m 2g(M + m 1) m 1M + m 2(m 1 + M) m 1M + m 2(m 1 + M) m 1m 2g Mm2g (c) (d) m 1M + m 2(m 1 + M) m 1M + m 2(m 1 + M) (a) 2.20 m/s2 (b) 27.4 N (a) 600 N (b) 1100 N (forward) tan θ 1 2mg mg mg (a) T1 = , T2 = = (b) θ2 = tan–1 sin θ1 sin θ2 sin tan –1 1 tan θ 2 n = (82.3 N) cos θ, a = (9.80 m/s2) 2 sin θ 1 © 2000 by Harcourt College Publishers. All rights reserved. Chapter 5 Solutions *5.1 For the same force F, acting on different masses F = m1a1 and F = m2a2 (a) m 1 a2 1 = = m2 a 1 3 (b) F = (m1 + m2)a = 4m1a = m1(3.00 m/s2) a = 0.750 m/s2 *5.2 5.3 F = 10.0 N, m = 2.00 kg F 10.0 N = = 5.00 m/s2 m 2.00 kg (a) a= (b) Fg = mg = (2.00 kg)(9.80 m/s2) = 19.6 N (c) a= 2F 2(10.0 N) = = 10.0 m/s2 m 2.00 kg m = 3.00 kg, a = (2.00i + 5.00j)m/s2 F = ma = (6.00i + 15.0j) N F = (6.00)2 + (15.0)2 N = 16.2 N 5.4 mtrain = 15,000,000 kg F = 750,000 N a= F 75.0 × 104 N = = 5.00 × 10–2 m/s2 m 15.0 × 106 kg vf = vi + at t= vi = 0 v f – v i (80.0 km/h)(1000 m/km)(1 h/3600 s) = a 5.00 × 10–2 m/s2 t = 444 s © 2000 by Harcourt College Publishers. All rights reserved. 2 5.5 Chapter 5 Solutions We suppose the barrel is horizontal. m = 5.00 × 10–3 kg, vf = 320 m/s, vi = 0, x = 0.820 m – ∆v – F x = ma = m ∆t (Eq. 5.2) Find ∆t from Eq. 2.2 ∆t = ∆x 0.820 m = = 5.13 × 10–3 s – 160 m/s v – (320 m/s) ∴ Fx = (5.00 × 10–3 kg) = 312 N (5.13 × 10 –3) Along with this force, which we assume is horizontal, exerted by the exploding gunpowder, the bullet feels a downward 49.0 mN force of gravity and an upward 49.0 mN force exerted by the barrel surface under it. 5.6 Fg = mg = 1.40 N, m = 0.143 kg vf = 32.0 m/s, vi = 0, ∆t = 0.0900 s – v = 16.0 m/s – ∆v a = = 356 m/s2 ∆t (a) – Distance x = v t = (16.0 m/s)(0.0900 s) = 1.44 m (b) ∑F = ma Fpitcher – 1.40 Nj = (0.143 kg)(356i m/s2) Fp = (50.9i + 1.40j) N 5.7 Fg = weight of ball = mg vrelease = v, time to accelerate = t a= ∆v v v = = i ∆t t t (a) v vt – Distance x = v t = t = 2 2 (b) Fp – Fg j = Fp = Fg v i g t F gv i + Fg j gt © 2000 by Harcourt College Publishers. All rights reserved. Chapter 5 Solutions *5.8 Fg = mg 1 pound = (0.453 592 37 kg)(32.1740 ft/s2) 12.0 in 0.0254 m = 4.45 N 1 ft 1 in. 5.9 m = 4.00 kg, vi = 3.00i m/s, v8 = (8.00i + 10.0j) m/s, t = 8.00 s a= ∆v (5.00i + 10.0j) = m/s2 t 8.00 F = ma = (2.50i + 5.00j) N F= 5.10 (a) (2.50)2 + (5.00)2 = 5.59 N Let the x-axis be in the original direction of the molecule's motion. vf = vi + at –670 m/s = 670 m/s + a(3.00 × 10–13 s) a = –4.47 × 1015 m/s2 (b) For the molecule ∑F = ma. Its weight is negligble. Fwall on molecule = 4.68 × 10–26 kg (–4.47 × 1015 m/s2) = –2.09 × 10–10 N – F molecule on wall = +2.09 × 10–10 N 2 5.11 (a) 2 2 F = ma and v f = v i + 2ax or a = 2 vf – v i 2x Therefore, 2 F=m 2 (v f – v i ) 2x F = (9.11 × 10–31 kg) [(7.00 × 105 m/s2)2 – (3.00 × 105 m/s)2] (2)(0.0500 m) = 3.64 × 10 –18 N © 2000 by Harcourt College Publishers. All rights reserved. 3 Chapter 5 Solutions 4 (b) The weight of the electron is Fg = mg = (9.11 × 10–31 kg)(9.80 m/s2) = 8.93 × 10 –30 N The accelerating force is 4.08 × 1011 times the weight of the electron. Goal Solution G: We should expect that only a very small force is required to accelerate an electron, but this force is probably much greater than the weight of the electron if the gravitational force can be neglected. O: Since this is simply a linear acceleration problem, we can use Newton’s second law to find the force as long as the electron does not approach relativistic speeds (much less than 3 × 108 m/s), which is certainly the case for this problem. We know the initial and final velocities, and the distance involved, so from these we can find the acceleration needed to determine the force. 2 2 A: From v f = v i + 2ax and ∑F = ma we can solve for the acceleration and the force. 2 a= (a) F = 2 (v f – v i ) 2x 2 and so ∑F = 2 m(vf – v i ) 2x (9.11 × 10–31 kg) [(7.00 × 105 m/s2) – (3.00 × 105 m/s)2] = 3.64 × 10–18 N (2)(0.0500 m) (b) The weight of the electron is Fg = mg = (9.11 × 10–31 kg)(9.80 m/s2) = 8.93 × 10–30 N The ratio of the accelerating force to the weight is F = 4.08 × 1011 Fg L: The force that causes the electron to accelerate is indeed a small fraction of a newton, but it is much greater than the gravitational force. For this reason, it is quite reasonable to ignore the weight of the electron in electric charge problems. 5.12 5.13 (a) Fg = mg = 120 lb = 4.448 (b) m= N (120 lb) = 534 N lb Fg 534 N = = 54.5 kg g 9.80 m/s2 Fg = mg = 900 N m= 900 N = 91.8 kg 9.80 m/s2 (Fg)on Jupiter = (91.8 kg)(25.9 m/s2) = 2.38 kN © 2000 by Harcourt College Publishers. All rights reserved. Chapter 5 Solutions *5.14 Imagine a quick trip by jet, on which you do not visit the rest room and your perspiration is just canceled out by a glass of tomato juice. By subtraction, (Fg)p = mgp and (Fg)C = mgC give ∆Fg = m(gP – gC). For a person whose mass is 88.7 kg, the change in weight is ∆Fg = (88.7)(9.8095 – 9.7808) = 2.55 N A precise balance scale, as in a doctor's office, reads the same in different locations because it compares you with the standard masses on its beams. A typical bathroom scale is not precise enough to reveal this difference. 5.15 (a) ∑F = F1 + F2 = (20.0i + 15.0j) N ∑F = ma, 20.0i + 15.0j = 5.00 a F2 a = (4.00i + 3.00 j) m/s2 or a = 5.00 m/s2 at θ = 36.9° F2 90.0° 60.0° m m F1 (b) F2x = 15.0 cos 60.0° = 7.50 N a F2y = 15.0 sin 60.0° = 13.0 N F2 = (7.50i + 13.0j) N ∑F = F1 + F2 = (27.5i + 13.0j) N = ma = 5.00a a = (5.50i + 2.60j) m/s2 5.16 We find acceleration: r – ri = vit + 4.20 m i – 3.30 m j = 0 + 1 2 at 2 1 a(1.20 s)2 = 0.720 s2 a 2 a = (5.83 i – 4.58j) m/s2 Now ∑F = ma becomes F2 = 2.80 kg (5.83i – 4.58j) m/s2 + (2.80 kg)9.80 m/s2 j F2 = (16.3i + 14.6j) N 5.17 (a) You and earth exert equal forces on each other: m yg = M ea e If your mass is 70.0 kg, © 2000 by Harcourt College Publishers. All rights reserved. F1 b 5 6 Chapter 5 Solutions ae = (70.0 kg)(9.80 m/s2) = ~ 10–22 m/s2 5.98 × 1024 kg © 2000 by Harcourt College Publishers. All rights reserved. Chapter 5 Solutions (b) You and planet move for equal times in x = 2xy = ay xe = 5.18 2xe ae ae my 70.0 kg x = x = (0.500 m) ~ 10–23 m a y y m e y 5.98 × 1024 kg (20.0)2 + (10.0 – 15.0)2 = 20.6 N F= a= 1 2 at . If the seat is 50.0 cm high, 2 F m a = 5.15 m/s2 at 14.0° S of E *5.19 Choose the x-axis forward. Then ∑Fx = max (2000 Ni) – (1800 Ni) = (1000 kg)a a = 0.200 m/s2i 5.20 1 2 1 at = 0 + (0.200 m/s2)(10.0 s) 2 = 10.0 m 2 2 (b) xf – xi = vit + (c) vf = vi + at = 0 + (0.200 m/s2i)(10.0 s) = 2.00 m/s i ∑F = ma reads (–2.00i + 2.00j + 5.00i – 3.00j – 45.0i) N = m(3.75 m/s2)â where â represents the direction of a (–42.0i – 1.00j) N = m(3.75 m/s2)â F= (42.0)2 + (1.00)2 N at Arctan 1.00 below the –x-axis 42.0 = m(3.75 m/s2)â F = 42.0 N at 181° = m(3.75 m/s2)â © 2000 by Harcourt College Publishers. All rights reserved. 7 8 Chapter 5 Solutions For the vectors to be equal, their magnitudes and their directions must be equal: (a) ∴ â is at 181° counterclockwise from the x-axis (b) m= (d) vf = vi + at = 0 + (3.75 m/s2 at 181°)10.0 s 42.0 N = 11.2 kg 3.75 m/s2 = 37.5 m/s at 181° = 37.5 m/s cos 181°i + 37.5 m/s sin 181°j v = (–37.5i – 0.893j) m/s (c) v = 37.52 + 0.8932 m/s = 37.5 m/s *5.21 (a) 15.0 lb up *5.22 vx = dx dy = 10t, vy = = 9t2 dt dt ax = dv x dvy = 10, ay = = 18t dt dt (b) 5.00 lb up (c) 0 At t = 2.00 s, ax = 10.0 m/s2, ay = 36.0 m/s2 Fx = max = (3.00 kg)(10.0 m/s2) = 30.0 N Fy = may = (3.00 kg)(36.0 m/s2) = 108 N F= 5.23 2 2 F x + F y = 112 N m = 1.00 kg 50.0 m α mg = 9.80 N 0.200 m T T 0.200 m tan α = 25.0 m mg α = 0.458° Balance forces, 2T sin α = mg T= 9.80 N = 613 N 2 sin α © 2000 by Harcourt College Publishers. All rights reserved. Chapter 5 Solutions 5.24 T3 = Fg (1) T1 sin θ1 + T2 sin θ2 = Fg (2) T1 cos θ1 = T2 cos θ2 (3) T1 9 T2 θt22 θt11 Eliminate T2 and solve for T1, T1 = Fg cos θ2 Fg cos θ2 = (sin θ1 cos θ2 + cos θ1 sin θ2) sin (θ 1 + θ 2) Fg T3 = Fg = 325 N 5.25 T1 = Fg cos 25.0° = 296 N sin 85.0° T2 = T1 cos θ1 cos 60.0° = (296 N) = 163 N cos θ2 cos 25.0° See the solution for T1 in Problem 5.24. T *5.26 (a) An explanation proceeding from fundamental physical principles will be best for the parents and for you. Consider forces on the bit of string touching the weight hanger as shown in the free-body diagram: θt Tx Horizontal Forces: ∑Fx = max ⇒ –Tx + T cos θ = 0 Vertical Forces: ∑Fy = may ⇒ –Fg + T sin θ = 0 Fg You need only the equation for the vertical forces to find that the tension in the string is given by T = Fg . The force the child feels gets smaller, changing from T to T cos θ, sin θ while the counterweight hangs on the string. On the other hand, the kite does not notice what you are doing and the tension in the main part of the string stays constant. You do not need a level, since you learned in physics lab to sight to a horizontal line in a building. Share with the parents your estimate of the experimental uncertainty, which you make by thinking critically about the measurement, by repeating trials, practicing in advance and looking for variations and improvements in technique, including using other observers. You will then be glad to have the parents themselves repeat your measurements. (b) T= Fg (0.132 kg)(9.80 m/s2) = = 1.79 N sin θ sin 46.3° © 2000 by Harcourt College Publishers. All rights reserved. 10 5.27 Chapter 5 Solutions (a) Isolate either mass T + mg = ma = 0 T T = mg The scale reads the tension T, so 49.0 N T = mg = 5.00 kg × 9.80 m/s2 = 49.0 N (b) Isolate each mass T2 T2 + 2T1 = 0 T2 = 2 T1 = 2mg T1 = 98.0 N (c) T1 y ∑F = n + T + mg = 0 n x T Take the component along the incline nx + Tx + mgx = 0 or θt==30.0° 30.0° 0 + T – mg sin 30.0° = 0 T = mg sin 30.0° = = 24.5 N 5.28 θt mg (5.00)(9.80) = 2 2 49.0 N Let R represent the horizontal force of air resistance. (a) ∑Fx = max becomes T sin 40.0° – R = 0 ∑Fy = may reads T cos 40.0° – Fg = 0 Then T = mg 6.08 × 103 N = = 7.93 × 103 N cos 40.0° cos 40.0° R = 7.93 × 103 N sin 40 = 5.10 × 103 N © 2000 by Harcourt College Publishers. All rights reserved. Chapter 5 Solutions (b) 11 The value of R will be the same. Now T sin 7.00° – R = 0 T= 5.10 × 103 N = 4.18 × 104 N sin 7.00° T cos 7.00° – mg = 0 m= (4.18 × 104 N) cos 7.00° = 4.24 × 103 kg 9.80 m/s2 mwater = 4.24 × 103 kg – 620 kg = 3.62 × 103 kg 5.29 Choosing a coordinate system with i East and j North. N (5.00 N)j + F1 = 10.0 N ∠ 30.0° = (5.00 N)j + (8.66 N)i ∴ F1 = 8.66 N (East) F2 a = 10.0 m/s 2 30.0° E 1.00 kg F1 Goal Solution G: The net force acting on the mass is ∑F = ma = (1 kg)(10 m/s2) = 10 N, so if we examine a diagram of the forces drawn to scale, we see that F1 ≈ 9 N directed to the east. O: We can find a more precise result by examining the forces in terms of vector components. For convenience, we choose directions east and north along i and j, respectively. A: a = [(10.0 cos 30.0°)i + (10.0 sin 30.0°)j] m/s2 = (8.66i + 5.00j) m/s2 From Newton's second law, ∑F = m a =(1.00 kg)[(8.66i + 5.00j) m/s2] = (8.66i + 5.00j) N And ∑F = F1 + F2. so F1 = ∑F – F2 = (8.66i + 5.00j – 5.00j) N = 8.66i N = 8.66 N to the east L: Our calculated answer agrees with the prediction from the force diagram. © 2000 by Harcourt College Publishers. All rights reserved. 12 5.30 Chapter 5 Solutions (a) The cart and mass accelerate horizontally. ∑Fy = may + T cos θ – mg = 0 ∑Fx = max + T sin θ = ma Substitute T = mg cos θ mg sin θ = mg tan θ = ma cos θ a = g tan θ (b) *5.31 a = (9.80 m/s2) tan23.0° = 4.16 m/s2 Let us call the forces exerted by each person F1 and F2. Thus, for pulling in the same direction, Newton's second law becomes F1 + F2 = (200 kg)(1.52 m/s2) or F1 + F2 = 304 N (1) When pulling in opposite directions, F1 – F2 = (200 kg)(–0.518 m/s2) or F1 – F2 = –104 N (2) Solving simultaneously, we find F1 = 100 N , and F2 = 204 N y 5.32 The two forces acting on the block are the normal force, n, and the weight, mg. If the block is considered to be a point mass and the x-axis is chosen to be parallel to the plane then the free body diagram will be as shown in the figure to the right. The angle θ is the angle of inclination of the plane. Applying Newton's second law for the accelerating system (and taking the direction of motion as the positive direction) we have n x t θ mg sin θ ∑Fy = n – mg cos θ = 0; n = mg cos θ ∑Fx = –mg sin θ = ma; a = –g sin θ (a) When θ = 15.0° a = –2.54 m/s2 © 2000 by Harcourt College Publishers. All rights reserved. θ t mg cos θ Chapter 5 Solutions (b) Starting from rest 2 2 v f = v i + 2ax vf = *5.33 2 2ax = (2)(–2.54 m/s2)(–2.00 m) = 3.18 m/s 2 v f = v i + 2ax Taking v = 0, vi = 5.00 m/s, and a = –g sin(20.0°) gives 0 = (5.00)2 – 2(9.80) sin (20.0°)x or, 5.34 x= 25.0 = 3.73 m 2(9.80) sin (20.0°) With m1 = 2.00 kg, m2 = 6.00 kg and θ = 55.0°, y′ T y n a T m2 m2g cos θ a x m1 θ t m2g sin θ m1 g m2g (a) x′ ∑Fx = m2g sin θ – T = m2a and T – m1g = m1a a= m 2g sin θ – m 1g = 3.57 m/s2 m1 + m2 (b) T = m1(a + g) = 26.7 N (c) Since vi = 0, vf = at = (3.57 m/s2)(2.00 s) = 7.14 m/s © 2000 by Harcourt College Publishers. All rights reserved. 13 14 5.35 Chapter 5 Solutions Applying Newton's second law to each block (motion along the x-axis). For m2: ∑Fx = F – T = m2a For m1: ∑Fx = T = m1a Solving these equations for a and T, we find a = F m1 + m2 and T= Fm 1 m1 + m2 a a n1 n2 T m1 T m1g *5.36 F m2 m2g First, consider the 3.00 kg rising mass. The forces on it are the tension, T, and its weight, 29.4 N. With the upward direction as positive, the second law becomes ∑Fy = may T – 29.4 N = (3.00 kg)a T + Rising Mass Falling Mass m1 = 3.00 kg m2 = 5.00 kg (1) The forces on the falling 5.00 kg mass are its weight and T, and its acceleration is the same as that of the rising mass. Calling the positive direction down for this mass, we have + (Fg)1 = 29.4 N ∑Fy = may 49 N – T = (5.00 kg)a (2) Equations (1) and (2) can be solved simultaneously to give (a) the tension as T = 36.8 N (b) and the acceleration as a = 2.45 m/s2 T © 2000 by Harcourt College Publishers. All rights reserved. (Fg)2 = 49 N Chapter 5 Solutions (c) Consider the 3.00 kg mass. We have y = v it + 5.37 T – m 1g = m 1a 1 2 1 at = 0 + (2.45 m/s2)(1.00 s) 2 = 1.23 m 2 2 (1) Forces acting on 2.00 kg block Fx – T = m2a (2) Forces acting on 8.00 kg block (a) Eliminate T and solve for a. a= (b) a > 0 for F x > m 1 g = 19.6 N Eliminate a and solve for T. T= (c) F x – m 1g m1 + m2 m1 (F + m2g) m1 + m2 x F x, N ax, m/s2 –100 –12.5 –78.4 –9.80 T = 0 for F x ≤ –m 2g = –78.4 N –50.0 –6.96 0 –1.96 50.0 3.04 100 8.04 ax (m/s2) 12 10 8 80 100 60 0 20 40 −20 −40 −60 −80 −100 6 4 2 Fx (N) −4 −6 −8 −10 −12 5.38 (a) Pulley P1 has acceleration a2. Since m1 moves twice the distance P1 moves in the same time, m1 has twice the acceleration of P1. i.e., a 1 = 2a 2 © 2000 by Harcourt College Publishers. All rights reserved. 15 16 Chapter 5 Solutions (b) From the figure, and using F = ma: m 2g – T 2 = m 2a2 (1) T 1 = m 1a1 = 2m 1a2 (2) T2 – 2T1 = 0 (3) T1 P1 T2 P2 m1 a1 T2 Equation (1) becomes m2g – 2T1 = m2a2 a2 m2 This equation combined with Equation (2) yields m2g 2T1 m 2 m1 + = m2g m1 2 T1 = (c) 5.39 m 1m 2 1 2 2m 1 + m 2 g and T2 = m 1m 2 m 1 + 14 m 2 g From the values of T1 and T2 we find that a1 = T1 m 2g = m1 2m 1 + 12 m 2 a2 = m 2g 1 a = 2 1 4m 1 + m 2 First, we will compute the needed accelerations: (1) Before it starts to move: ay = 0. (2) During the first 0.800 s: ay = (3) While moving at constant velocity: ay = 0. (4) During the last 1.50 s: ay = vy – viy 1.20 m/s – 0 = = 1.50 m/s2. t 0.800 s vy – viy 0 – 1.20 m/s = = –0.800 m/s2. t 1.50 s Newton's second law is: T = 706 N + (72.0 kg)ay. (a) When ay = 0, T = 706 N (b) When ay = 1.50 m/s2, T = 814 N (c) When ay = 0, T = 706 N (d) When ay = –0.800 m/s2, T = 648 N © 2000 by Harcourt College Publishers. All rights reserved. Chapter 5 Solutions Goal Solution G: Based on sensations experienced riding in an elevator, we expect that the man should feel slightly heavier when the elevator first starts to ascend, lighter when it comes to a stop, and his normal weight when the elevator is not accelerating. His apparent weight is registered by the spring scale beneath his feet, so the scale force should correspond to the force he feels through his legs (Newton’s third law). O: We should draw free body diagrams for each part of the elevator trip and apply Newton’s second law to find the scale force. The acceleration can be found from the change in speed divided by the elapsed time. A: Consider the free-body diagram of the man shown below. The force F is the upward force exerted on the man by the scale, and his weight is Fg = mg = (72.0 kg)(9.80 m/s2) = 706 N With + y defined to be up, Newton’s second law gives ∑Fy = +Fs – Fg = ma So the upward scale force is Fs = 706 N + (72.0 kg) [Equation 1] Where a is the acceleration the man experiences as the elevator changes speed. ( a ) Before the elevator starts moving, the acceleration of the elevator is zero (a = 0) so Equation 1 gives the force exerted by the scale on the man as 706 N (upward). Thus, the man exerts a downward force of 706 N on the scale. (b) During the first 0.800 s of motion, the man’s acceleration is a= ∆v (+1.20 m/s – 0) = = 1.50 m/s2 ∆t 0.800 s Substituting a into Equation 1 then gives: Fs = 706 N + (72.0 kg)(+1.50 m/s2) = 814 N (c) While the elevator is traveling upward at constant speed, the acceleration is zero and Equation 1 again gives a scale force Fs = 706 N © 2000 by Harcourt College Publishers. All rights reserved. 17 Chapter 5 Solutions 18 (d) During the last 1.50 s, the elevator starts with an upward velocity of 1.20 m/s, and comes to rest with an acceleration a= ∆v 0 – (+1.20 m/s) = = –0.800 m/s2 ∆t 1.50 s Fs = 706 N + (72.0 kg)(–0.800 m/s2) = 648 N L: The calculated scale forces are consistent with our predictions. This problem could be extended to a couple of extreme cases. If the acceleration of the elevator were +9.8 m/s2, then the man would feel twice as heavy, and if a = –9.8 m/s2 (free fall), then he would feel “weightless”, even though his true weight (Fg= mg) would remain the same. *5.40 From Newton's third law, the forward force of the ground on the sprinter equals the magnitude of the friction force the sprinter exerts on the ground. If the sprinter's shoe is not to slip on the ground, this is a static friction force and its maximum magnitude is fs, max = µsmg. n From Newton's second law applied to the sprinter, fs, max = µsmg = mamax where amax is the maximum forward acceleration the sprinter can achieve. From this, the acceleration is seen to be amax = µsg. Note that the mass has canceled out. fs If µs = 0.800, amax = 0.800(9.80 m/s2) = 7.84 m/s2 independent of the mass 5.41 Fg = mg For equilibrium: f = F and n = Fg n Also, f = µn i.e., µ = f F = n Fg F f µs = 75.0 N (25.0)(9.80) N and µk = 60.0 N (25.0)(9.80) N µs = 0.306 µk = 0.245 © 2000 by Harcourt College Publishers. All rights reserved. Fg Chapter 5 Solutions *5.42 F = µn = ma and in this case the normal force n = mg; therefore, F = µmg = ma or µ = a g The acceleration is found from (v f – v i) (80.0 mi/h)(0.447 (m/s)/(mi/h)) = = 4.47 m/s2 t 8.00 s a= Substituting this value into the expression for µ we find µ= 5.43 4.47 m/s2 = 0.456 9.80 m/s2 vi = 50.0 mi/h = 22.4 m/s 2 (a) vi (22.4 m/s)2 x= = = 256 m 2µg 2(0.100)(9.80 m/s2) (b) x= 2 5.44 vi (22.4 m/s)2 = = 42.7 m 2µg 2(0.600)(9.80 m/s2) msuitcase = 20.0 kg, F = 35.0 N (a) F cos θ = 20.0 N cos θ = (b) n 20.0 = 0.571, 35.0 F θ = 55.2° n = Fg – F sin θ = [196 – 35.0(0.821)] N Fg n = 167 N 5.45 m = 3.00 kg, θ = 30.0°, x = 2.00 m, t = 1.50 s (a) 1 x = 2 at 2 1 2.00 m = 2 a(1.50 s)2 → a = 4.00 = 1.78 m/s2 (1.50) 2 ΣF = n + f + mg = ma Along x: 0 – f + mg sin 30.0° = ma → f = m(g sin 30.0° – a) Along y: n + 0 – mg sin 30.0° = 0 → n = mg cos 30.0° (b) θt f µk = f m(g sin 30.0° – a) a = = tan 30.0° – = 0.368 mg cos 30.0° g(cos 30.0°) n © 2000 by Harcourt College Publishers. All rights reserved. 19 20 Chapter 5 Solutions (c) f = m(g sin 30.0° – a) = (3.00)(9.80 sin 30.0° – 1.78) = 9.37 N (d) v f = v i + 2a(xf – xi) where xf – xi = 2.00 m 2 2 2 v f = 0 + 2(1.78)(2.00) = 7.11 m2/s2 vf = *5.46 7.11 m2/s2 = 2.67 m/s –f + mg sin θ = 0 and +n – mg cos θ = 0 with f = µn yield µs = tan θc = tan(36.0°) = 0.727 µk = tan θc = tan(30.0°) = 0.577 5.47 F Fg = 60.0 N θt θ = 15.0° φ n φ = 35.0° fk θt F = 25.0 N mg = Fg (a) The sled is in equilibrium on the plane. Resolving along the plane: F cos(φ – θ) = mg sin θ + fk. Resolving ⊥ plane: n + F sin(φ – θ) = mg cos θ. Also, fk = µkn F cos(φ – θ) – mg sin θ = µk [mg cos θ – F sin(φ – θ)] 25.0 cos 20.0° – 60.0 sin 15.0° = µk (60.0 cos 15.0° – 25.0 sin 20.0°) µk = 0.161 (b) Resolving ⊥ to the plane: n = mg cos θ. Along the plane we have ΣF = ma. mg sin θ – fk = ma Also, fk = µkn = µkmg cos θ. © 2000 by Harcourt College Publishers. All rights reserved. Chapter 5 Solutions So along the plane we have mg sin θ – µk mg cos θ = ma a = g(sin θ – µk cos θ) = (9.80 m/s2)(sin 15.0° – 0.161 cos 15.0°) = 1.01 m/s2 *5.48 mg sin 5.00° – f = max and f = µmg cos 5.00° ∴ g sin 5.00° – µg cos 5.00° = ax ax = g(sin 5.00° – µ cos 5.00°) = – 0.903 m/s2 From Equation 2.12, 2 2 v f – v i = 2ax –(20.0)2 = –2(0.903)x x = 221 m n f x mg θ 5.49 T – ff = 5.00a (for 5.00 kg mass) 9.00g – T = 9.00a 5.00 kg (for 9.00 kg mass) Adding these two equations gives: 9.00(9.80) – 0.200(5.00)(9.80) = 14.0a a = 5.60 m/s2 ∴ T = 5.00(5.60) + 0.200(5.00)(9.80) = 37.8 N © 2000 by Harcourt College Publishers. All rights reserved. 9.00 kg 21 22 5.50 Chapter 5 Solutions Let a represent the positive magnitude of the acceleration –aj of m l, of the acceleration –ai of m 2, and of the acceleration +aj of m 3. Call T12 the tension in the left rope and T23 the tension in the cord on the right. n T12 For m1, ΣFy = ma y +T12 – m1g = – m1a For m2, ΣFx = ma x –T12 + µkn + T23 = –m2a and ΣFy = ma y n – m2g = 0 for m3, ΣFy = ma y T23 – m3g = +m3a T23 f = µk n m2g we have three simultaneous equations T12 T23 m1g m3g –T12 + 39.2 N = (4.00 kg)a +T12 – 0.350(9.80 N) – T23 = (1.00 kg)a +T23 – 19.6 N = (2.00 kg)a (a) Add them up: +39.2 N – 3.43 N – 19.6 N = (7.00 kg)a a = 2.31 m/s2, down for m 1, left for m 2, and up for m 3 (b) Now –T12 + 39.2 N = 4.00 kg(2.31 m/s2) T 12 = 30.0 N and T23 – 19.6 N = 2.00 kg(2.31 m/s2) T 23 = 24.2 N 5.51 (a) n1 n2 T m1 T F m2 f1 = µk n1 f2 = µk n 2 m1g 118 N m2g 176 N © 2000 by Harcourt College Publishers. All rights reserved. 68.0 N Chapter 5 Solutions (b) 68.0 – T – µm2g= m2a (Block #2) T – µm1g = m1a (Block #1) Adding, 68.0 – µ(m1 + m2)g = (m1 + m2)a a= 68.0 – µg = 1.29 m/s2 (m 1 + m 2 ) T = m1a + µm1g = 27.2 N 5.52 (a) The rope makes angle Arctan 10.0 cm = 14.0° 40.0 cm ΣFy = ma y + 10.0 N sin 14.0° + n – 2.20 kg(9.80 m/s2) = 0 n = 19.1 N fk = µkn = 0.400(19.1 N) = 7.65 N ΣFx = ma x + 10.0 N cos 14.0° – 7.65 N = (2.20 kg)a a = 0.931 m/s2 (b) ΣFx = ma x 10.0 N cos θ – fk = 0 10.0 N cos θ = fk = µkn = 0.400[2.20 kg(9.80 m/s2) – 10.0 N sin θ] 10.0 N 1 – sin2 θ = 8.62 N – 4.00 N sin θ 100 – 100 sin2 θ = 74.4 – 69.0 sin θ + 16.0 sin2 θ –116 sin2 θ + 69.0 sin θ + 25.6 = 0 sin θ = –69.0 ± (69.0)2 – 4(25.6)(–116) 2(–116) sin θ = – 0.259 or 0.854 θ = –15.0° or 58.6° The negative root would refer to the pulley below the block. We choose tan 58.6° = 10.0 cm x x = 6.10 cm © 2000 by Harcourt College Publishers. All rights reserved. 23 24 Chapter 5 Solutions *5.53 (Case 1, impending upward motion) n P cos 50.0° Setting ∑Fx = 0: P cos 50.0° – n = 0 fs, max = µsn = µsP cos 50.0° fs, max = µs n = 0.250(0.643)P = 0.161P P sin 50.0° mg Setting ∑Fy = 0: fs, max = µs n P sin 50.0° – 0.161P – (3.00)(9.80) = 0 n P cos 50.0° Pmax = 48.6 N (Case 2, impending downward motion) P sin 50.0° fs, max = 0.161P as in Case 1. Setting ∑Fy = 0: P sin 50.0° + 0.161P – (3.00)(9.80) = 0 Pmin = 31.7 N 5.54 ∑F = ma gives the object's acceleration a= ∑F (8.00i – 4.00t j) N = m 2.00 kg a = (4.00 m/s2)i – (2.00 m/s3)t j = dv dt Its velocity is t v ⌠ a dt ⌠ dv = v – vi = v – 0 = ⌡ ⌡ 0 vi t v=⌠ ⌡ [(4.00 m/s2)i – (2.00 m/s3)t j] dt 0 v = (4.00t m/s2)i – (1.00t2 m/s3)j (a) We require v = 15.0 m/sv 2 = 225 m2/s 2 16.0t2 m2/s4 + 1.00t4 m2/s 6 = 225 m2/s 2 1.00t4 + 16.0 s2t2 – 225 s4 = 0 t2 = –16.0 ± (16.0)2 – 4(–225) = 9.00 s2 2.00 t = 3.00 s © 2000 by Harcourt College Publishers. All rights reserved. mg Chapter 5 Solutions © 2000 by Harcourt College Publishers. All rights reserved. 25 26 Chapter 5 Solutions Take ri = 0 at t = 0. The position is t t 0 0 ⌠ [(4.00t m/s2)i – (1.00t2 m/s3)j]dt r=⌠ ⌡ v dt = ⌡ r = (4.00 m/s2) t2 t3 i – (1.00 m/s3) j 2 3 at t = 3 s we evaluate 5.55 (c) r = (18.0i – 9.00j) m (b) So r = (18.0)2 + (9.00)2 m = 20.1 m (a) T T T T n n 320 N (b) 160 N First consider Pat and the chair as the system. Note that two ropes support the system, and T = 250 N in each rope. Applying ΣF = ma 2T – 480 = ma where Solving for a gives a = (c) 480 N m= 480 = 49.0 kg 9.80 (500 – 480) = 0.408 m/s2 49.0 ΣF (on Pat ) = n + T – 320 = ma where m= 320 = 32.7 kg 9.80 n = ma + 320 – T = 32.7(0.408) + 320 – 250 = 83.3 N © 2000 by Harcourt College Publishers. All rights reserved. Chapter 5 Solutions *5.56 (a) 27 n1 F = ma 18.0 = (2.00 + 3.00 + 4.00)a F a = 2.00 m/s2 (b) P m1 The force on each block can be found by knowing mass and acceleration: m1 g ΣF1 = m1a = 2.00(2.00) = 4.00 N n2 ΣF2 = m2a = 3.00(2.00) = 6.00 N P ΣF3 = m3a = 4.00(2.00) = 8.00 N (c) Q m2 Therefore, ΣF1 = 4.00 N = F – P P = 14.0 N m2 g n3 ΣF2 = 6.00 N = P – Q Q Q = 800 N m3 m3g *5.57 We find the diver's impact speed by anaylzing his free-fall motion: 2 2 v f = v i + 2ax = 02 + 2(–9.80 m/s2)(–10.0 m) vf = –14.0 m/s Now for the 2.00 s of stopping, we have vf = vi + at 0 = –14.0 m/s + a(2.00 s), a = +7.00 m/s2 ∑Fy = ma Call the force exerted by the water on the driver R. +R – (70.0 kg)(9.80 m/s2) = (70.0 kg)(7.00 m/s2) R = 1.18 kN © 2000 by Harcourt College Publishers. All rights reserved. 28 Chapter 5 Solutions 5.58 a a n1 = 2mg cos θ n2 = mg cos θ T1 2m m T1 2mg sin θ 2mg cos θ f1 T2 T2 a M mg sin θ mg cos θ f2 Mg Applying Newton's second law to each object gives: (1) and (3) T1 = f1 + 2m(g sin θ + a) (2) T2 – T1 = f2 + m(g sin θ + a) T2 = M(g – a) Parts (a) and (b): Equilibrium (⇒ a = 0) and frictionless incline (⇒ f1 = f2 = 0) Under these conditions, the equations reduce to (1) T1 = 2mg sin θ (2) T2 – T1 = mg sin θ and (3') T2 = Mg Substituting (1') and (3') into equation (2') then gives M = 3m sin θ , so equation (3') becomes T2 = 3mg sin θ . Parts (c) and (d): M = 6m sin θ(double the value found above), and f1 = f2 = 0. With these conditions present, the equations become T1 = 2m(g sin θ + a) T2 – T1 = m(g sin θ + a) Solved simultaneously, these yield a = 1 + sin θ 1 + 2 sin θ T1 = 4mg sin θ and and T2 = 6m sin θ(g – a) g sin θ , 1 + 2 sin θ 1 + sin θ 1 + 2 sin θ T2 = 6mg sin θ Part (e): Equilibrium (⇒ a = 0) and impending motion up the incline so M = Mmax while f1 = 2µs mg cos θ and f2 = µs mg cos θ, both directed down the incline. Under these conditions, the equations become T1 = 2mg (sin θ + µs cos θ), T2 – T1 = mg (sin θ + µs cos θ), and T2 = Mmax g which yield M max = 3m(sin θ + µs cos θ) . Part (f): Equilibrium (⇒ a = 0) and impending motion down the incline so M = Mmin while f1 = 2µs mg cos θ and f2 = µs mg cos θ, both directed up the incline. Under these conditions, the equations are T1 = 2mg (sin θ – µs cos θ), T2 – T1 = mg (sin θ – µs cos θ), and T2 = Mmin g which yield Mmin = 3m(sin θ – µs cos θ) . © 2000 by Harcourt College Publishers. All rights reserved. Chapter 5 Solutions Part (g): T2, max – T2, min = Mmaxg – Mming = 6mgµs cos θ © 2000 by Harcourt College Publishers. All rights reserved. 29 30 5.59 Chapter 5 Solutions (a) First, we note that F = T1. Next, we focus on the mass M and write T5 = Mg. Next, we focus on the bottom pulley and write T5 = T2 + T3. Finally, we focus on the top pulley and write T4 = T1 + T2 + T3. Since the pulleys are massless and frictionless, T1 = T3, and T2 = T3. From this information, we have T5 = 2 T2 , Mg so T2 = . 2 Then T 1 = T 2 = T 3 = Mg 3 Mg , and T4 = 2 2 T4 T1 T2 T3 T5 , and T5 = Mg M 5.60 Mg 2 (b) Since F = T1, we have F = (a) ΣF = F1 + F2 = (–9.00i + 3.00j) N Acceleration a = axi + ayj = F ΣF (–9.00i + 3.00j) N = = (–4.50i + 1.50j) m/s2 m 2.00 kg Velocity v = vxi + vyj = vi + at = at v = (– 4.50i + 1.50j)(m/s2)(10 s) = (– 45.0i + 15.0j) m/s (b) The direction of motion makes angle θ with the x-direction. θ = tan–1 v y 15.0 m/s = tan–1 – v x 45.0 m/s θ = –18.4° + 180° = 162° from +x-axis (c) Displacement: x-displacement = x – xi = vxit + 1 1 a t2 = (– 4.50 m/s2)(10.0 s) 2 = –225 m 2 x 2 y-displacement = y – yi = vyit + 1 1 ayt2 = (+1.50 m/s2)(10.0 s) 2 = +75.0 m 2 2 ∆r = (–225i + 75.0j) m (d) Position: ≡ r = r1 + ∆r r = (–2.00i + 4.00j) + (–225i + 75.0j) = (–227i + 79.0j) m © 2000 by Harcourt College Publishers. All rights reserved. Chapter 5 Solutions 5.61 (a) The crate is in equilibrium. Let the normal force acting on it be n and the friction force, fs. Resolving vertically: n = Fg + P sin θ Horizontally: P cos θ = fs But fs ≤ µsn i.e., P cos θ ≤ µs (Fg + P sin θ) or P(cos θ – µs sin θ) ≤ µsFg Divide by cos θ: P(1 – µs tan θ) ≤ µsFg sec θ Then Pminimum = (b) P= 0.400(100 N)sec θ 1 – 0.400 tan θ θ (deg) P (N) 5.62 (a) µsFg sec θ 1 – µ s tan θ 0.00 40.0 15.0 46.4 Following Example 5.6 30.0 60.1 45.0 94.3 60.0 260 a = g sin θ = (9.80 m/s2) sin 30.0° a = 4.90 m/s2 (b) The block slides distance x on the incline, with sin 30.0° = 0.500 m/x x = 1.00 m 2 2 v f = v i + 2a(xf – xi) = 0 + 2(4.90 m/s2)(1.00 m) vf = 3.13 m/s after time ts = Now in free fall yf – yi = vyit + 2xf 2(1.00 m) = = 0.639 s vf 3.13 m/s 1 a t2 2 y 1 –2.00 m = (–3.13 m/s) sin 30.0°t – 2 (9.80 m/s2) t2 (4.90 m/s2)t2 + (1.56 m/s)t – 2.00 m = 0 t= –1.56 m/s ± (1.56 m/s)2 – 4(4.90 m/s2)(–2.00 m) 9.80 m/s2 t = 0.499 s, the other root being unphysical. (c) x = vxt = [(3.13 m/s) cos 30.0°] (0.499 s) = 1.35 m (d) total time = ts + t = 0.639 s + 0.499 s = 1.14 s © 2000 by Harcourt College Publishers. All rights reserved. 31 32 Chapter 5 Solutions (e) The mass of the block makes no difference. © 2000 by Harcourt College Publishers. All rights reserved. Chapter 5 Solutions *5.63 33 With motion impending, n + T sin θ – mg = 0 T sin θ n f = µs(mg – T sin θ) and T cos θ – µs mg + µs T sin θ = 0 f µ s mg T= cos θ + µs sin θ so To minimize T, we maximize cos θ + µs sin θ T cos θ m Fg d (cos θ + µs sin θ) = 0 = –sin θ + µs cos θ dθ (a) θ = Arctan µs = Arctan 0.350 = 19.3° (b) T= (0.350)(1.30 kg)(9.80 m/s2) = 4.21 N cos 19.3° + 0.350 sin 19.3° 5.64 n1 = m1g f1 m1 n2 = m2g cos θ f2 m2 m1g mm22ggcos costθ m2g sin θ θt For the system to start to move when released, the force tending to move m2 down the incline, m2g sin θ, must exceed the maximum friction force which can retard the motion: fmax = f1, max + f2, max = µs, 1n1 + µs, 2n2 = µs, 1m1g + µs, 2m2g cos θ From Table 5.2, µs, 1 = 0.610 (aluminum on steel) and µs, 2 = 0.530 (copper on steel). With m1 = 2.00 kg, m2 = 6.00 kg, θ = 30.0°, the maximum friction force is found to be fmax = 38.9 N. This exceeds the force tending to cause the system to move, m2g sin θ = (6.00)(9.80) sin 30.0° = 29.4 N. Hence, the system will not start to move when released . The friction forces increase in magnitudes until the total friction force retarding the motion, f = f1 + f2, equals the force tending to set the system in motion. That is until f = m 2g sin θ = 29.4 N . © 2000 by Harcourt College Publishers. All rights reserved. 34 5.65 Chapter 5 Solutions (a) First, draw a free-body diagram, (Fig. 1) of the top block. n1 = 19.6 N Top Block 2.00 kg F = 10.0 N f = µkn1 mg = 19.6 N Since ay = 0, n1 = 19.6 N and fk = µkn1 = (0.300)(19.6) = 5.88 N ∑Fx = maT gives 10.0 N – 5.88 N = (2.00 kg)aT, or aT = 2.06 m/s2 (for top block) Now draw a free-body diagram (Fig. 2) of the bottom block and observe that n1 n2 f Bottom Block 8.00 kg Mg ∑Fx = MaB gives f = 5.88 N = (8.00 kg)aB, or aB = 0.753 m/s2 (for the bottom block) In time t, the distance each block moves (starting from rest) is dT = 1 a t2 2 T and dB = 1 a t2 2 B For the top block to reach the right edge of the bottom block, it is necessary that dT = dB + L or (See Figure 3) dT L dB 1 1 (2.06 m/s2) t2 = (0.735 m/s2) t2 + 3.00 m 2 2 which gives: t = 2.13 s © 2000 by Harcourt College Publishers. All rights reserved. Initial position of left edges of both blocks Chapter 5 Solutions (b) 35 From above, 1 dB = (0.735 m/s2)(2.13 s) 2 = 1.67 m 2 5.66 t2(s2) t(s) 0 1.02 1.53 2.01 2.64 3.30 3.75 0 1.040 2.341 4.040 6.970 10.89 14.06 From x = Acceleration determination for a cart on an incline x(m) 0 0.100 0.200 0.350 0.500 0.750 1.00 1.5 y = 0.0714x 2 R = 0.9919 Distance, m 1 0.5 1 2 at 2 0 0 the slope of a graph of x versus t2 is 5 10 15 Squared time, seconds squared 1 a, 2 and a = 2 × slope = 2(0.0714 m/s2) = 0.143 m/s2 From a' = g sin θ, we obtain a' = (9.80 m/s2) 1.774 = 0.137 m/s2, different by 4% 127.1 The difference is accounted for by the uncertainty in the data, which we may estimate from the third point as 0.350 – (0.0714)(4.04) = 18% 0.350 5.67 (a) n1 = 19.6 N n1 = 19.6 N n2 fs 2.00 kg 5.00 kg fs m1g = 19.6 N F fk m2g = 49 N The force of static friction between the blocks accelerates the 2.00 kg block. © 2000 by Harcourt College Publishers. All rights reserved. 36 Chapter 5 Solutions (b) ΣF = ma, for both blocks together F – µn2 = ma, F – (0.200)[(5.00 + 2.00)(9.80)] = (5.00 + 2.00)3.00 Therefore F = 34.7 N (c) f = µ1(2.00)(9.80) = m 1a = 2.00(3.00) Therefore µ = 0.306 5.68 (a) 5.00 kg f1 10.0 kg f1 5.00 kg F = 45.0 N 10.0 kg f2 T 49.0 N 98.0 N f1 and n1 appear in both diagrams as action-reaction pairs (b) 5.00 kg: ΣFx = ma n1 = m1g = (5.00)(9.80) = 49.0 N f1 –T = 0 T = f1 = µmg = (0.200)(5.00)(9.80) = 9.80 N 10.0 kg: ΣFx = ma n2 n1 n1 ΣFy = 0 45.0 – f1 – f2 = 10.0a n2 – n1 – 98.0 = 0 f2 = µn2 = µ(n1 + 98.0) = (0.20)(49.0 + 98.0) = 29.4 N 45 – 9.80 – 29.4 = 10.0a a = 0.580 m/s2 © 2000 by Harcourt College Publishers. All rights reserved. 45.0 N Chapter 5 Solutions 5.69 37 ΣF = ma For m1: T = m1a a = m2g m1 For m2: T – m2g = 0 For all 3 blocks: F = (M + m1 + m2)a = (M + m 1 + m 2) m 2g m1 n1 m1 F M T m1 m2 m1g T n2 n F m2 M + m1 + m2 m2 g (Fg)total Goal Solution Draw separate free-body diagrams for blocks m1 and m2. Remembering that normal forces are always perpendicular to the contacting surface, and always push on a body, draw n1 and n2 as shown. Note that m2 should be in contact with the cart, and therefore does have a normal force from the cart. Remembering that ropes always pull on bodies in the direction of the rope, draw the tension force T. Finally, draw the gravitational force on each block, which always points downwards. m1 F M m2 © 2000 by Harcourt College Publishers. All rights reserved. Chapter 5 Solutions 38 G: What can keep m2 from falling? Only tension in the cord connecting it with m1 . This tension pulls forward on m1 to accelerate that mass. This acceleration should be proportional to m2 and to g and inversely proportional to m1, so perhaps a = (m2/m1)g We should also expect the applied force to be proportional to the total mass of the system. O: Use ΣF = ma and the free-body diagrams above. A: For m2, For m1, T – m2g = 0 or T = m2g T = m 1a a= or Substituting for T, we have a = T m1 m 2g m1 For all 3 blocks, F = (M + m1 + m2) a. Therefore, F = (M + m1 + m2) m 2g m1 L: Even though this problem did not have a numerical solution, we were still able to rationalize the algebraic form of the solution. This technique does not always work, especially for complex situations, but often we can think through a problem to see if an equation for the solution makes sense based on the physical principles we know. 5.70 (1) m1(a – A) = T ⇒ a = T/m1 + A (2) MA = Rx = T ⇒ A = T/M (3) m2a = m2g – T ⇒ T = m2(g – a) (a) a−A T A Substitute the value for a from (1) into (3) and solve for T ; T = m2[g – (T/m1 + A)] Substitute for A from (2); T = m2 g – (b) m 1M T T + = m 2g m M m M + m 1 1 2(m 1 + M) Solve (3) for a and substitute value of T a = T m1 m 2g(M + m 1) m 1M + m 2(M + m 1) © 2000 by Harcourt College Publishers. All rights reserved. M m2 a Chapter 5 Solutions (c) From (2), A = T/M ; Substitute the value of T A= (d) *5.71 (a) 39 a–A = m 1m 2g m 1M + m 2(m 1 + M) Mm2g m 1M + m 2(m 1 + M) Motion impending n = 49.0 N n = 49.0 N fs1 P 5.00 kg 15.0 kg fs1 fs2 Fg = 49.0 N fs1 = µn = 14.7 N (b) P = fs1 + fs2 = 14.7 N + 98.0 N = 113 N (c) Once motion starts, kinetic friction acts. 196 N 147 N fs2 = 0.500(196 N) = 98.0 N 112.7 N – 0.100(49.0 N) – 0.400(196 N) = 15.0 kg a2 a2 = 1.96 m/s2 0.100(49.0 N) = 5.00 kg a1 a1 = 0.980 m/s2 5.72 Since it has a larger mass, we expect the 8.00 kg block to move down the plane. The acceleration for both blocks should have the same magnitude since they are joined together by a non-stretching string. ΣF1 = m1a1: –m1g sin 35.0° + T = m1a ΣF2 = m2a2: –m2g sin 35.0° + T = –m2a 8.00 kg 3.50 kg and –(3.50)(9.80) sin 35.0° + T = 3.50a 35° –(8.00)(9.80) sin 35.0° + T = – 8.00a T = 27.4 N a = 2.20 m/s2 © 2000 by Harcourt College Publishers. All rights reserved. 35° 40 5.73 Chapter 5 Solutions ∑F 1 = m 1a: –m1g sin 35.0° – fk, 1 + T = m1a (1) –(3.50)(9.80) sin 35.0° – µk(3.50)(9.80) cos 35.0° + T = (3.50)(1.50) ∑F 2 = m 2a: +m2g sin 35.0° – fk, 2 – T = m2a (2) +(8.00)(9.80) sin 35.0° – µk(8.00)(9.80) cos 35.0° – T = (8.00)(1.50) Solving equations (1) and (2) simultaneously gives 5.74 (a) µk = 0.0871 (b) T = 27.4 N The forces acting on the sled are (a) T – Ff = ma T – 500 N = (100 kg)(1.00 m/s2) T = 600 N (b) Frictional force pushes the horse forward. f – T = mhorsea f – 600 N = (500 kg)(1.00 m/s2) f = 1100 N (c) f – Ff = 600 N Σm = 100 kg + 500 kg a= ΣF 600 N = = 1.00 m/s2 Σm 600 kg © 2000 by Harcourt College Publishers. All rights reserved. Chapter 5 Solutions 5.75 41 mg sin θ = m(5.00 m/s2) θ = 30.7° T = mg cos θ = (0.100)(9.80) cos 30.7° T = 0.843 N y T θt a = 5.00 m/s 2 x mg 5.76 (a) Apply Newton's 2nd law to two points where butterflies are attached on either half of mobile (other half the same, by symmetry) (1) T2 cos θ2 – T1 cos θ1 = 0 D l θt11 θ2 l (2) T1 sin θ1 – T2 sin θ2 – mg = 0 (3) T2 cos θ2 – T3 = 0 (4) T2 sin θ2 – mg = 0 2mg sin θ1 Substitute (3) into (1) for T2 cos θ2 T3 – T1 cos θ1 = 0, T3 = T1 cos θ1 Substitute value of T1 ; From Eq. (4), T2 = T3 = 2mg l l l L = 5l Substituting (3) into (1) for T2 sin θ2 T1 sin θ1 – mg – mg = 0 Then T 1 = θt1 θ2 cos θ1 2mg = sin θ1 tan θ 1 mg sin θ 2 © 2000 by Harcourt College Publishers. All rights reserved. 42 Chapter 5 Solutions (b) We must find θ2 and substitute for θ2: T2 = mg sin tan –1 1 2 tan θ 1 divide (4) by (3); T2 sin θ2 mg mg = ⇒ tan θ2 = T2 cos θ2 T3 T3 Substitute value of T3 ⇒ tan θ2 = mg tan θ1 2mg tan θ 1 2 θ 2 = tan –1 (c) D is the total horizontal displacement of each string D = 2 l cos θ1 + 2 l cos θ2 + l and L = 5 l L 1 D = 2 cos θ1 + 2 cos tan –1 2 tan θ 1 + 1 5 5.77 If all the weight is on the rear wheels, (a) F = ma mgµs = ma But ∆x = µs = (b) at2 gµst2 2∆x = , so µs = 2 2 2 gt 2(0.250 mi)(1609 m/mi) = 3.34 (9.80 m/s2)(4.96 s)2 Time would increase, as the wheels would skid and only kinetic friction would act; or perhaps the car would flip over. © 2000 by Harcourt College Publishers. All rights reserved. Chapter 5 Solutions 5.78 ∑F y = ma y: n – mg cos θ = 0, or 43 y n = (8.40)(9.80) cos θ n n = (82.3 N) cos θ ∑F x = ma x: mg sin θ = ma, or a = g sin θ m mg sin θ a = (9.80 m/s2) sin θ θ (deg) n (N) a (m/s2) 0.00 82.3 0.00 5.00 82.0 0.854 10.0 81.1 1.70 15.0 79.5 2.54 20.0 77.4 3.35 25.0 74.6 4.14 30.0 71.3 4.90 35.0 67.4 5.62 40.0 63.1 6.30 45.0 58.2 6.93 50.0 52.9 7.51 55.0 47.2 8.03 60.0 41.2 8.49 65.0 34.8 8.88 70.0 28.2 9.21 75.0 21.3 9.47 80.0 14.3 9.65 85.0 7.17 9.76 90.0 0.00 9.80 At 0˚, the normal force is the full weight and the acceleration is zero. At 90˚, the mass is in free fall next to the vertical incline. x mg cos θ n (N) 80 60 40 20 θ (deg) 0 0 20 40 60 80 100 40 60 80 100 2 a (m/s ) 10 8 6 4 2 θ (deg) 0 0 20 © 2000 by Harcourt College Publishers. All rights reserved. 44 Chapter 5 Solutions © 2000 by Harcourt College Publishers. All rights reserved. Chapter 6 Even Answers 2. 4. 6. 8. 10. 12. 14. 16. (a) 1100 N (b) 2.04 times her weight (a) 8.32 × 10–8 N (b) 9.13 × 1022 m/s2 (a) 5.59 × 103 m/s (b) 239 min (c) 735 N (a) 0.105 m/s (b) 1.10 × 10–2 m/s2 0.996g (a) (–0.163i + 0.233j) m/s2 (b) 6.53 m/s (c) (–0.181i + 0.181j) m/s2 (a) 149 N (b) 10.4 m/s (a) 1.33 m/s2 (b) 1.79 m/s2 at 47.9˚ inward 18. (a) v = 20. 22. 24. 58. 60. 8.88 N (a) 8.62 m (b) Mg , downward (c) 8.45 m/s2 (a) 3.60 m/s2 (b) zero (c) Observer in car (non-inertial) claims an 18.0 N force toward the left and an 18.0 N force toward the right. An inertial observer (outside car) claims only an 18.0 N force toward the right. (a) 1.41 h (b) 17.1 93.8 N (a) 6.27 m/s2 downward (b) 784 N upward (c) 283 N upward (a) 53.8 m/s (b) 148 m 1.40 –0.212 m/s2 (a) 0.980 m/s (a) 7.70 × 10–4 kg/m (b) 0.998 N (c) ≈ 49 m, ≈ 6.3 s, ≈ 27 m/s (b) 81.8 m (c) 15.9˚ (a) 6.67 × 103 N (b) 20.3 m/s 6.56 × 1015 rev/s g(cos φ tan θ – sin φ) (a) m2g (b) m2g (c) (m2/m1)gR 780 N (a) 967 lb (b) -647 lb (pilot must be strapped in) (c) vary speed and radius of path so that v2 = gR 62.2 rev/min 2.14 rev/min 62. 64. (a) πRg (b) mπg (a) 8.04 s (b) 379 m/s (c) 1.19 × 10–2 m/s (d) 9.55 cm 66. (a) vmin = 26. 28. 30. 32. 34. 36. 40. 42. 44. 46. 48. 50. 52. 54. 56. 68. 70. R 2T – g (b) n = 2T m Rg(tan θ – µ s) Rg(tan θ + µ s) , vmax = (b) µs = tan θ 1 + µ s tan θ 1 – µ s tan θ (c) 8.57 m/s ≤ v ≤ 16.6 m/s (a) 0.0132 m/s (b) 1.03 m/s (c) 6.87 m/s (a) 78.3 m/s (b) 11.1 s (c) 121 m © 2000 by Harcourt College Publishers. All rights reserved. 2 Chapter 6 Even Answers © 2000 by Harcourt College Publishers. All rights reserved. Chapter 6 Solutions 6.1 6.2 (a) 200 m Average speed = v– = = 8.00 m/s 25.0 s (b) F= mv2 200 m where r = = 31.8 m r 2π F= (1.50 kg)(8.00 m/s)2 = 3.02 N 31.8 m (a) ΣFx = ma x T= (b) mv2 55.0 kg (4.00 m/s)2 = = 1100 N r 0.800 m The tension is larger than her weight by 1100 N = 2.04 times (55.0 kg)(9.80 m/s2) 6.3 m = 3.00 kg: r = 0.800 m. The string will break if the tension exceeds the weight corresponding to 25.0 kg, so Tmax = Mg = 25.0 × 9.80 = 245 N When the 3.00 kg mass rotates in a horizontal circle, the tension provides the centripetal force, so mv2 (3.00)v2 T= = r 0.800 Then v2 = (0.800Tmax) rT 0.800T 0.800 × 245 = ≤ = = 65.3 m2/s2 m 3.00 3.00 3.00 and 0 < v < 65.3 or 0 < v < 8.08 m/s © 2000 by Harcourt College Publishers. All rights reserved. Goal Solution The string will break if the tension T exceeds the test weight it can support, Tmax = mg = (25.0 kg)(9.80 m/s2) = 245 N As the 3.00-kg mass rotates in a horizontal circle, the tension provides the central force. © 2000 by Harcourt College Publishers. All rights reserved. Chapter 6 Solutions From ∑F = ma, T = mv2 r rTmax = m Then, v ≤ (0.800 m)(245 N) = 8.08 m/s (3.00 kg) So the mass can have speeds between 0 and 8.08 m/s. ◊ 6.4 6.5 (a) F= mv2 (9.11 × 10–31 kg)(2.20 × 106 m/s)2 = = 8.32 × 10 –8 N inward r 0.530 × 10–10 m (b) a= v 2 (2.20 × 106 m/s)2 = = 9.13 × 1022 m/s2 inward r 0.530 × 10–10 m Neglecting relativistic effects. F = ma c = F = (2 × 1.661 × 10–27 kg) 6.6 (a) We require that mv 2 r (2.998 × 107 m/s)2 = 6.22 × 10 –12 N (0.480 m) GmMe mv2 MeG = but g = 2 r2 r R e In this case r = 2Re, therefore, v= 6.7 g v2 = or v = 4 2Re gRe 2 (9.80 m/s2)(6.37 × 106 m) = 5.59 × 103 m/s 2 (b) T= 2πr (2π)(2)(6.37 × 106 m) = = 239 min v 5.59 × 103 m/s (c) F= GmMe mg (300 kg)(9.80 m/s2) = = = 735 N (2R e) 2 4 4 The orbit radius is r = 1.70 × 106 m + 100 km = 1.80 × 106 m. Now using the information in Example 6.6, GMmms ms22π2r2 = = m sa r2 rT2 (a) a= GMm (6.67 × 10–11)(7.40 × 1022) = = 1.52 m/s2 r2 (1.80 × 106 m) 2 (b) a= v2 ,v= r (c) v= 2πr 2π(1.80 × 106) ,T= = 6820 s T 1.66 × 103 (1.52 m/s2)(1.80 × 106 m) = 1.66 km/s © 2000 by Harcourt College Publishers. All rights reserved. 3 4 6.8 Chapter 6 Solutions (a) Speed = distance/time. If the radius of the hand of the clock is r then v= 2πr ⇒ vT = 2πr T rm = rs ∴ Tmvm = Tsvs where vm = 1.75 × 10–3 m/s, Tm = (60.0 × 60.0)s and Ts = 60.0 s. Tm 3.60 × 103 s vm = (1.75 × 10–3 m/s) = 0.105 m/s Ts 60.0 s vs = (b) v= 2πr vT (0.105 m/s)(60.0 s) for the second hand, r = = = 1.00 m T 2π 2π Then ar = 6.9 (a) (b) v2 (0.105 m/s)2 = = 1.10 × 10–2 m/s2 r 1.00 m static friction ma i = f i + n j + mg(–j) ΣFy = 0 = n – mg thus n = mg and ΣFr = m Then µ = *6.10 6.11 v2 = f = µn = µmg r v2 (50.0 cm/s)2 = = 0.0850 rg (30.0 cm)(980 cm/s2) 86.5 km 1 h 1000 m 2 h 3600 s 1 km v 2 1 g = 0.966g a= = r 61.0 m 9.80 m/s2 n = mg since ay = 0 n ar The centripetal force is the frictional force f. From Newton's second law mv 2 f = ma r = r f mg © 2000 by Harcourt College Publishers. All rights reserved. Chapter 6 Solutions But the friction condition is f ≤ µs n i.e., mv 2 ≤ µs mg r v≤ µs rg = (0.600)(35.0 m)(9.80 m/s2) v ≤ 14.3 m/s 6.12 (b) v= 235 m = 6.53 m/s 36.0 s The radius is given by 1 2π r = 235 m 4 r = 150 m (a) ar = = v 2 toward center r (6.53 m/s)2 at 35.0° north of west 150 m = (0.285 m/s2)(cos 35.0°(– i) + sin 35.0° j) = –0.233 m/s2 i + 0.163 m/s2 j (c) –a = (vf – vi) t = (6.53 m/s j – 6.53 m/s i) 36.0 s = – 0.181 m/s2 i + 0.181 m/s2 j © 2000 by Harcourt College Publishers. All rights reserved. 5 6 6.13 Chapter 6 Solutions T cos 5.00° = mg = (80.0 kg)(9.80 m/s2) (a) T = 787 N T = (68.6 N)i + (784 N)j (b) T T sin 5.00° = mar 5.00° ar = 0.857 m/s2 y x mg 6.14 (a) The reaction force n1 represents n1 the apparent weight of the woman F = ma i.e., mg – n1 = mv2 r , so n1 = mg – mv2 v a r n1 = 600 – 600 (9.00)2 = 149 N 9.80 11.0 (b) If n1 = 0, mg = This gives v = mv2 r rg = (11.0 m)(9.80 m/s2) = 10.4 m/s © 2000 by Harcourt College Publishers. All rights reserved. mg Chapter 6 Solutions 6.15 Let the tension at the lowest point be T. F = ma mv2 r T – mg = mar = T = m g + v 2 (8.00 m/s)2 = (85.0 kg) 9.80 m/s2 + = 1.38 kN > 1000 N r 10.0 m He doesn't make it across the river because the vine breaks. T ar mg 6.16 (a) (b) ar = v 2 (4.00 m/s)2 = = 1.33 m/s2 r 12.0 m 2 2 v r = 12.0 m a= a r + aT a= (1.33)2 + (1.20)2 = 1.79 m/s2 at an angle θ = tan-1 ar = 47.9° inward a T 6.17 M = 40.0 kg, R = 3.00 m, T = 350 N (a) ΣF = 2T – Mg = T Mv2 R T v2 = (2T – Mg) R M v2 = [700 – (40.0)(9.80)] 3.00 = 23.1(m2/s2) 40.0 Mg child + seat v = 4.81 m/s © 2000 by Harcourt College Publishers. All rights reserved. 7 Chapter 6 Solutions 8 (b) n – Mg = F = n = Mg + Mv2 R Mg child alone Mv2 23.1 = 40.0 9.80 + = 700 N R 3.00 n Goal Solution G: If the tension in each chain is 350 N at the lowest point, then the force of the seat on the child should just be twice this force or 700 N. The child’s speed is not as easy to determine, but somewhere between 0 and 10 m/s would be reasonable for the situation described. O: We should first draw a free body diagram that shows the forces acting on the seat and apply Newton’s laws to solve the problem. A: We can see from the diagram that the only forces acting on the system of child+seat are the tension in the two chains and the weight of the boy: ∑F = 2T – mg = ma where a = v2 is the centripetal acceleration r F = Fnet = 2(350 N) – (40.0 kg)(9.80 m/s2) = 308 N upwards v= F maxr = m (308 N)(3.00 m) = 4.81 m/s ◊ 40.0 kg The child feels a normal force exerted by the seat equal to the total tension in the chains. n = 2(350 N) = 700 N (upwards) ◊ L: Our answers agree with our predictions. It may seem strange that there is a net upward force on the boy yet he does not move upwards. We must remember that a net force causes an acceleration, but not necessarily a motion in the direction of the force. In this case, the acceleration is due to a change in the direction of the motion. It is also interesting to note that the boy feels about twice as heavy as normal, so he is experiencing an acceleration of about 2g’s. 6.18 (a) Consider the forces acting on the system consisting of the child and the seat: ∑Fy = may ⇒ 2T – mg = m v2 R v2 = R 2T – g m v= R 2T – g m © 2000 by Harcourt College Publishers. All rights reserved. Chapter 6 Solutions (b) Consider the forces acting on the child alone: ∑Fy = may ⇒ n = m g + v 2 R and from above, v2 = R 2T – g , so m n = m g + 6.19 ΣFy = 2T – g = 2T m mv2 = mg + n r But n = 0 at this minimum speed condition, so mv 2 = mg ⇒ v = r 6.20 (9.80 m/s2)(1.00 m) = 3.13 m/s At the top of the vertical circle, T=m 6.21 gr = v2 – mg R (4.00)2 – (0.400)(9.80) = 8.88 N 0.500 or T = (0.400) (a) v = 20.0 m/s, n = force of track on roller coaster, and R = 10.0 m. ΣF = Mv2 = n – Mg R From this we find n = Mg + Mv2 (500 kg)(20.0 m/s2) = (500 kg)(9.80 m/s2) + R 10.0 m n = 4900 N + 20,000 N = 2.49 × 104 N B C r 15 m 10 m A © 2000 by Harcourt College Publishers. All rights reserved. 9 10 Chapter 6 Solutions (b) At B, n – Mg = – Mv2 R The max speed at B corresponds to n=0 2 Mvmax –Mg = – ⇒ vmax = R 6.22 (a) ar = r= (b) Rg = 15.0(9.80) = 12.1 m/s v2 r v2 (13.0 m/s)2 = = 8.62 m ar 2(9.80 m/s2) Let n be the force exerted by rail. Newton's law gives Mg + n = Mv2 r n = M v2 – g = M(2g – g) = Mg, downward r (c) ar = v2 (13.0 m/s)2 = = 8.45 m/s2 r 20.0 m If the force by the rail is n1, then n1 + Mg = Mv2 = Mar r n1 = M(ar – g) which is < 0, since ar = 8.45 m/s2 Thus, the normal force would have to point away from the center of the curve. Unless they have belts, the riders will fall from the cars. To be safe we must require n1 to be positive. Then ar > g. We need v2 > g or v > r rg = (20.0 m)(9.80 m/s2) v > 14.0 m/s 6.23 v= (a) 2πr 2π(3.00 m) = = 1.57 m/s T (12.0 s) a= v 2 (1.57 m/s)2 = = 0.822 m/s2 r (3.00 m) © 2000 by Harcourt College Publishers. All rights reserved. Chapter 6 Solutions (b) 11 For no sliding motion, ff = ma = (45.0 kg)(0.822 m/s2) = 37.0 N 6.24 37.0 N = 0.0839 (45.0 kg)(9.80 m/s2) (c) ff = µmg, µ = (a) ΣFx = Ma, a = (b) If v = const, a = 0, so T = 0 (This is also an equilibrium situation.) (c) Someone in the car (noninertial observer) claims that the forces on the mass along x are T and a fictitious force (–Ma). Someone at rest outside the car (inertial observer) claims that T is the only force on M in the x-direction. T 18.0 N = = 3.60 m/s2 M 5.00 kg to the right. 5.00 kg 6.25 ∑Fx = T sin θ = max (1) ∑Fx = T cos θ – mg = 0 or T cos θ = mg (2) (a) Dividing (1) by (2) gives tan θ = ax g θ = tan–1 ax 3.00 = tan–1 = 17.0° g 9.80 (b) From (1), T = ma x (0.500 kg)(3.00 m/s2) = = 5.12 N sin θ sin 17.0° © 2000 by Harcourt College Publishers. All rights reserved. 12 Chapter 6 Solutions Goal Solution G: If the horizontal acceleration were zero, then the angle would be 0, and if a = g, then the angle would be 45°, but since the acceleration is 3.00 m/s2, a reasonable estimate of the angle is about 20°. Similarly, the tension in the string should be slightly more than the weight of the object, which is about 5 N. O: We will apply Newton’s second law to solve the problem. A: The only forces acting on the suspended object are the force of gravity mg and the force of tension T, as shown in the free-body diagram. Applying Newton's second law in the x and y directions, T cos θ T sin θ mg ∑Fx = T sin θ = ma (1) ∑Fy = T cos θ – mg = 0 or T cos θ = mg (2) ( a ) Dividing equation (1) by (2) gives tan θ = a 3.00 m/s2 = = 0.306 g 9.80 m/s2 Solving for θ, θ = 17.0° (b) From Equation (1), T= ma (0.500 kg)(3.00 m/s2) = = 5.12 N sin θ sin (17.0°) L: Our answers agree with our original estimates. This problem is very similar to Prob. 5.30, so the same concept seems to apply to various situations. © 2000 by Harcourt College Publishers. All rights reserved. Chapter 6 Solutions 6.26 (a) ∑Fr = mar mg = g= T= (b) 6.27 13 mv2 m 2πR 2 = R R T 4π2R T2 4 π 2R = 2π g 6.37 × 106 m = 5.07 × 103 = 1.41 h 9.80 m/s2 2πR vnew Tnew Tcurrent 24.0 h speed increase factor = = = = = 17.1 vcurrent 2πR Tnew 1.41 h Tcurrent Fmax = Fg + ma = 591 N Fmin = Fg – ma = 391 N (a) Adding, 2Fg = 982 N, Fg = 491 N (b) Since Fg = mg, m = (c) Subtracting the above equations, 2ma = 200 N *6.28 491 N = 50.1 kg 9.80 m/s2 ∴ a = 2.00 m/s2 In an inertial reference frame, the girl is accelerating horizontally inward at v 2 (5.70 m/s)2 = = 13.5 m/s2 r 2.40 m In her own non-inertial frame, her head feels a horizontally outward fictitious force equal to its mass times this acceleration. Together this force and the weight of her head add to have a magnitude equal to the mass of her head times an acceleration of g2 + (v2/r)2 = (9.80)2 + (13.5)2 m/s2 = 16.7 m/s2 This is larger than g by a factor of 16.7 = 1.71. 9.80 Thus, the force required to lift her head is larger by this factor, or the required force is F = 1.71(55.0 N) = 93.8 N © 2000 by Harcourt College Publishers. All rights reserved. 14 6.29 Chapter 6 Solutions 4 π 2R e cos 35.0° = 0.0276 m/s2 T2 ar = N (anet)y = 9.80 – (ar)y = 9.78 m/s2 35.0° (anet)x = 0.0158 m/s2 θ = arctan (exaggerated size) ax = 0.0927° ay θ t gO anet 35.0° Equator 2 *6.30 m = 80.0 kg, vT = 50.0 m/s, mg = (a) At v = 30.0 m/s a=g– (b) DρAvT DρA mg ∴ = 2 = 0.314 2 2 vT DρAv2/2 (0.314)(30.0)2 = 9.80 – = 6.27 m/s2 downward m 80.0 At v = 50.0 m/s, terminal velocity has been reached. ΣFy = 0 = mg – R ⇒ R = mg = (80.0 kg)(9.80 m/s2) = 784 N directed up (c) At v = 30.0 m/s D ρ Av 2 = (0.314)(30.0)2 = 283 N 2 6.31 (a) upward a = g – bv When v = vT, a = 0 and g = bvT. b= g vT The Styrofoam falls 1.50 m at constant speed vT in 5.00 s. Thus, vT = Then b = y 1.50 m = = 0.300 m/s t 5.00 s 9.80 m/s2 = 32.7 s–1 0.300 m/s © 2000 by Harcourt College Publishers. All rights reserved. ar Chapter 6 Solutions (b) At t = 0, v = 0 and a = g = 9.80 m/s2 down (c) When v = 0.150 m/s, 15 a = g – bv = 9.80 m/s2 – (32.7 s–1)(0.150 m/s) = 4.90 m/s2 down *6.32 (a) ρ= m 1 2 ; A = 0.0201 m2; R = ρADv t = mg V 2 4 m = ρV = (0.830 g/cm3) π (8.00 cm)3 = 1.78 kg 3 Assuming a drag coefficient of D = 0.500 for this spherical object, 2(1.78 kg)(9.80 m/s2) = 53.8 m/s 0.500(1.20 kg/m3)(0.0201 m2) vt = (b) 2 2 v f = v i + 2gh = 0 + 2gh 2 vf (53.8 m/s)2 h= = = 148 m 2g 2(9.80 m/s2) *6.33 Since the upward velocity is constant, the resultant force on the ball is zero. Thus, the upward applied force equals the sum of the gravitational and drag forces (both downward): F = mg + bv. The mass of the copper ball is m= 4πρr3 4 kg = π 8.92 × 103 3 (2.00 × 10–2 m) 3 = 0.299 kg 3 3 m The applied force is then F = mg + bv = (0.299)(9.80) + (0.950)(9.00 × 10–2) = 3.01 N 6.34 ∑Fy = may +T cos 40.0° – mg = 0 T= 40.0 m/s (620 kg)(9.80 m/s2) = 7.93 × 103 N cos 40.0° 20.0 m 40.0° ∑Fx = max 620 kg –R + T sin 40.0° = 0 1 R = (7.93 × 103 N) sin 40.0° = 5.10 × 103 N = 2 DρAv2 D= 2R 2(5.10 × 103 N)(kg m/s2/N) = = 1.40 2 ρ Av (1.20 kg/m2)3.80 m2 (40.0 m/s)2 © 2000 by Harcourt College Publishers. All rights reserved. 16 6.35 Chapter 6 Solutions (a) At terminal velocity, R = vtb = mg ∴b= (b) mg (3.00 × 10–3 kg)(9.80 m/s2) = = 1.47 N ⋅ s/m vt (2.00 × 10–2 m/s) From Equation 6.5, the velocity on the bead is v = vt (1 – e–bt/m) v = 0.630 vt when e–bt/m = 0.370 or at time t = – m ln(0.370) = 2.04 × 10–3 s b (c) At terminal velocity, R = vtb = mg = 2.94 × 10 –2 N *6.36 The resistive force is R= 1 1 DρAv2 = (0.250)(1.20 kg/m3)(2.20 m2)(27.8 m/s) 2 2 2 R = 255 N a = –R/m = –(255 N)/(1200 kg) = –0.212 m/s2 6.37 (a) v(t) = v ie –ct v(20.0 s) = 5.00 = vie–20.0c, vi = 10.0 m/s So 5.00 = 10.0e–20.0c and –20.0c = ln 1 2 ln2 1 c=– 20.0 = 3.47 × 10 –2 s–1 © 2000 by Harcourt College Publishers. All rights reserved. Chapter 6 Solutions (b) 17 At t = 40.0 s v = (10.0 m/s)e–40.0c = (10.0 m/s)(0.250) = 2.50 m/s v = v ie–ct (c) a= 6.38 dv = – cvie–ct = – cv dt ΣFx = ma x –kmv 2 = ma x = m t ⌠ –k ⌡ 0 v ⌠ dt = ⌡ dv dt v –2 dv vf v –k(t – 0) = v= *6.39 v –1 –1 v f =– 1 1 + v vf vf (1 + ktv f) 1 DρAv2, we estimate that D = 1.00, ρ = 1.20 kg/m3, A = (0.100 m)(0.160 m) = 1.60 × 10–2 m2 2 and v = 27.0 m/s. The resistance force is then In R = 1 R = (1.00)(1.20 kg/m3)(1.60 × 10–2 m2)(27.0 m/s) 2 = 7.00 N 2 *6.40 or R ~ 10 1 N (a) At v = vt , a = 0, –mg – bvt = 0 vt = –mg (3.00 × 10–3 kg)(9.80 m/s2) =– = –0.980 m/s b 3.00 × 10–2 kg/s © 2000 by Harcourt College Publishers. All rights reserved. 18 Chapter 6 Solutions (b) t(s) x(m) v(m/s) F(mN) a(m/s2) 0 2 0 –29.4 –9.8 0.005 2 –0.049 –27.93 –9.31 0.01 1.999755 –0.09555 –26.534 –8.8445 0.015 1.9993 –0.13977 –25.2 –8.40 . . . we list the result after each tenth iteration 0.5 1.990 –0.393 –17.6 –5.87 0.1 1.965 –0.629 –10.5 –3.51 0.15 1.930 –0.770 –6.31 –2.10 0.2 1.889 –0.854 –3.78 –1.26 0.25 1.845 –0.904 –2.26 –0.754 0.3 1.799 –0.935 –1.35 –0.451 0.35 1.752 –0.953 –0.811 –0.270 0.4 1.704 –0.964 –0.486 –0.162 0.45 1.65 –0.970 –0.291 –0.0969 0.5 1.61 –0.974 –0.174 –0.0580 0.55 1.56 –0.977 –0.110 –0.0347 0.6 1.51 –0.978 –0.0624 –0.0208 0.65 1.46 –0.979 –0.0374 –0.0125 Terminal velocity is never reached. The leaf is at 99.9% of vt after 0.67 s. The fall to the ground takes about 2.14 s. Repeating with ∆t = 0.001 s, we find the fall takes 2.14 s. *6.41 (a) 2 When v = vt, a = 0, ΣF = –mg + Cv t = 0 vt = – mg =– C (4.80 × 10–4 kg)(9.80 m/s2) = –13.7 m/s 2.50 × 10–5 kg/m © 2000 by Harcourt College Publishers. All rights reserved. Chapter 6 Solutions (b) t(s) x(m) v(m/s) F(mN) a(m/s2) 0 0 0 – 4.704 –9.8 0.2 0 –1.96 – 4.608 –9.5999 0.4 –0.392 –3.88 – 4.3276 –9.0159 0.6 –1.168 –5.6832 –3.8965 –8.1178 0.8 –2.30 –7.3068 –3.3693 –7.0193 1.0 –3.77 –8.7107 –2.8071 –5.8481 1.2 –5.51 –9.8803 –2.2635 –4.7156 1.4 –7.48 –10.823 –1.7753 –3.6986 1.6 –9.65 –11.563 –1.3616 –2.8366 1.8 –11.96 –12.13 –1.03 –2.14 2 –14.4 –12.56 –0.762 –1.59 ... listing results after each fifth step 3 –27.4 –13.49 –0.154 –0.321 4 –41.0 –13.67 –0.0291 –0.0606 5 –54.7 –13.71 –0.00542 –0.0113 The hailstone reaches 99.95% of vt after 5.0 s, 99.99% of vt after 6.0 s, 99.999% of vt after 7.4 s. 6.42 (a) 2 At terminal velocity, ∑F = 0 = –mg + Cv t . C= (b) mg 2 vt = (0.142 kg)(9.80 m/s2) = 7.70 × 10–4 kg/m (42.5 m/s)2 Cv2 = (7.70 × 10–4 kg/m)(36.0 m/s)2 = 0.998 N © 2000 by Harcourt College Publishers. All rights reserved. 19 20 Chapter 6 Solutions (c) Elapsed Time (s) 0.00000 Altitude (m) 0.00000 Speed (m/s) 36.00000 Resistance Force (N) -0.99849 Net Force (N) –2.39009 Acceleration (m/s2) –16.83158 0.05000 1.75792 35.15842 –0.95235 –2.34395 –16.50667 2.95000 48.62327 0.82494 –0.00052 –1.39212 –9.80369 3.00000 48.64000 0.33476 –0.00009 –1.39169 –9.80061 3.05000 48.63224 –0.15527 0.00002 –1.39158 –9.79987 6.25000 1.25085 –26.85297 0.55555 –0.83605 –5.88769 6.30000 –0.10652 –27.14736 0.56780 –0.82380 –5.80144 … … Maximum height is about 49 m . It returns to the ground after about 6.3 s 6.43 (a) with a speed of approximately 27 m/s . 2 At constant velocity ΣF = 0 = –mg + Cv t (50.0 kg)(9.80 m/s2) = – 49.5 m/s 0.200 kg/m vt = – mg =– C vt = – (50.0 kg)(9.80 m/s) = – 4.95 m/s 20.0 kg/m with chute closed and with chute open. (b) time(s) height(m) velocity(m/s) 0 1000 0 1 995 – 9.7 2 980 –18.6 4 929 – 32.7 7 812 – 43.7 10 674 – 47.7 10.1 671 – 16.7 10.3 669 – 8.02 11 665 – 5.09 12 659 – 4.95 50 471 – 4.95 100 224 – 4.95 145 0 – 4.95 © 2000 by Harcourt College Publishers. All rights reserved. Chapter 6 Solutions 6.44 (a) time(s) x(m) y(m) 0 0 0 0.100 7.81 5.43 0.200 14.9 10.2 0.400 27.1 18.3 1.00 51.9 32.7 1.92 70.0 38.5 2.00 70.9 38.5 4.00 80.4 26.7 5.00 81.4 17.7 6.85 81.8 0 (b) range = 81.8 m (c) with θ we find range 30.0° 86.410 m 35.0° 81.8 m 25.0° 90.181 m 20.0° 92.874 m 15.0° 93.812 m 10.0° 90.965 m 17.0° 93.732 m 16.0° 93.8398 m 15.5° 93.829 m 15.8° 93.839 m 16.1° 93.838 m 15.9° 93.8402 m So we have maximum range at θ = 15.9° *6.45 (a) At terminal speed, ∑F = –mg + Cv2 = 0. Thus, C= mg (0.0460 kg)(9.80 m/s2) = = 2.33 × 10–4 kg/m v2 (44.0 m/s)2 © 2000 by Harcourt College Publishers. All rights reserved. 21 22 Chapter 6 Solutions (b) We set up a spreadsheet to calculate the motion, try different initial speeds, and home in on 53 m/s as that required for horizontal range of 155 m, thus: Time t (s) 0.0000 0.0027 … 2.5016 2.5043 2.5069 … 3.4238 3.4265 3.4291 … 5.1516 5.1543 (c) v= 2 2 v x + vy (m/s) 53.3000 53.2574 tan –1 (v y /v x ) (deg) 31.0000 30.9822 –9.8000 –9.8000 –9.8000 28.9375 28.9263 28.9151 0.0466 –0.0048 –0.0563 –8.8905 –8.9154 –8.9403 –9.3999 –9.3977 –9.3954 26.9984 26.9984 26.9984 –19.2262 –19.2822 –19.3382 0.0059 –23.3087 –0.0559 –23.3274 –7.0498 –7.0454 31.2692 31.2792 –48.1954 –48.2262 x (m) 0.0000 0.1211 vx ax (m/s) (m/s2) 45.6870 –10.5659 45.6590 –10.5529 y (m) 0.0000 0.0727 vy ay (m/s) (m/s2) 27.4515 –13.6146 27.4155 –13.6046 90.1946 90.2713 90.3480 28.9375 28.9263 28.9150 –4.2388 –4.2355 –4.2322 32.5024 32.5024 32.5024 0.0235 –0.0024 –0.0284 115.2298 115.2974 115.3649 25.4926 25.4839 25.4751 –3.2896 –3.2874 –3.2851 28.3972 28.3736 28.3500 154.9968 155.0520 20.8438 20.8380 –2.1992 –2.1980 Similarly, the initial speed is 42 m/s . The motion proceeds thus: Time x t (s) (m) 0.0000 0.0000 0.0035 0.1006 … 2.7405 66.3078 2.7440 66.3797 2.7475 66.4516 … 3.1465 74.4805 3.1500 74.5495 3.1535 74.6185 … 5.6770 118.9697 5.6805 119.0248 v= 2 2 v x + vy (m/s) 42.1500 42.1026 tan –1 (v y /v x ) (deg) 47.0000 46.9671 –9.8000 –9.8000 –9.8000 20.5485 20.5410 20.5335 0.0725 –0.0231 –0.1188 –3.9423 –3.9764 –4.0104 –9.7213 –9.7200 –9.7186 20.1058 20.1058 20.1058 –11.3077 –11.4067 –11.5056 0.0465 –25.2600 –0.0419 –25.2830 –6.5701 –6.5642 29.7623 29.7795 –58.0731 –58.1037 vx (m/s) 28.7462 28.7316 ax (m/s2) –4.1829 –4.1787 y (m) 0.0000 0.1079 vy ay (m/s) (m/s2) 30.8266 –14.6103 30.7754 –14.5943 20.5484 20.5410 20.5335 –2.1374 –2.1358 –2.1343 39.4854 39.4855 39.4855 0.0260 –0.0083 –0.0426 19.7156 19.7087 19.7018 –1.9676 –1.9662 –1.9649 38.6963 38.6825 38.6686 15.7394 15.7350 –1.2540 –1.2533 The trajectory in (c) reaches maximum height 39 m, as opposed to 33 m in (b). In both, the ball reaches maximum height when it has covered about 57% of its range. Its speed is a minimum somewhat later. The impact speeds are both about 30 m/s. © 2000 by Harcourt College Publishers. All rights reserved. Chapter 6 Solutions 6.46 (a) ΣFy = may = mv2 down R +n – 1800 kg (9.80 m/s2) = –(1800 kg) (16.0 m/s)2 = –1.10 × 104 N 42.0 m n = 6.67 × 103 N (b) 0 – mg = v= 6.47 (a) –mv2 r gr = ΣFy = may = mg – n = mv 2 R When n = 0, mg = Then, v = 6.48 mv2 R mv2 R n = mg – (b) 9.80 m/s2 (42.0 m) = 20.3 m/s F=m v2 r v= rF = m mv2 R gR (5.30 × 10 –11 m)(8.20 × 10 –8 N) = 2.18 × 106 m/s 9.11 × 10–31 kg frequency = (2.18 × 106 m/s) 1 rev = 6.56 × 1015 rev/s 2 π(5.30 × 10–11 m) © 2000 by Harcourt College Publishers. All rights reserved. 23 24 6.49 Chapter 6 Solutions (a) While the car negotiates the curve, the accelerometer is at the angle θ. Horizontally: T sin θ = mv2 r Vertically: T cos θ = mg θt T a where r is the radius of the curve, and v is the speed of the car. m v2 By division tan θ = . Then rg ar = v2 = g tan θ r ar = (9.80 m/s2) tan 15.0° ar = 2.63 m/s2 v2 (23.0 m/s)2 = = 201 m ar 2.63 m/s2 (b) r= (c) v2 = rg tan θ = (201 m)(9.80 m/s2) tan 9.00° v = 17.7 m/s 6.50 Take x-axis up the hill ∑Fx = max + T sin θ – mg sin φ = ma a = (T/m) sin θ – g sin φ ∑Fy = may + T cos θ – mg cos φ = 0 T = mg cos φ/cos θ a = g cos φ sin θ/cos θ – g sin φ a = g(cos φ tan θ – sin φ ) © 2000 by Harcourt College Publishers. All rights reserved. mg Chapter 6 Solutions 6.51 (a) 25 Since the 1.00 kg mass is in equilibrium, we have for the tension in the string, T = mg = (1.00)(9.80) = 9.80 N (b) The centripetal force is provided by the tension in the string. Hence, Fr = T = 9.80 N 6.52 mpuckv2 , we have v = r rF r = mpuck (c) Using Fr = (a) Since the mass m2 is in equilibrium, (1.00)(9.80) = 6.26 m/s 0.250 ∑Fy = T – m2g = 0 or (b) T = m 2g The tension in the string provides the required centripetal force for the puck. Thus, Fr = T = m 2g 6.53 m1v2 , we have v = R RF r = m1 m 2 gR m 1 (c) From Fr = (a) Since the centripetal acceleration of a person is downward (toward the axis of the earth), it is equivalent to the effect of a falling elevator. Therefore, F'g = Fg – (b) mv2 or F g > F'g r At the poles v = 0, and F'g = Fg = mg = (75.0)(9.80) = 735 N down At the equator, F'g = Fg – mar = 735 N – (75.0)(0.0337) N = 732 N down © 2000 by Harcourt College Publishers. All rights reserved. Chapter 6 Solutions 26 Goal Solution G: Since the centripetal acceleration is a small fraction (~0.3%) of g, we should expect that a person would have an apparent weight that is just slightly less at the equator than at the poles due to the rotation of the Earth. O: We will apply Newton’s second law and the equation for centripetal acceleration. A: ( a ) Let n represent the force exerted on the person by a scale, which is the "apparent weight." The true weight is mg. Summing up forces on the object in the direction towards the Earth's center gives mg – n = mac where ac = (1) v2 = 0.0337 m/s2 Rx is the centripetal acceleration directed toward the center of the Earth. Thus, we see that n = m(g – ac ) < mg or mg = n + mac > n ◊ (2) (b) If m = 75.0 kg, ac = 0.0337 m/s2, and g = 9.800 m/s2 , at the Equator: n = m(g – ac) = (75.0 kg)(9.800 m/s2 – 0.0337 m/s2) = 732.5 N ◊ at the Poles: n = mg = (75.0 kg)(9.800 m/s2) = 735.0 N ◊ (ac = 0) L: As we expected, the person does appear to weigh about 0.3% less at the equator than the poles. We might extend this problem to consider the effect of the earth’s bulge on a person’s weight. Since the earth is fatter at the equator than the poles, g is less than 9.80 m/s2 at the equator and slightly more at the poles, but the difference is not as significant as from the centripetal acceleration. (Can you prove this?) 6.54 ∑Fx = max ⇒ Tx = m v2 (20.4 m/s)2 =m = m (166 m/s2) r (2.50 m) ∑Fy = may ⇒ Ty – mg = 0 or Ty = mg = m(9.80 m/s2) The total tension in the string is T= Thus, m = 2 2 Tx + Ty = m (166)2 + (9.80)2 = 50.0 N 50.0 N (166)2 + (9.80)2 m/s2 = 0.300 kg © 2000 by Harcourt College Publishers. All rights reserved. Chapter 6 Solutions When the string is at the breaking point, Tx = m v2 (51.0 m/s2) = (0.300 kg) = 780 N r (1.00 m) and Ty = mg = (0.300 kg)(9.80 m/s2) = 2.94 N Hence, T = 6.55 2 2 Tx + Ty = (780)2 + (2.94)2 N = 780 N Let the angle that the wedge makes with the horizontal be θ. The equations for the mass m are mg = n cos θ and n sin θ = θt n mv2 r L where r = L cos θ. mg n cos θ Eliminating n gives = tan θ = n cos θ mg L cos θ θt mv 2 Therefore v2 = Lg cos θ tan θ = Lg sin θ v= 6.56 (a) gL sin θ v = 300 mi/h 88.0 ft/s = 440 ft/s 60.0 mi/h At the lowest point, his seat exerts an upward force; therefore, his weight seems to increase. His apparent weight is F'g = mg + m (b) v2 160 (440)2 = 160 + = 967 lb r 32.0 1200 At the highest point, the force of the seat on the pilot is directed down and v2 F'g = mg – m = – 647 lb r Since the plane is upside down, the seat exerts this downward force. (c) 6.57 mv2 . If we vary the aircraft's R and v such that the above is R true, then the pilot feels weightless. When F'g = 0, then mg = Call the proportionality constant k : ar = k/r2 v2/r = k/r2 © 2000 by Harcourt College Publishers. All rights reserved. 27 28 Chapter 6 Solutions (a) v = k1/2r–1/2 so v ∝ r –1/2 © 2000 by Harcourt College Publishers. All rights reserved. Chapter 6 Solutions (b) 29 v = 2π r/T = k1/2r–1/2 T= 2π r (k 1/2 r –1/2 ) 2π 3/2 r k 1 / 2 = T ∝ r3/2 6.58 For the block to remain stationary, ∑Fy = 0 and ∑Fx = mar. n1 = (mp + mb)g f ≤ µs1n1 = µs1 (mp + mb)g so At the point of slipping, the required centripetal force equals the maximum friction force: mb g ∴ (mp + mb) or vmax = 2 vmax r µs1rg = = µs1 (mp + mb)g mp g n1 (0.750)(0.120)(9.80) = 0.939 m/s fp For the penny to remain stationary on the block: ∑Fy = 0 ⇒ n2 – mpg = 0 and ∑Fx = mar ⇒ fp = mp or n2 = mpg f v2 r mb g mp g n2 When the penny is about to slip on the block, fp = fp, max = µs2n2 2 or vmax µs2m pg = m p r fp vmax = µs2rg = (0.520)(0.120)(9.80) = 0.782 m/s This is less than the maximum speed for the block, so the penny slips before the block starts to slip. The maximum rotation frequency is Max rpm = 6.59 v= vmax 1 rev 60 s = (0.782 m/s) = 62.2 rev/min 2πr 2 π (0.120 m) 1 min 2π r 2π (9.00 m) = = 3.77 m/s T (15.0 s) v2 = 1.58 m/s2 r (a) ar = (b) Flow = m(g + ar) = 455 N © 2000 by Harcourt College Publishers. All rights reserved. mp g 30 *6.60 Chapter 6 Solutions (c) Fhi = m(g – ar) = 329 N (d) Fmed = m g 2 + a r = 397 N at θ = tan-1 2 ar 1.58 = = 9.15° inward g 9.80 Standing on the inner surface of the rim, and moving with it, each person will feel a normal force exerted by the rim. This inward force supplies the centripetal force to cause the 3.00 m/s2 centripetal acceleration: ar = v= v2 r arr = (3.00 m/s2)(60.0 m) = 13.4 m/s The period of rotation comes from v = T= 2π r T 2π r 2π(60.0 m) = = 28.1 s v 13.4 m/s so the frequency of rotation is f= 6.61 (a) 1 1 1 60 s = = = 2.14 rev/min T 28.1 s 28.1 s 1 min The mass at the end of the chain is in vertical equilibrium. Thus T cos θ = mg mv2 Horizontally T sin θ = mar = r r = (2.50 sin θ + 4.00) m T l = 2.50 m R = 4.00 m θt r r = (2.50 sin 28.0° + 4.00) m = 5.17 m Then ar = v2 . 5.17 m By division tan θ = mg ar v2 = g 5.17g v2 = 5.17g tan θ = (5.17)(9.80)(tan 28.0°) m2/s2 v = 5.19 m/s (b) T cos θ = mg T= mg (50.0 kg)(9.80 m/s2) = = 555 N cos θ cos 28.0° © 2000 by Harcourt College Publishers. All rights reserved. Chapter 6 Solutions 6.62 (a) The putty, when dislodged, rises and returns to the original level in time t. To find t, we 2v use vf = vi + at: i.e., –v = +v – gt or t = where v is the speed of a point on the rim of the g wheel. If R is the radius of the wheel, v = Thus, v2 = πRg and v = (b) 2πR 2v 2πR , so t = = t g v πRg The putty is dislodged when F, the force holding it to the wheel is F= mv2 = m πg R f 6.63 (a) mv2 n= R f – mg = 0 f = µs n v= T= (b) 2πR T 4 π 2R µ s g n T = 2.54 s # *6.64 rev 1 rev 60 s rev = = 23.6 min 2.54 s min min mg Let the x–axis point eastward, the y-axis upward, and the z-axis point southward. 2 (a) v i sin 2θi The range is Z = g The initial speed of the ball is therefore vi = gZ = sin 2θi (9.80)(285) = 53.0 m/s sin 96.0° The time the ball is in the air is found from ∆y = viyt + 1 a t2 as 2 y 0 = (53.0 m/s)(sin 48.0°)t – (4.90 m/s2)t2 giving t = 8.04 s (b) 31 vix = 2πRe cos φi 2π(6.37 × 106 m) cos 35.0° = = 379 m/s 86400 s 86400 s © 2000 by Harcourt College Publishers. All rights reserved. 32 Chapter 6 Solutions © 2000 by Harcourt College Publishers. All rights reserved. Chapter 6 Solutions (c) 33 360˚ of latitude corresponds to a distance of 2πRe, so 285 m is a change in latitude of ∆φ = S 285 m (360°) = 2.56 × 10–3 degrees (360°) = 2 π R e 2π(6.37 × 106 m) The final latitude is then φf = φi – ∆φ = 35.0° – 0.00256° = 34.9974°. The cup is moving eastward at a speed vfx = 2πRe cos φf , which is larger than the 86400 s eastward velocity of the tee by ∆vx = vfx – vfi = 2πR e 2πR e [cos φf – cos φi] = [cos(φi – ∆φ) – cos φi] 86400 s 86400 s 2πR e [cos φi cos ∆φ + sin φi sin ∆φ – cos φi] 86400 s 2πR e Since ∆φ is such a small angle, cos ∆φ ≈ 1 and ∆vx ≈ sin φi sin ∆φ. 86400 s = ∆v x ≈ (d) *6.65 2 π (6.37 × 106 m) sin 35.0° sin 0.00256° = 1.19 × 10–2 m/s 86400 s ∆x = (∆vx)t = (1.19 × 10–2 m/s)(8.04 s) = 0.0955 m = 9.55 cm v2 , both m and r are unknown but remain constant. Therefore, ∑F is proportional to v2 r 18.0 2 and increases by a factor of as v increases from 14.0 m/s to 18.0 m/s. The total force at 14.0 the higher speed is then In ∑F = m ∑Ffast = 18.0 2 (130 N) = 215 N 14.0 Symbolically, write ∑Fslow = m m (14.0 m/s) 2 and ∑Ffast = (18.0 m/s) 2. r r Dividing gives ∑Ffast 18.0 2 = , or ∑Fslow 14.0 ∑Ffast = 18.0 2 18.0 2 ∑Fslow = (130 N) = 215 N 14.0 14.0 This force must be horizontally inward to produce the driver’s centripetal acceleration. © 2000 by Harcourt College Publishers. All rights reserved. 34 6.66 Chapter 6 Solutions (a) If the car is about to slip down the incline, f is directed up the incline. ∑Fy = n cos θ + f sin θ – mg = 0 where f = µsn gives n= mg cos θ (1 + µ s tan θ) and Then, ∑Fx = n sin θ – f cos θ = m 2 vmin R f= θt n f µsmg cos θ (1 + µ s tan θ) θt mg yields n cos θ vmin = Rg(tan θ – µ s) 1 + µ s tan θ f sin θ f cos θ n sin θ When the car is about to slip up the incline, f is directed down the incline. Then, ∑Fy = n cos θ – f sin θ – mg = 0 with f = µsn yields n= mg cos θ(1 – µs tan θ) and f= µsmg cos θ(1 – µs tan θ) mg θt In this case, ∑Fx = n sin θ + f cos θ = m v max = (b) (c) If vmin = 2 vmax R n , which gives Rg(tan θ + µ s) 1 – µ s tan θ Rg(tan θ – µ s) = 0, then µ s = tan θ . 1 + µ s tan θ vmin = (100 m)(9.80 m/s2)(tan 10.0° – 0.100) = 8.57 m/s 1 + (0.100) tan 10.0° vmax = (100 m)(9.80 m/s2)(tan 10.0° + 0.100) = 16.6 m/s 1 – (0.100)tan 10.0° f θ t mg n cos θ n sin θ f cos θ f sin θ mg © 2000 by Harcourt College Publishers. All rights reserved. Chapter 6 Solutions *6.67 (a) 35 The bead moves in a circle with radius r = R sin θ at a speed of v= 2πr 2πR sin θ = T T The normal force has an inward radial component of n sin θ and upward component of n cos θ. ∑Fy = may ⇒ n cos θ – mg = 0 or n= θt R mg cos θ v2 mg m 2πR sin θ 2 becomes sin θ = , which reduces to r R sin θ T cos θ g sin θ 4π2R sin θ gT2 = . This has two solutions: (1) sin θ = 0 ⇒ θ = 0°, and (2) cos θ = 2 . 2 cos θ T 4π R Then ∑Fx = n sin θ = m If R = 15.0 cm and T = 0.450 s, the second solution yields cos θ = (9.80 m/s2)(0.450 s)2 = 0.335 4π2(0.150 m) and θ = 70.4° Thus, in this case, the bead can ride at two positions θ = 70.4° and θ = 0° . (b) At this slower rotation, solution (2) above becomes cos θ = (9.80 m/s2)(0.850 s)2 = 1.20, which is impossible. 4π2(0.150 m) n θt In this case, the bead can ride only at the bottom of the loop, θ = 0° . The loop’s rotation must be faster than a certain threshold value in order for the bead to move away from the lowest position. 6.68 At terminal velocity, the accelerating force of gravity is balanced by frictional drag: mg = arv + br2v2 (a) mg = 3.10 × 10–9 v + 0.870 × 10–10 v2 4 For water, m = ρV = (1000 kg/m8) π (10–5) 3 3 4.11 × 10–11 = (3.10 × 10–9)v + (0.870 × 10–10)v2 Assuming v is small, ignore the second term: v = 0.0132 m/s © 2000 by Harcourt College Publishers. All rights reserved. mg 36 Chapter 6 Solutions (b) mg = 3.10 × 10–8 v + 0.870 × 10–8 v2 Here we cannot ignore the second term because the coefficients are of nearly equal magnitude. 4.11 × 10–8 = (3.10 × 10–8)v + (0.870 × 10–8)v2 v= (c) (3.10)2 + 4(0.870)(4.11) = 1.03 m/s 2(0.870) –3.10 ± mg = 3.10 × 10–7v + 0.870 × 10–6 v2 Assuming v > 1 m/s, and ignoring the first term: 4.11 × 10–5 = 0.870 × 10–6 v2 v = 6.87 m/s 6.69 ∑Fy = Ly – Ty – mg = L cos 20.0° – T sin 20.0° – 7.35 N = may = 0 ∑Fx = Lx + Tx = L sin 20.0° + T cos 20.0° = m v2 r L v2 (35.0 m/s)2 m = (0.750 kg) = 16.3 N r [(60.0 m) cos 20.0°] 20.0° ∴ L sin 20.0° + T cos 20.0° = 16.3 N L cos 20.0° – T sin 20.0° = 7.35 N L+T 20.0° T cos 20.0° 16.3 N = sin 20.0° sin 20.0° mg L–T sin 20.0° 7.35 N = cos 20.0° cos 20.0° T (cot 20.0° + tan 20.0°) = 16.3 N 7.35 N – sin 20.0° cos 20.0° T(3.11) = 39.8 N T = 12.8 N © 2000 by Harcourt College Publishers. All rights reserved. Chapter 6 Solutions 6.70 v= mg –bt 1 – exp b m At t → ∞, v → vT = mg b At t = 5.54 s, 0.500vt = vt 1 – exp –b(5.54 s) 9.00 kg exp –b(5.54 s) = 0.500 9.00 kg –b(5.54 s) = ln 0.500 = –0.693 9.00 kg b= 9.00 kg(0.693) = 1.13 m/s 5.54 s (a) vt = mg 9.00 kg(9.80 m/s2) = = 78.3 m/s b 1.13 kg/s (b) 0.750vt = vt 1 – exp –1.13t 9.00 s exp –1.13t = 0.250 9.00 s t= (c) 9.00 (ln 0.250) s = 11.1 s –1.13 dx mg bt = 1 – exp – dt b m x t ⌠ mg –bt ⌠ dx = ⌡ b 1 – exp m dt ⌡ x0 0 x – x0 = = mgt m 2g –bt t + exp 2 b b m 0 mgt m 2g –bt + exp – 1 2 b b m © 2000 by Harcourt College Publishers. All rights reserved. 37 38 Chapter 6 Solutions At t = 5.54 s, x = (9.00 kg)(9.80 m/s2) 5.54 s (9.00 kg)2(9.80 m/s2) + [exp(–0.693) – 1] 1.13 kg/s (1.13 kg/s)2 x = 434 m + (626 m)(–0.500) = 121 m 6.71 (a) t (s) 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.0 11.0 12.0 13.0 14.0 15.0 16.0 17.0 18.0 19.0 20.0 d (m) 4.88 18.9 42.1 73.8 112 154 199 246 296 347 399 452 505 558 611 664 717 770 823 876 d (m) 900 800 700 600 500 400 300 200 100 t (s) 0 0 (c) 2 4 6 8 10 12 14 16 18 A straight line fits the points from t = 11.0 s to 20.0 s quite precisely. Its slope is the terminal speed. vt = slope = 876 m – 399 m = 53.0 m/s 20.0 s – 11.0 s © 2000 by Harcourt College Publishers. All rights reserved. 20 Chapter 6 Solutions © 2000 by Harcourt College Publishers. All rights reserved. 39 Chapter 7 Even Answers 2. 4. 6. 8. 12. 14. 16. 18. 20. 22. 24. 26. 28. 30. 32. 34. 36. 38. 40. 42. 44. 46. 48. 50. 52. 54. 56. 58. 60. 62. 64. 66. 68. 70. 72. 76. 1.59 × 103 J (a) 79.4 N (b) 1.49 kJ (c) –1.49 kJ (a) 329 J (b) 0 (c) 0 (d) –185 J (e) 144 J 28.9 16.0 5.33 W (a) graph is a straight line passing through points (2 m, 0 N) and (3 m, 8 N) (b) -12.0 J 50.0 J (a) 575 N/m (b) 46.0 J (a) 9.00 kJ (b) 11.7 kJ, larger by 29.6% 3W kg/s2 (a) 33.8 J (b) 135 J (a) 2.00 m/s (b) 200 N (a) 1.94 m/s (b) 3.35 m/s (c) 3.87 m/s (a) 4.56 kJ (b) 6.34 kN (c) 422 km/s2 (d) 6.34 kN 0.116 m (a) 4.10 × 10–18 J (b) 1.14 × 10–17 N (c) 1.25 × 1013 m/s2 (d) 2.40 × 10–7 s 1.25 m/s ~ 104 W 685 bundles (a) 20.6 kJ (b) 686 W (0.919 hp) $46.2 5.92 km/L (a) 7.38 × 10–13 J (b) 94.5% (a) 4.38 × 1011 J (b) 4.38 × 1011 J 2.92 m/s Ax Ay Az (a) cos α = , cos β = , cos γ = A A A mgnhh s mgvh (a) (b) v + nh s v + nh s 7.37 N/m 57.7 W (b) 2kL2 + kA 2 – 2kL A 2 + L 2 (b) 125 N/m (c) 13.1 N (a) –5.60 J (b) 0.152 (c) 2.28 rev –1.37 × 10–21 J (b) Consider the power input when a constant force F is used to push an object of weight w distance d across a rough horizontal floor, at constant speed, in time t. Then b= µk. © 2000 by Harcourt College Publishers. All rights reserved. 2 Chapter 7 Even Answers © 2000 by Harcourt College Publishers. All rights reserved. Chapter 7 Solutions *7.1 W = Fd = (5000 N)(3.00 km) = 15.0 MJ *7.2 The component of force along the direction of motion is F cos θ = (35.0 N) cos 25.0° = 31.7 N The work done by this force is W = (F cos θ)d = (31.7 N)(50.0 m) = 1.59 × 103 J 7.3 7.4 (a) W = mgh = (3.35 × 10–5)(9.80)(100) J = 3.28 × 10–2 J (b) Since R = mg, Wair resistance = –3.28 × 10–2 J (a) ΣFy = F sin θ + n – mg = 0 d = 20.0 m n = mg – F sin θ n ΣFx = F cos θ – µkn = 0 F F cos θ n= µk fR = µkn mg – F sin θ = ∴ 18.0 kg 20.0° θt==20.0° F cos θ µk F= µkmg µk sin θ + cos θ F= (0.500)(18.0)(9.80) = 79.4 N 0.500 sin 20.0° + cos 20.0° (b) WF = Fd cos θ = (79.4 N)(20.0 m) cos 20.0° = 1.49 kJ (c) fk = F cos θ = 74.6 N mg Wf = fk d cos θ = (74.6 N)(20.0 m) cos 180° = –1.49 kJ 7.5 (a) W = Fd cos θ = (16.0 N)(2.20 m) cos 25.0° = 31.9 J (b) and (c) (d) The normal force and the weight are both at 90° to the motion. Both do 0 work. ∑W = 31.9 J + 0 + 0 = 31.9 J © 2000 by Harcourt College Publishers. All rights reserved. 2 7.6 Chapter 7 Solutions ∑Fy = may d = 5.00 m n + (70.0 N) sin 20.0° – 147 N = 0 n n = 123 N 70.0 N sin 20.0° fk = µkn = 0.300 (123 N) = 36.9 N (a) fk W = Fd cos θ 70.0 N cos 20.0° 15.0 kg = (70.0 N)(5.00 m) cos 20.0° = 329 J 7.7 (b) W = Fd cos θ = (123 N)(5.00 m) cos 90.0° = 0 J (c) W = Fd cos θ = (147 N)(5.00 m) cos 90.0° = 0 (d) W = Fd cos θ = (36.9 N)(5.00 m) cos 180° = –185 J (e) ∆K = Kf – Ki = ∑W = 329 J – 185 J = +144 J W = mg(∆y) = mg(l – l cos θ) = (80.0 kg)(9.80 m/s2)(12.0 m)(1 – cos 60.0°) = 4.70 kJ l 60° ∆y 7.8 A = 5.00; B = 9.00; θ = 50.0° A · B = AB cos θ = (5.00)(9.00) cos 50.0° = 28.9 7.9 A · B = AB cos θ = 7.00(4.00) cos (130° – 70.0°) = 14.0 7.10 A · B = (Axi + Ay j + Azk) · (Bxi + By j + Bzk) A · B = AxBx (i · i) + AxBy (i · j) + AxBz (i · k) + AyBx (j · i) + AyBy (j · j) + AyBz (j · k) + AzBx (k · i) + AzBy (k · j) + AzBz (k · k) © 2000 by Harcourt College Publishers. All rights reserved. mg = 147 N Chapter 7 Solutions A · B = A xB x + A yB y + A zB z © 2000 by Harcourt College Publishers. All rights reserved. 3 4 7.11 7.12 Chapter 7 Solutions (a) W = F · d = Fxx + Fyy = (6.00)(3.00) N · m + (–2.00)(1.00) N · m = 16.0 J (b) θ = cos–1 F·d = cos–1 Fd 16 [(6.00)2 + (–2.00)2][(3.00)2 + (1.00)2] = 36.9° A – B = (3.00i + j – k) – (–i + 2.00j + 5.00k) A – B = 4.00i – j – 6.00k C · (A – B) = (2.00j – 3.00k) · (4.00i – j – 6.00k) = 0 + (–2.00) + (+18.0) = 16.0 7.13 (a) A = 3.00i – 2.00j θ = cos–1 (b) B = 4.00i – 4.00j A·B 12.0 + 8.00 = cos–1 = 11.3° AB (13.0)(32.0) B = 3.00i – 4.00j + 2.00k cos θ = A = –2.00i + 4.00j A·B –6.00 – 16.0 = AB (20.0)(29.0) θ = 156° (c) A = i – 2.00j + 2.00k B = 3.00j + 4.00k θ = cos–1 A · B – 6.00 + 8.00 = cos–1 = 82.3° A B 9.00 · 25.0 *7.14 We must first find the angle between the two vectors. It is: y θ = 360° – 118° – 90.0° – 132° = 20.0° 118° Then x F ⋅ v = Fv cos θ = (32.8 N)(0.173 m/s) cos 20.0° F = 32.8 N 132° θ t or F ⋅ v = 5.33 N⋅m J = 5.33 = 5.33 W s s © 2000 by Harcourt College Publishers. All rights reserved. v = 17.3 cm/s Chapter 7 Solutions 7.15 5 f ⌠ Fdx = area under curve from xi to xf W=⌡ i (a) xi = 0 Fx (N) B 6 xf = 8.00 m 4 W = area of triangle ABC = 1 AC × altitude, 2 2 xi = 8.00 m C 2 −2 4 6 −4 1 W 0→8 = 2 × 8.00 m × 6.00 N = 24.0 J (b) A 0 8 E 10 12 x (m) D xf = 10.0 m W = area of ∆CDE = 1 CE × altitude, 2 W 8→10 = 12 × (2.00 m) × (–3 .00 N) = –3.00 J (c) *7.16 W 0→10 = W 0→8 + W 8→10 = 24.0 + (–3.00) = 21.0 J Fx = (8x – 16) N (a) Fx (N) 20 (3, 8) 10 0 1 −10 2 3 4 x (m) −20 (b) 7.17 W net = –(2.00 m)(16.0 N) (1.00 m)(8.00 N) + = –12.0 J 2 2 W = ∫ Fx dx and W equals the area under the Force-Displacement Curve (a) For the region 0 ≤ x ≤ 5.00 m, Fx (N) 3 2 (3.00 N)(5.00 m) W= = 7.50 J 2 (b) 1 0 0 2 4 For the region 5.00 ≤ x ≤ 10.0, W = (3.00 N)(5.00 m) = 15.0 J © 2000 by Harcourt College Publishers. All rights reserved. 6 8 10 12 14 16 x (m) 6 Chapter 7 Solutions (c) For the region 10.0 ≤ x ≤ 15.0, W= (d) (3.00 N)(5.00 m) = 7.50 J 2 For the region 0 ≤ x ≤ 15.0 W = (7.50 + 7.50 + 15.0) J = 30.0 J 7.18 f 5m i 0 ⌠ F · ds = ⌠ W=⌡ ⌡ (4xi + 3yj) N · dxi 5m 5m ⌠ = 50.0 J ⌡0 (4 N/m) x dx + 0 = (4 N/m)x2/2 0 *7.19 7.20 k= F Mg (4.00)(9.80) N = = = 1.57 × 103 N/m y y 2.50 × 10–2 m For 1.50 kg mass y = (b) Work = 1 ky 2 2 Work = 1 (1.57 × 103 N · m)(4.00 × 10–2 m) 2 = 1.25 J 2 (a) Spring constant is given by F = kx k= (b) 7.21 mg (1.50)(9.80) = = 0.938 cm k 1.57 × 103 (a) F (230 N) = = 575 N/m x (0.400 m) 1 Work = Favg x = (230 N)(0.400 m) = 46.0 J 2 Compare an initial picture of the rolling car with a final picture with both springs compressed Ki + ∑W = Kf F (N) 1500 1000 500 d (cm) 0 0 10 20 30 40 50 © 2000 by Harcourt College Publishers. All rights reserved. Chapter 7 Solutions Use equation 7.11. Ki + 1 1 2 2 2 2 k (x – x1f ) + k2 (x2i – x2f ) = Kf 2 2 1 1i 1 1 1 2 mv i + 0 – (1600 N/m)(0.500 m) 2 + 0 – (3400 N/m)(0.200 m) 2 = 0 2 2 2 1 2 (6000 kg) v i – 200 J – 68.0 J = 0 2 vi = 7.22 (a) 2 × 268 J/6000 kg = 0.299 m/s f ⌠ F · ds W=⌡ i 0.600 m ⌠ W=⌡ 0 (15000 N + 10000 x N/m – 25000 x2 N/m2) dx cos 0° W = 15,000x + 10,000x2 25,000x3 0.600 – 2 3 0 W = 9.00 kJ + 1.80 kJ – 1.80 kJ = 9.00 kJ (b) Similarly, W = (15.0 kN)(1.00 m) + (10.0 kN/m)(1.00 m)2 2 – (25.0 kN/m2)(1.00 m)3 3 W = 11.7 kJ , larger by 29.6% 7.23 4.00 J = 1 k(0.100 m)2 2 ∴ k = 800 N/m and to stretch the spring to 0.200 m requires ∆W = 1 (800)(0.200) 2 – 4.00 J = 12.0 J 2 Goal Solution G: We know that the force required to stretch a spring is proportional to the distance the spring is stretched, and since the work required is proportional to the force and to the distance, then W ∝ x2. This means if the extension of the spring is doubled, the work will increase by a factor of 4, so that for x = 20 cm, W = 16 J, requiring 12 J of additional work. O: Let’s confirm our answer using Hooke’s law and the definition of work. © 2000 by Harcourt College Publishers. All rights reserved. 7 Chapter 7 Solutions 8 A: The linear spring force relation is given by Hooke’s law: Fs = –kx Integrating with respect to x, we find the work done by the spring is: xy xy xx xx ⌠ (–kx)dx = – ⌠ Fs dx = ⌡ Ws = ⌡ 1 2 2 k (x f – xi ) 2 However, we want the work done on the spring, which is W = –Ws = 1 2 2 k(x f – xi ) 2 We know the work for the first 10 cm, so we can find the force constant: k= 2W 0–10 2 x0–10 = 2(4.00 J) = 800 N/m (0.100 m)2 Substituting for k, xi and xf, the extra work for the next step of extension is W= 1 (800 N/m) [(0.200 m)2 – (0.100 m)2] = 12.0 J 2 L: Our calculated answer agrees with our prediction. It is helpful to remember that the force required to stretch a spring is proportional to the distance the spring is extended, but the work is proportional to the square of the extension. 7.24 7.25 W= 1 kd 2 2 ∴k= 2W d2 ∆W = 1 1 k(2d)2 – kd 2 2 2 ∆W = 3 kd2 = 3W 2 (a) The radius to the mass makes angle θ with the horizontal, so its weight makes angle θ with the negative side of the x-axis, when we take the x-axis in the direction of motion tangent to the cylinder. x F n t θ R ∑Fx = max θ t F – mg cos θ = 0 F = mg cos θ © 2000 by Harcourt College Publishers. All rights reserved. mg Chapter 7 Solutions f ⌠ F ⋅ ds W=⌡ (b) i We use radian measure to express the next bit of displacement as ds = r dθ in terms of the next bit of angle moved through: π/2 ⌠ W=⌡ mg cos θ Rdθ = mgR sin θ π/2 0 0 W = mgR (1 – 0) = mgR *7.26 7.27 F N kg ⋅ m/s2 kg = = = 2 m s x m [k] = 1 (0.600 kg)(2.00 m/s) 2 = 1.20 J 2 (a) KA = (b) 1 2 mvB = KB 2 vB = (c) 2K B = m (2)(7.50) = 5.00 m/s 0.600 ∑W = ∆K = KB – KA = 1 2 2 m(vB – vA ) 2 = 7.50 J – 1.20 J = 6.30 J *7.28 7.29 (a) K= 1 1 mv2 = (0.300 kg)(15 .0 m/s) 2 = 33.8 J 2 2 (b) K= 1 1 (0.300)(30.0) 2 = (0.300)(15.0) 2 (4) = 4(33.8) = 135 J 2 2 vi = (6.00i – 2.00j) m/s (a) vi = Ki = (b) 2 2 v i x + v iy = 40.0 m/s 1 1 2 mv i = (3.00 kg)(40.0 m2/s2) = 60.0 J 2 2 v = 8.00i + 4.00j v2 = v · v = 64.0 + 16.0 = 80.0 m2/s2 ∆K = K – Ki = 1 3.00 2 m(v2 – v i ) = (80.0) – 60.0 = 60.0 J 2 2 © 2000 by Harcourt College Publishers. All rights reserved. 9 10 7.30 Chapter 7 Solutions (a) ∆K = ∑W 1 (2500 kg) v2 = 5000 J 2 v = 2.00 m/s (b) W=F·d 5000 J = F(25.0 m) F = 200 N 7.31 (a) ∆K = 1 mv2 – 0 = ∑W, so 2 v2 = 2W/m 7.32 v= 2W/m (b) W = F ⋅ d = Fxd ⇒ Fx = W/d (a) ∆K = Kf – Ki = vf = (b) (c) 2(7.50 J) = 1.94 m/s 4.00 kg 1 2 mv f – 0 = ∑W = (area under curve from x = 0 to x = 10.0 m) 2 2(area) = m ∆K = Kf – Ki = vf = 1 2 mv f – 0 = ∑W = (area under curve from x = 0 to x = 5.00 m) 2 2(area) = m ∆K = Kf – Ki = vf = *7.33 and 2(22.5 J) = 3.35 m/s 4.00 kg 1 2 mv f – 0 = ∑W = (area under curve from x = 0 to x = 15.0 m) 2 2(area) = m 2(30.0 J) = 3.87 m/s 4.00 kg d = 5.00 m ∑Fy = may n – 392 N = 0 n n = 392 N fk = µkn = 0.300(392 N) = 118 N fk (a) WF = Fd cos θ = (130)(5.00) cos 0° = 650 J (b) Wfk = fk d cos θ = (118)(5.00) cos 180° = –588 J F = 130 N 40.0 kg mg = 392 N © 2000 by Harcourt College Publishers. All rights reserved. Chapter 7 Solutions (c) Wn = nd cos θ = (392)(5.00) cos 90° = 0 (d) Wg = mg cos θ = (392)(5.00) cos (–90°) = 0 (e) ∆K = Kf – Ki = ∑W 11 1 2 mv f – 0 = 650 J – 588 J + 0 + 0 = 62.0 J 2 7.34 2K f = m (f) vf = (a) Ki + ∑W = Kf = 2(62.0 J) = 1.76 m/s 40.0 kg 1 2 mv f 2 1 0 + ∑W = (15.0 × 10–3 kg)(780 m/s) 2 = 4.56 kJ 2 W 4.56 × 103 J = = 6.34 kN d cos θ (0.720 m) cos 0° (b) F= (c) vf – v i (780 m/s)2 – 0 a= = = 422 km/s2 2x 2(0.720 m) (d) ∑F = ma = (15 × 10–3 kg)(422 × 103 m/s2) = 6.34 kN (a) Wg = mgl cos (90.0° + θ) = (10.0 kg)(9.80 m/s2)(5.00 m) cos 110° = – 168 J (b) fk = µkn = µkmg cos θ 2 7.35 2 F = 100 N l Wf = – lfk = lµkmg cos θ cos 180° vi Wf = – (5.00 m)(0.400)(10.0)(9.80) cos 20.0° = – 184 J (c) W F = Fl = (100)(5.00) = 500 J (d) ∆K = ∑W = W F + W f + W g = 148 J (e) ∆K = vf = θt 1 1 2 2 mv f – mv i 2 2 2(∆K) 2 + vi = m 2(148) + (1.50)2 10.0 = 5.65 m/s © 2000 by Harcourt College Publishers. All rights reserved. Chapter 7 Solutions 12 7.36 ∑W = ∆K = 0 L d ⌠ kx dx = 0 ⌠ ⌡0 mg sin 35.0° dl – ⌡ 0 mg sin 35.0° (L) = d= = 7.37 1 kd 2 2 2 mg sin 35.0°(L) k 2(12.0 kg)(9.80 m/s2) sin 35.0° (3.00 m) = 0.116 m 3.00 × 104 N/m vi = 2.00 m/s µk = 0.100 ∑W = ∆K – fk x = 0 – 1 2 mv i 2 – µkmgx = – 1 2 mv i 2 2 vi (2.00 m/s)2 x= = = 2.04 m 2µk g 2(0.100)(9.80) Goal Solution G: Since the sled’s initial speed of 2 m/s (~ 4 mph) is reasonable for a moderate kick, we might expect the sled to travel several meters before coming to rest. O: We could solve this problem using Newton’s second law, but we are asked to use the workkinetic energy theorem: W = Kf – Ki, where the only work done on the sled after the kick results from the friction between the sled and ice. (The weight and normal force both act at 90° to the motion, and therefore do no work on the sled.) A: The work due to friction is W = –fk d where fk = µkmg. Since the final kinetic energy is zero, W = ∆K = 0 – Ki = – 2 Solving for the distance d = 1 2 mv i 2 2 mv i vi (2.00 m)2 = = = 2.04 m 2µk mg 2µk g 2(0.100)(9.80 m/s2) L: The distance agrees with the prediction. It is interesting that the distance does not depend on the mass and is proportional to the square of the initial velocity. This means that a small car and a massive truck should be able to stop within the same distance if they both skid to a stop from the same initial speed. Also, doubling the speed requires 4 times as much stopping distance, which is consistent with advice given by transportation safety officers who suggest at least a 2 second gap between vehicles (as opposed to a fixed distance of 100 feet). © 2000 by Harcourt College Publishers. All rights reserved. Chapter 7 Solutions 7.38 (a) vf = 0.01c = 10–2(3.00 × 108 m/s) = 3.00 × 106 m/s Kf = (b) 1 1 2 mv f = (9.11 × 10–31 kg)(3.00 × 106 m/s) 2 = 4.10 × 10–18 J 2 2 Ki + Fd cos θ = Kf 0 + F(0.360 m) cos 0° = 4.10 × 10–18 N · m F = 1.14 × 10 –17 N a= (d) 1 xf – xi = (v i+ v f) t 2 t= 7.39 ∑F 1.14 × 10 –17 N = = 1.25 × 1013 m/s2 m 9.11 × 10–31 kg (c) (a) 2(xf – xi) 2(0.360 m) = = 2.40 × 10–7 s (v i + v f) (3.00 × 106 m/s) ∑W = ∆K ⇒ fd cos θ = 1 1 2 2 mv f – mv i 2 2 1 f(4.00 × 10–2 m) cos 180° = 0 – (5.00 × 10–3 kg)(600 m/s) 2 2 f = 2.25 × 104 N (b) 7.40 t= d 4.00 × 10–2 m = = 1.33 × 10–4 s – [0 + 600 m/s]/2 v ∑W = ∆K m 1gh – m 2gh = vf = 2(m 1 – m 2)gh 2(0.300 – 0.200)(9.80)(0.400) m2 = m1 + m2 0.300 + 0.200 s2 vf = 1.57 m/s = 1.25 m/s (a) Ws = 2 7.41 1 2 (m + m2) v f – 0 2 1 1 1 2 1 2 kx i – kx f = (500)(5.00 × 10–2) 2 – 0 = 0.625 J 2 2 2 ∑W = 1 1 1 2 2 2 mv f – mv i = mv f – 0 2 2 2 so vf = 2(∑W) = m 2(0.625) m/s = 0.791 m/s 2.00 © 2000 by Harcourt College Publishers. All rights reserved. 13 14 Chapter 7 Solutions (b) ∑W = Ws + Wf = 0.625 J + (–µkmgd) = 0.625 J – (0.350)(2.00)(9.80)(5.00 × 10–2) J = 0.282 J vf = *7.42 2(∑W) = m 2(0.282) m/s = 0.531 m/s 2.00 A 1300-kg car speeds up from rest to 55.0 mi/h = 24.6 m/s in 15.0 s. The output work of the engine becomes its final kinetic energy, 1 (1300 kg)(24.6 m/s) 2 = 390 kJ 2 with power 390000 J ~ 10 4 W , around 30 horsepower. 15.0 s W mgh (700 N)(10.0 m) = = = 875 W t t 8.00 s 7.43 Power = 7.44 Efficiency = e = useful energy output/total energy input. The force required to lift n bundles of shingles is their weight, nmg. 7.45 *7.46 e= n mgh cos 0° Pt n= eP t (0.700)(746 W)(7200 s) kg · m2 = × 3 = 685 bundles 2 mgh (70.0 kg)(9.80 m/s )(8.00 m) s ·W Pa = fa v ⇒ fa = (a) Pa 2.24 × 104 = = 830 N v 27.0 ∑W = ∆K, but ∆K = 0 because he moves at constant speed. The skier rises a vertical distance of (60.0 m) sin 30.0° = 30.0 m. Thus, Win = –Wg = (70.0 kg)g(30.0 m) = 2.06 × 104 J = 20.6 kJ (b) The time to travel 60.0 m at a constant speed of 2.00 m/s is 30.0 s. Thus, Pinput = W 2.06 × 104 J = = 686 W ∆t 30.0 s = 0.919 hp © 2000 by Harcourt College Publishers. All rights reserved. Chapter 7 Solutions 7.47 (a) The distance moved upward in the first 3.00 s is 0 + 1.75 m/s – ∆y = v t = (3.00 s) = 2.63 m 2 W= 1 1 2 1 2 2 1 2 mv f – 2 mv i + mgyf – mgyi = mv f – 2 mv i + mg(∆y) 2 2 W= 1 (650 kg)(1.75 m/s) 2 – 0 + (650 kg)g(2.63 m) = 1.77 × 104 J 2 – Also, W = P t so (b) – W 1.77 × 104 J P = = = 5.91 × 10 3 W = 7.92 hp t 3.00 s When moving upward at constant speed (v = 1.75 m/s), the applied force equals the weight = (650 kg)(9.80 m/s2) = 6.37 × 103 N. Therefore, P = Fv = (6.37 × 103 N)(1.75 m/s) = 1.11 × 10 4 W = 14.9 hp *7.48 energy = power × time For the 28.0 W bulb: Energy used = (28.0 W)(1.00 × 104 h) = 280 kilowatt ⋅ hrs total cost = $17.00 + (280 kWh)($0.080/kWh) = $39.40 For the 100 W bulb: Energy used = (100 W)(1.00 × 104 h) = 1.00 × 103 kilowatt ⋅ hrs # bulb used = 1.00 × 104 h = 13.3 750 h/bulb total cost = 13.3($0.420) + (1.00 × 103 kWh)($0.080/kWh) = $85.60 Savings with energy-efficient bulb = $85.60 – $39.40 = $46.2 © 2000 by Harcourt College Publishers. All rights reserved. 15 16 7.49 Chapter 7 Solutions (a) fuel needed = = (b) (c) 7.50 1 2 2 2 mv f – 12 mv i useful energy per gallon 1 2 (900 kg)(24.6 m/s)2 (0.150)(1.34 × 108 J/gal) = 1 2 2 mv f – 0 eff. × (energy content of fuel) = 1.35 × 10 –2 gal 73.8 1 gal 55.0 mi 1.00 h 1.34 × 108 J (0.150) = 8.08 kW 38.0 mi 1.00 h 3600 s 1 gal power = At a speed of 26.8 m/s (60.0 mph), the car described in Table 7.2 delivers a power of P1 = 18.3 kW to the wheels. If an additional load of 350 kg is added to the car, a larger output power of P2 = P1 + (power input to move 350 kg at speed v) will be required. The additional power output needed to move 350 kg at speed v is: ∆P out = (∆f)v = (µ rmg)v Assuming a coefficient of rolling friction of µr = 0.0160, the power output now needed from the engine is P2 = P1 + (0.0160)(350 kg)(9.80 m/s2)(26.8 m/s) = 18.3 kW + 1.47 kW With the assumption of constant efficiency of the engine, the input power must increase by the same factor as the output power. Thus, the fuel economy must decrease by this factor: P1 (fuel economy)2 = P (fuel economy) 1 = 18.3 km 6.40 18.3 + 1.47 L 2 or (fuel economy)2 = 5.92 km L © 2000 by Harcourt College Publishers. All rights reserved. Chapter 7 Solutions 7.51 17 When the car of Table 7.2 is traveling at 26.8 m/s (60.0 mph), the engine delivers a power of P1 = 18.3 kW to the wheels. When the air conditioner is turned on, an additional output power of ∆P = 1.54 kW is needed. The total power output now required is P2 = P1 + ∆P = 18.3 kW + 1.54 kW Assuming a constant efficiency of the engine, the fuel economy must decrease by the same factor as the power output increases. The expected fuel economy with the air conditioner on is therefore P1 (fuel economy)2 = P (fuel economy) 1 = 18.3 km 6.40 18.3 + 1.54 L 2 7.52 or (fuel economy)2 = 5.90 (a) K= 1 1 – (v/c)2 km L 1 – 1 mc2 = 1 – (0.995)2 – 1 (9.11 × 10–31)(2.998 × 108) 2 K = 7.38 × 10–13 J (b) Classically, K= 1 1 mv2 = (9.11 × 10–31 kg) [(0.995)(2.998 × 108 m/s)]2 = 4.05 × 10–14 J 2 2 This differs from the relativistic result by 7.38 × 10–13 J – 4.05 × 10–14 J 100% = 94.5% 7.38 × 10–13 J % error = 7.53 ∑W = Kf – Ki = 1 1 – (v f/c)2 or ∑W = 1 (a) ∑W = 1 1 – (vf 1– /c)2 1 – (0.750)2 – 1 mc2 – 1 1 – (v i/c)2 – 1 mc2 mc2 1 – (v i/c)2 – 1 1– (0.500)2 (1.673 × 10–27 kg)(2.998 × 108 m/s) 2 ∑W = 5.37 × 10–11 J (b) ∑W = 1 1– (0.995)2 – 1 (1.673 × 10–27 kg)(2.998 × 108 m/s) 2 1 – (0.500)2 ∑W = 1.33 × 10–9 J © 2000 by Harcourt College Publishers. All rights reserved. Chapter 7 Solutions 18 Goal Solution G: Since particle accelerators have typical maximum energies on the order of GeV (1eV = 1.60 × 10–19 J), we could expect the work required to be ~10–10 J. O: The work-energy theorem is W = Kf – Ki which for relativistic speeds (v ~ c) is: W = 1 mc2 – mc2 2 2 1 – v i /c 1 2 v f /c2 1– 1 A: (a) W = – 1 (1.67 × 10–27 kg)(3.00 × 108 m/s) 2 1 – (0.750)2 – 1 1 – (0.500)2 W = (0.512 – 0.155)(1.50 10–10 J) = 5.37 10–11 J (b) E = 1 1 – (0.995)2 – 1 (1.50 × 10–10 J) ◊ – 1 (1.50 × 10–10 J) – (1.155 – 1)(1.50 × 10–10 J) W = (9.01 – 0.155)(1.50 10-10 J) = 1.33 10-9 J ◊ L: Even though these energies may seem like small numbers, we must remember that the proton has very small mass, so these input energies are comparable to the rest mass energy of the proton (938 MeV = 1.50 × 10–10 J). To produce a speed higher by 33%, the answer to part (b) is 25 times larger than the answer to part (a). Even with arbitrarily large accelerating energies, the particle will never reach or exceed the speed of light. This is a consequence of special relativity, which will be examined more closely in a later chapter. *7.54 (a) Using the classical equation, K= (b) 1 1 mv2 = (78.0 kg)(1.06 × 105 m/s) 2 = 4.38 × 1011 J 2 2 Using the relativistic equation, K= 1 1 – (v/c)2 – 1 mc2 1 = 1 – (1.06 × 105/2.998 × 108)2 – 1 (78.0 kg)(2.998 × 108 m/s) 2 K = 4.38 × 1011 J © 2000 by Harcourt College Publishers. All rights reserved. Chapter 7 Solutions 19 When (v/c) << 1, the binomial series expansion gives 1 [1 – (v/c)2]–1/2 ≈ 1 + (v/c) 2 2 Thus, [1 – (v/c)2]–1/2 – 1 ≈ (v/c)2 1 1 (v/c) 2 mc2 = mv 2. That 2 2 is, in the limit of speeds much smaller than the speed of light, the relativistic and classical expressions yield the same results. and the relativistic expression for kinetic energy becomes K ≈ *7.55 At start, v = (40.0 m/s) cos 30.0°i + (40.0 m/s) sin 30.0°j At apex, v = (40.0 m/s) cos 30.0°i + 0j = 34.6i m/s and K = *7.56 1 1 mv2 = (0.150 kg)(34.6 m/s) 2 = 90.0 J 2 2 Concentration of Energy output = (0.600 J/kg · step)(60.0 kg) 1 step J = 24.0 1.50 m m F = 24.0 J N · m 1 = 24.0 N m J P = Fv 70.0 W = (24.0 N)v v = 2.92 m/s 7.57 The work-kinetic energy theorem is Ki + ∑W = Kf The total work is equal to the work by the constant total force: 1 1 2 2 mv i + (ΣF) · (r – ri) = mv f 2 2 1 1 2 2 mv i + ma · (r – ri) = mv f 2 2 2 2 v i + 2a · (r – r i) = v f © 2000 by Harcourt College Publishers. All rights reserved. 20 7.58 Chapter 7 Solutions (a) A ⋅ i = (A)(1) cos α. But also, A ⋅ i = Ax. Thus, (A)(1) cos α = Ax or cos α = Similarly, cos β = 7.59 Ax A Ay A and cos γ = Az A where A = Ax + Ay + Az 2 2 2 (b) cos2 α + cos2 β + cos2 γ = A x 2 A y 2 A z 2 A2 + + = =1 A A A A2 (a) x = t + 2.00t3 therefore, (b) a= v= dx = 1 + 6.00t2 dt K= 1 1 mv2 = (4.00)(1 + 6.00t2) 2 2 2 = (2 .00 + 24.0t2 + 72.0t4) J dv = (12.0t) m/s2 dt F = ma = 4.00(12.0t) = (48.0t) N *7.60 (c) P = Fv = (48.0t)(1 + 6.00t2) = (48.0t + 288t3) W (d) ⌠ W=⌡ (a) 2.00 0 2.00 ⌠ P dt = ⌡ 0 (48.0t + 288t3) dt = 1250 J The work done by the traveler is mghsN where N is the number of steps he climbs during the ride. N = (time on escalator)(n) where (time on escalator) = h , and vertical velocity of person vertical velocity of person = v + nhs © 2000 by Harcourt College Publishers. All rights reserved. Chapter 7 Solutions Then, N = nh and the work done by the person becomes v + nh s Wperson = (b) mgnhh s v + nh s The work done by the escalator is W e = (power)(time) = [(force exerted)(speed)](time) = mgvt where t = h as above. Thus, v + nh s mgvh v + nh s We = As a check, the total work done on the person’s body must add up to mgh, the work an elevator would do in lifting him. It does add up as follows: ∑W = W person + W e = 7.61 xf ⌠ W=⌡ xi mgnhhs mgh(nh s + v) mgvh + = = mgh v + nh s v + nh s v + nh s xf ⌠ (– kx + βx3) dx F dx = ⌡ 0 2 4 –kx f βxf + 2 4 W= –kx 2 βx4 xf + 2 4 0 W= (–10.0 N/m)(0.100 m)2 (100 N/m3)(0.100 m)4 + 2 4 = W = – 5.00 × 10–2 J + 2.50 × 10–3 J = – 4.75 × 10–2 J *7.62 ∑Fx = max ⇒ kx = ma k= m a (4.70 × 10–3 kg)0.800(9.80 m/s2) = = 7.37 N/m x 0.500 × 10–2 m n a Fs m m mg © 2000 by Harcourt College Publishers. All rights reserved. 21 Chapter 7 Solutions 22 7.63 Consider the work done on the pile driver from the time it starts from rest until it comes to rest at the end of the fall. ∑W = ∆K ⇒ Wgravity + Wbeam = so 1 1 2 2 mv f – mv i 2 2 – (mg)(h + d) cos 0° + (F )(d) cos 180° = 0 – 0 – (mg)(h + d) (2100 kg)(9.80 m/s2)(5.12 m) Thus, F = = = 8.78 × 105 N d 0.120 m Goal Solution G: Anyone who has hit their thumb with a hammer knows that the resulting force is greater than just the weight of the hammer, so we should also expect the force of the pile driver to be greater than its weight: F > mg ~20 kN. The force on the pile driver will be directed upwards. O: The average force stopping the driver can be found from the work that results from the gravitational force starting its motion. The initial and final kinetic energies are zero. A: Choose the initial point when the mass is elevated and the final point when it comes to rest again 5.12 m below. Two forces do work on the pile driver: gravity and the normal force exerted by the beam on the pile driver. Wnet = Kf – Ki so that mgsw cos 0 +nsn cos 180 = 0 where m = 2 100 kg, sw = 5.12 m, and sn = 0.120 m. In this situation, the weight vector is in the direction of motion and the beam exerts a force on the pile driver opposite the direction of motion. (2100 kg) (9.80 m/s2) (5.12 m) – n (0.120 m) = 0 Solve for n. n = 1.05 × 105 J = 878 kN (upwards) ◊ 0.120 m L: The normal force is larger than 20 kN as we expected, and is actually about 43 times greater than the weight of the pile driver, which is why this machine is so effective. © 2000 by Harcourt College Publishers. All rights reserved. Chapter 7 Solutions Additional Calculation: Show that the work done by gravity on an object can be represented by mgh, where h is the vertical height that the object falls. Apply your results to the problem above. By the figure, where d is the path of the object, and h is the height that the object falls, d θ dy h h = |dy| = d cos θ Since F = mg, mgh = Fd cos θ = F·d In this problem, mgh = n(dn), or (2100 kg)(9.80 m/s2)(5.12 m) = n(0.120 m) and n = 878 kN 7.64 Let b represent the proportionality constant of air drag fa to speed: f a Let fr represent the other frictional forces. Take x-axis along each roadway. For the gentle hill ∑F x = ma x – bv – fr + mg sin 2.00° = 0 – b(4.00 m/s) – fr + 25.7 N = 0 For the steeper hill –b(8.00 m/s) – fr + 51.3 N = 0 Subtracting, b(4.00 m/s) = 25.6 N b = 6.40 N · s/m and then fr = 0.0313 N. Now at 3.00 m/s the vehicle must pull her with force bv + fr = (6.40 N · s/m)(3.00 m/s) + 0.0313 N = 19.2 N and with power P = F · v = 19.2 N(3.00 m/s) cos 0° = 57.7 W © 2000 by Harcourt College Publishers. All rights reserved. = bv 23 24 7.65 7.66 Chapter 7 Solutions (a) P = Fv = F(vi + at) = F 0 + (b) P= (a) The new length of each spring is F t = m F 2 t m (20.0 N) 2 (3.00 s) = 240 W 5.00 kg its extension is x2 + L2 , so x2 + L2 – L and the force it k L exerts is k ( x2 + L2 – L) toward its fixed end. The y components of the two spring forces add to zero. Their x components add to A m L k F = –2ik ( x2 + L2 – L)x/ x 2 + L 2 F = –2kxi (1 – L/ x 2 + L 2 ) (b) f ⌠ Fx dx W=⌡ i 0 ⌠ –2kx (1 – L/ x 2 + L 2 )dx W=⌡ A 0 0 A A ⌠ x dx + kL ⌡ ⌠ (x2 + L2) –1/2 2x dx W = –2k ⌡ W = –2k x2 0 (x 2 + L 2)1/2 0 + kL 2 A (1/2) A W = –0 + kA2 + 2kL2 – 2kL A 2 + L 2 W = 2kL 2 + kA 2 – 2k L A 2 + L 2 7.67 (a) F1 = (25.0 N)(cos 35.0° i + sin 35.0° j) = (20.5i + 14.3j) N F2 = (42.0 N)(cos 150° i + sin 150° j) = (–36.4 i + 21.0 j) N (b) ∑F = F1 + F2 = (–15.9i + 35.3j) N (c) a= (d) v = vi + at = (4.00i + 2.50j) m/s + (–3.18i + 7.07j)(m/s2)(3.00 s) ∑F = (–3.18i + 7.07j) m/s2 m v = (–5.54i + 23.7j) m/s © 2000 by Harcourt College Publishers. All rights reserved. (top view) x Chapter 7 Solutions (e) r = ri + vit + 25 1 2 at 2 1 r = 0 + (4.00i + 2.50j)(m/s)(3.00 s) + (–3.18i + 7.07j)(m/s2)(3.00 s) 2 2 d = r = (–2.30i + 39.3j) m (f) Kf = 1 1 2 mv f = (5.00 kg) [(5.54)2 + (23.7)2](m/s)2 = 1.48 kJ 2 2 (g) Kf = 1 1 2 mv i + ∑F ⋅ d = (5.00 kg) [(4.00)2 + (2.50)2](m/s)2 2 2 + [(–15.9 N)(–2.30 m) + (35.3 N)(39.3 m)] = 55.6 J + 1426 J = 1.48 kJ 7.68 (a) F (N) 25 20 15 10 5 L (mm) 0 0 (b) 50 100 150 200 F (N) 2.00 L (mm) 15.0 F (N) 14.0 L (mm) 112 4.00 32.0 16.0 126 6.00 49.0 18.0 149 8.00 64.0 20.0 175 10.0 79.0 22.0 190 12.0 98.0 A straight line fits the first eight points, and the origin. By least-square fitting, its slope is 0.125 N/mm ± 2% = 125 N/m ± 2%. In F = kx, the spring constant is k = F/x, the same as the slope of the F-versus-x graph. (c) F = kx = (125 N/m)(0.105 m) = 13.1 N © 2000 by Harcourt College Publishers. All rights reserved. 26 7.69 Chapter 7 Solutions (a) ∑W = ∆K Ws + Wg = 0 1 (1.40 × 103 N/m) × (0.100 m)2 – (0.200 kg)(9.80)(sin 60.0°)x = 0 2 x = 4.12 m (b) ∑W = ∆K Ws + Wg + Wf = 0 1 (1.40 × 103 N/m) × (0.100)2 – [(0.200)(9.80)(sin 60.0°) 2 + (0.200)(9.80)(0.400)(cos 60.0°)]x = 0 x = 3.35 m *7.70 1 1 2 2 m(v f – v i ) = (0.400 kg) [(6.00)2 – (8.00)2] (m/s)2 = –5.60 J 2 2 (a) W = ∆K = (b) W = fd cos 180° = –µkmg(2πr) –5.60 J = –µk(0.400 kg)(9.80 m/s2)2π(1.50 m) Thus, µk = 0.152 (c) After N revolutions, the mass comes to rest and Kf = 0. Thus, W = ∆K = 0 – Ki = – 1 2 mv i 2 or –µk mg [N(2πr)] = – 1 2 mv i 2 This gives N= 1 2 2 mv i µ k mg(2π r) = 1 2 (8.00 m/s)2 (0.152)(9.80 m/s2)2π(1.50 m) = 2.28 rev © 2000 by Harcourt College Publishers. All rights reserved. Chapter 7 Solutions 7.71 1 N 1.20 (5.00 cm)(0.0500 m) 2 cm = (0.100 kg)(9.80 m/s2)(0.0500 m) sin 10.0° + 1 (0.100 kg) v2 2 0.150 J = 8.51 × 10–3 J + (0.0500 kg)v2 v= 0.141 = 1.68 m/s 0.0500 10.0° 7.72 If positive F represents an outward force, (same direction as r), then f r i ri f 13 –13 7 –7 W=⌡ ⌠ F ⋅ ds = ⌠ ⌡ (2F 0σ r – F 0σ r ) dr W= +2F 0σ 13r–12 F0σ7r–6 rf – (–12) (–6) r i W= –F 0σ 13(rf – ri 6 –12 –12 –6 W = 1.03 × 10–77 [r f ) –6 + –6 F0σ7(r f – ri ) F0σ7 –6 F0σ13 –12 –6 –12 = [rf – ri ] – [rf – ri ] 6 6 6 –6 – r i ] – 1.89 × 10–134 [r –12 f –r –12 i ] W = 1.03 × 10–77 [1.88 × 10–6 – 2.44 × 10–6] 10+60 – 1.89 × 10–134 [3.54 × 10–12 – 5.96 × 10–8] 10120 W = –2.49 × 10–21 J + 1.12 × 10–21 J = – 1.37 × 10–21 J 7.73 (a) ∑W = ∆K m 2gh – µm1gh = v= = 1 2 (m + m 2)(v2 – v i ) 2 1 2gh(m 2 – µm 1 ) (m 1 + m 2 ) 2(9.80)(20.0)[0.400 – (0.200)(0.250)] = 14.5 m/s (0.400 + 0.250) © 2000 by Harcourt College Publishers. All rights reserved. 27 28 Chapter 7 Solutions (b) W f + W g = ∆K = 0 –µ(∆m 1 + m 1)gh + m 2gh = 0 µ(∆m1 + m1) = m2 ∆m1 = (c) m2 0.400 kg – m1 = – 0.250 kg = 1.75 kg µ 0.200 W f + W g = ∆K = 0 –µm 1gh + (m 2 – ∆m 2)gh = 0 ∆m2 = m2 – µm1 = 0.400 kg – (0.200)(0.250 kg) = 0.350 kg 7.74 P ∆t = W = ∆K = (∆m)v 2 2 v∆t A The density is v ∆m ∆m ρ= = vol A ∆x Substituting this into the first equation and solving for P, since ∆x =v ∆t for a constant speed, we get P= ρ Av 3 2 Also, since P = Fv, F= ρ Av 2 2 23.7 7.75 375dx ⌠ We evaluate ⌡ 12.8 x3 + 3.75x by calculating 375(0.100) 375(0.100) 375(0.100) + +... = 0.806 3 3 (12.8) + 3.75(12.8) (12.9) + 3.75(12.9) (23.6)3 + 3.75(23.6) and 375(0.100) 375(0.100) 375(0.100) + +... = 0.791 (12.9)3 + 3.75(12.9) (13.0)3 + 3.75(13.0) (23.7)3 + 3.75(23.7) The answer must be between these two values. We may find it more precisely by using a value for ∆x smaller than 0.100. Thus, we find the integral to be 0.799 N ⋅ m . © 2000 by Harcourt College Publishers. All rights reserved. Chapter 7 Solutions *7.76 (a) The suggested equation P t = bwd implies all of the following cases: (1) Pt=b (3) P w (2d) 2 t d = bw 2 2 and (2) P t w = b d 2 2 (4) P t = b w d 2 2 These are all of the proportionalities Aristotle lists. v = constant d n fk = µkn F w (b) For one example, consider a horizontal force F pushing an object of weight w at constant velocity across a horizontal floor with which the object has coefficient of friction µk. ∑F = ma implies that: +n – w = 0 and F – µkn = 0 so that F = µkw As the object moves a distance d , the agent exerting the force does work W = Fd cos θ = Fd cos 0° = µkwd and puts out power P = W/t This yields the equation P t = µkwd which represents Aristotle’s theory with b = µk. Our theory is more general than Aristotle’s. Ours can also describe accelerated motion. © 2000 by Harcourt College Publishers. All rights reserved. 29 Chapter 8 Even Answers 14. 16. 18. 20. (a) 80.0 J (b) 10.7 J (c) 0 (b) 35.0 J (a) 22.0 J, 40.0 J (b) Yes, ∆K + ∆U ≠ 0 (a) –9.00 J, No (conservative force) (b) 3.39 m/s (c) 9.00 J (a) 19.8 m/s (b) 294 J (c) (30.0i – 39.6j) m/s kx 2 d= –x 2mg sin θ 1.92 m/s (a) 0.537 m/s (b) 0.0588 m 1.84 m 914 N/m 22. (a) 24. 26. 28. 30. 32. 34. 36. 38. 40. 42. 44. 46. 48. 52. 54. 56. 58. 60. 40.8˚ (a) 14.0 m/s (b) 31.3 m/s (c) 24.2 m/s (d) 44.9 m 2.06 kN 26.5 m/s 3.68 m/s 168 J (a) 24.5 m/s (b) Yes (c) 206 m (d) unrealistic (a) 0.381 m (b) 0.143 m (c) 0.371 m 44.1 kW (7 – 9x2y)i – 3x3j See Instructor’s Manual (a) stable (b) neutral (c) unstable (a) 8.19 × 10–14 J (b) 3.60 × 10–8 J (c) 1.80 × 1014 J (d) 5.38 × 1041 J (a) 0.588 J (b) 0.588 J (c) 2.42 m/s (d) UC = 0.392 J, KC = 0.196 J 33.4 kW (44.8 hp) (a) 100 J (b) 0.410 J (c) 2.84 m/s (d) –9.80 mm (e) 2.85 m/s 0.115 (a) (3x2 – 4x – 3)i (b) x = 1.87 and –0.535 (c) x = –0.535 (stable), and x = 1.87 (unstable) (a) 0.378 m (b) 2.30 m/s (c) 1.08 m (b) 7.42 m/s h (4 sin2 θ + 1) 5 100.6˚ at h = 2H/3 or at h = R, whichever is smaller 3.92 kJ 2. 4. 6. 8. 10. 12. 62. 64. 66. 68. 72. 74. 2(m 1 – m 2 )g h 2m 1 h (b) m1 + m2 m1 + m2 © 2000 by Harcourt College Publishers. All rights reserved. 2 Chapter 8 Even Answers © 2000 by Harcourt College Publishers. All rights reserved. Chapter 8 Solutions *8.1 (a) With our choice for the zero level for potential energy at point B, UB = 0 . At point A, the potential energy is given by UA = mgy where y is the vertical height above zero level. With 135 ft = 41.1 m this height is found as: y = (41.1 m) sin 40.0° = 26.4 m Thus, UA = (1000 kg)(9.80 m/s2)(26.4 m) = The change in potential energy as it moves from A to B is UB – UA = 0 – 2.59 × 105 J = (b) With our choice of the zero level at point A, we have UA = 0 . The potential energy at B is given by UB = mgy where y is the vertical distance of point B below point A. In part (a), we found the magnitude of this distance to be 26.5 m. Because this distance is now below the zero reference level, it is a negative number. Thus, UB = (1000 kg)(9.80 m/s2)(–26.5 m) = The change in potential energy in going from A to B is UB – UA = –2.59 × 105 J – 0 = © 2000 by Harcourt College Publishers. All rights reserved. 2 Chapter 8 Solutions *8.2 (a) We take the zero level of potential energy at the lowest point of the arc. When the string is held horizontal initially, the initial position is 2.00 m above the zero level. Thus, Ug = mgy = (40.0 N)(2.00 m) = 80.0 J (b) From the sketch, we see that at an angle of 30.0° the ball is at a vertical height of (2.00 m)(1 – cos 30.0°) above the lowest point of the arc. Thus, Ug = mgy = (40.0 N)(2.00 m)(1 – cos 30.0°) = 10.7 J (c) The zero level has been selected at the lowest point of the arc. Therefore, U g = 0 this location. 8.3 Fg = mg = (4.00 kg)(9.80 m/s2) = 39.2 N (a) Work along OAC = work along OA + work along AC = Fg(OA) cos 90.0° + Fg(AC) cos 180° = (39.2 N)(5.00 m)(0) + (39.2 N)(5.00 m)(–1) = –196 J (b) W along OBC = W along OB + W along BC = (39.2 N)(5.00 m) cos 180° + (39.2 N)(5.00 m) cos 90.0° = –196 J (c) Work along OC = Fg(OC) cos 135° = (39.2 N)(5.00 × 2 m) – 1 = –196 J 2 The results should all be the same, since gravitational forces are conservative. 8.4 (a) W= and if the force is constant, this can be written as ⌠ ds = W=F⋅⌡ © 2000 by Harcourt College Publishers. All rights reserved. at Chapter 8 Solutions (b) 5.00 m ⌠ =⌠ ⌡ (3i + 4j) ⋅ (dx i + dy j) = (3.00 N) ⌡ W= 5.00 m W = (3.00 N) x 0 5.00 m + (4.00 N)y 0 0 5.00 m ⌠ dx + (4.00 N) ⌡ 0 = 15.0 J + 20.0 J = 35.0 J The same calculation applies for all paths. 8.5 (a) 5.00 m ⌠ WOA = ⌡ 0 5.00 m ⌠ dxi ⋅ (2yi + x2j) = ⌡ 0 2y dx and since along this path, y = 0 WOA = 0 5.00 m ⌠ WAC = ⌡ 0 5.00 m ⌠ dy j ⋅ (2yi + x2j) = ⌡ 0 x2 dy . For x = 5.00 m, WAC = 125 J and WOAC = 0 + 125 = 125 J (b) 5.00 m ⌠ ⋅ (2yi + x2j) = ⌡ WOB = 0 x2 dy since along this path, x = 0 WOB = 0 5.00 m ⌠ WBC = ⌡ 0 5.00 m ⌠ dxi ⋅ (2yi + x2j) = ⌡ 0 2y dx since y = 5.00 m, WBC = 50.0 J and WOBC = 0 + 50.0 = 50.0 J (c) ⌠ (dxi + dyj) ⋅ (2yi + x2j) = ⌡ ⌠ (2y dx + x 2 dy) WOC = ⌡ Since x = y along OC, 5.00 m ⌠ WOC = ⌡ 8.6 0 (2x + x2)dx = 66.7 J (d) F is non-conservative (a) Uf = Ki – Kf + Ui since the work done is path dependent. Uf = 30.0 – 18.0 + 10.0 = 22.0 J E = 40.0 J 8.7 (b) Yes, ∆E = ∆K + ∆U; for conservative forces ∆K + ∆U = 0. (a) ⌠ W = ∨ Fx dx = ⌡ 5.00 m 1 (2x + 4)dx = 5.00 m 2x2 + 4x 2 1 © 2000 by Harcourt College Publishers. All rights reserved. dy 3 4 Chapter 8 Solutions = 25.0 + 20.0 – 1.00 – 4.00 = 40.0 J © 2000 by Harcourt College Publishers. All rights reserved. Chapter 8 Solutions (b) ∆K + ∆U = 0 ∆U = –∆K = –W = – 40.0 J 2 (c) mv1 ∆K = K f – 2 2 mv1 Kf = ∆K + = 62.5 J 2 8.8 (a) F = (3.00i + 5.00j) N m = 4.00 kg r = (2.00i – 3.00j) m W = 3.00(2.00) + 5.00(–3.00) = –9.00 J The result does not depend on the path since the force is conservative. (b) W = ∆K –9.00 = so 8.9 v= 32.0 – 9.00 = 3.39 m/s 2.00 (c) ∆U = –W = 9.00 J (a) U=–⌠ ⌡ (–Ax + Bx2)dx = (b) ⌠ ∆U = – ⌡ x 0 3.00 m ∆K = 8.10 4.00v2 (4.00) 2 – 4.00 2 2 (a) F dx = 2.00 m Ax 2 Bx 3 – 2 3 A[(3.00)2 – (2.00)2] B[(3.00)3 – (2.00)3] 5.00 19.0 – = A– B 2 3 2 3 – 5.00 A + 19.0 B 3 2 Energy is conserved between point P and the apex of the trajectory. Since the horizontal component of velocity is constant, 1 1 1 1 2 2 2 2 mv i = mv i x + mv iy = mv ix 2 2 2 2 + mgh © 2000 by Harcourt College Publishers. All rights reserved. 5 6 Chapter 8 Solutions viy = = 19.8 m/s © 2000 by Harcourt College Publishers. All rights reserved. Chapter 8 Solutions (b) ∆K &P ∅ B = Wg = mg(60.0 m) = (0.500 kg)(9.80 m/s2)(60.0 m) = 294 J (c) Now let the final point be point B. vix = vfx = 30.0 m/s ∆K &P ∅ B = 2 v fy = 1 1 2 2 mv fy – mv iy = 294 J 2 2 2 2 (294) + viy = 1176 + 392 m vfy = –39.6 m/s vB = (30.0 m/s)i – (39.6 m/s)j 8.11 mgh = 1 kx2 2 (3.00 kg)(9.80 m/s2)(d + 0.200 m)sin 30.0° = 1 400(0.200 m)2 2 14.7d + 2.94 = 8.00 d = 0.344 m 8.12 Choose the zero point of gravitational potential energy at the level where the mass comes to rest. Then because the incline is frictionless, we have EB = EA ⇒ KB + UgB + UsB = KA + UgA + UsA or 0 + mg(d + x) sin θ + 0 = 0 + 0 + Solving for d gives d= 1 kx2 2 kx 2 –x 2mg sin θ © 2000 by Harcourt College Publishers. All rights reserved. 7 8 8.13 Chapter 8 Solutions (a) (∆K)A∅B = ΣW = W g = mg∆h = mg(5.00 – 3.20) 1 1 2 2 mvB – mvA = m(9.80)(1.80) 2 2 vB = 5.94 m/s Similarly, v C = (b) *8.14 2 v A + 2g(5.00 – 2.00) Wg&A ∅ C = mg(3.00 m) = 147 J Ki + Ui + ∆E = Kf + Uf 0 + m(9.80 m/s2)(2.00 m – 2.00 m cos 25.0°) = vf = 8.15 = 7.67 m/s (2)(9.80 m/s2)(0.187 m) 1 2 mv f + 0 2 = 1.92 m/s Ui + Ki = Uf + Kf mgh + 0 = mg(2R) + 1 mv2 2 g(3.50 R) = 2 g(R) + 1 2 v 2 v= 3.00 g R ·F = m v2 R n + mg = m v2 R n = m v2 3.00 g R – g = m – g R R n = 2.00 mg n = 2.00 (5.00 ∞ 10–3 kg)(9.80 m/s2) n = 0.0980 N downward © 2000 by Harcourt College Publishers. All rights reserved. Chapter 8 Solutions Goal Solution G: Since the bead is released above the top of the loop, it will have enough potential energy to reach point A and still have excess kinetic energy. The energy of the bead at the top will be proportional to h and g. If it is moving relatively slowly, the track will exert an upward force on the bead, but if it is whipping around fast, the normal force will push it toward the center of the loop. O: The speed at the top can be found from the conservation of energy, and the normal force can be found from Newton’s second law. A: We define the bottom of the loop as the zero level for the gravitational potential energy. Since vi = 0, Ei = Ki + Ui = 0 + mgh = mg(3.50R) The total energy of the bead at point A can be written as EA = KA + UA = 1 2 mvA + mg(2R) 2 Since mechanical energy is conserved, Ei = EA, and we get 1 2 mvA + mg(2R) = mg(3.50R) 2 2 v A = 3.00gR or vA = 3.00gR © 2000 by Harcourt College Publishers. All rights reserved. 9 Chapter 8 Solutions 10 To find the normal force at the top, we may construct a free-body diagram as shown, where we assume that n is downward, like mg. Newton's second law gives F = mac, where ac is the centripetal acceleration. 2 mvA m(3.00gR) n + mg = = = 3.00mg R R n = 3.00mg – mg = 2.00mg n = 2.00(5.00 ∞ 10–3 kg)(9.80 m/s2) = 0.0980 N downward L: Our answer represents the speed at point A as proportional to the square root of the product of g and R, but we must not think that simply increasing the diameter of the loop will increase the speed of the bead at the top. In general, the speed will increase with increasing release height, which for this problem was defined in terms of the radius. The normal force may seem small, but it is twice the weight of the bead. 8.16 (a) At the equilibrium position for the mass, the tension in the spring equals the weight of the mass. Thus, elongation of the spring when the mass is at equilibrium is: kxo = mg ⇒ xo = mg (0.120)(9.80) = = 0.0294 m k 40.0 The mass moves with maximum speed as it passes through the equilibrium position. Use energy conservation, taking Ug = 0 at the initial position of the mass, to find this speed: Kf + Ugf + Usf = Ki + Ugi + Usi 1 1 2 2 mvmax + mg(–xo) + kx0 = 0 + 0 + 0 2 2 2 vmax = kx0 2gx 0 – = m 2(9.80)(0.0294) – (40.0)(0.0294)2 = 0.537 m/s 0.120 © 2000 by Harcourt College Publishers. All rights reserved. Chapter 8 Solutions (b) When the mass comes to rest, Kf = 0. Therefore, Kf + Ugf + Usf = Ki + Ugi + Usi becomes 0 + mg(–x) + x= 8.17 1 kx2 = 0 + 0 + 0 which becomes 2 2mg = 2x0 = 0.0588 m k From conservation of energy, Ugf = Usi, or (0.250 kg)(9.80 m/s2)h = (1/2)(5000 N/m)(0.100 m)2 This gives a maximum height h = 10.2 m *8.18 From leaving ground to highest point Ki + Ui = Kf + Uf 1 m(6.00 m/s)2 + 0 = 0 + m(9.80 m/s2)y 2 The mass makes no difference: *8.19 ∴y= (6.00 m/s)2 = 1.84 m (2)(9.80 m/s2) (a) 1 mv2 = mgh 2 v= 2gh = 19.8 m/s (b) E = mgh = 78.4 J (c) K10 + U10 = 78.4 J K10 = 39.2 J U10 = 39.2 J K10 = 1.00 U 10 © 2000 by Harcourt College Publishers. All rights reserved. 11 12 8.20 Chapter 8 Solutions Choose y = 0 at the river. Then yi = 36.0 m, yf = 4.00 m, the jumper falls 32.0 m, and the cord stretches 7.00 m. Between balloon and bottom, Ki + Ugi + Usi = Kf + Ugf + Usf 1 2 0 + mgyi + 0 = 0 + mgyf + 2 kx f (700 N)(36.0 m) = (700 N)(4.00 m) + k= 8.21 1 k(7.00 m)2 2 22400 J = 914 N/m 24.5 m2 Using conservation of energy (a) (5.00 kg)g(4.00 m) = (3.00 kg)g(4.00 m) + v= (b) 1 (5.00 + 3.00) v2 2 19.6 = 4.43 m/s 1 (3.00) v2 = mg ∆y = 3.00g ∆y 2 ∆y = 1.00 m ymax = 4.00 m + ∆y = 5.00 m 8.22 m1 > m2 (a) (b) m 1gh = 1 (m + m2) v2 + m2gh 2 1 v= 2(m 1 – m 2 )g h (m 1 + m 2 ) Since m2 has kinetic energy 1 m 2v 2, it will rise an additional height ∆h determined 2 from m 2g ∆h = 1 m 2v 2 2 or from (a), ∆h = v2 2g = (m 1 – m 2)h (m 1 + m 2 ) The total height m 2 reaches is h + ∆h = 2m 1 h m1 + m2 © 2000 by Harcourt College Publishers. All rights reserved. Chapter 8 Solutions © 2000 by Harcourt College Publishers. All rights reserved. 13 14 8.23 Chapter 8 Solutions (a) Ki + Ugi = Kf + Ugf 1 1 2 2 mv i + 0 = mv f + mgyf 2 2 1 1 1 2 2 2 mv xi + mv yi = mv xf + mgyf 2 2 2 But vxi = vxf, so for the first ball 2 vyi yf = 2g = (1000 sin 37.0°)2 = 2(9.80) and for the second yf = (b) (1000)2 = 2(9.80) The total energy of each is constant with value 1 (20.0 kg)(1000 m/s) 2 8.24 2 = In the swing down to the breaking point, energy is conserved: mgr cos θ = 1 mv2 2 at the breaking point consider radial forces · Fr = mar + Tmax – mg cos θ = m Eliminate v2 r v2 = 2 g cos θ r Tmax – mg cos θ = 2 mg cos θ Tmax = 3 mg cos θ θ = Arc cos Tmax 44.5 N = Arc cos 3(2.00 kg)(9.80 m/s2) 3 mg θ = 40.8° © 2000 by Harcourt College Publishers. All rights reserved. Chapter 8 Solutions *8.25 (a) 15 The force needed to hang on is equal to the force F the trapeze bar exerts on the performer. From the free-body diagram for the performer’s body, as shown, v2 F – mg cos θ = m l or v2 F = mg cos θ + m l Apply conservation of mechanical energy between the starting point and any later point: 1 mg(l – l cos θi) = mg(l – l cos θ) + mv2 2 Solve for mv2/l and substitute into the force equation to obtain F = mg(3 cos θ – 2 cos θi) (b) At the bottom of the swing, θ = 0° so F = mg(3 – 2 cos θi). F = 2mg = mg(3 – 2 cos θi), which gives θi = 60.0° 2 *8.26 (a) At point 3, ∑Fy = may gives n + mg = m v3 . R For apparent weightlessness, n = 0. This gives v3 = (b) Rg = (20.0)(9.80) = 14.0 m/s Now, from conservation of energy applied between points 1 and 3, 1 1 2 2 mv1 + mgy1 = mv3 + mgy3 2 2 so v1 = 2 v 3 + 2g(y 3 – y 1) = (14.0)2 + 2(9.80)(40.0) = 31.3 m/s © 2000 by Harcourt College Publishers. All rights reserved. 16 Chapter 8 Solutions (c) The total energy is the same at points 1 and 2: 1 1 2 2 mv1 + mgy1 = mv2 + mgy2, which yields 2 2 2 v2 = (d) v 1 + 2g(y 1 – y 2) = (31.3)2 + 2(9.80)(–20.0) = 24.2 m/s Between points 1 and 4: 1 1 2 2 mv1 + mgy1 = mv4 + mgy4, giving 2 2 2 2 v1 – v 4 (31.3)2 – (10.0)2 H = y4 – y1 = = 2g 2(9.80) = 44.9 m *8.27 The force of tension and subsequent force of compression in the rod do no work on the ball, since they are perpendicular to each step of displacement. Consider energy conservation between the instant just after you strike the ball and the instant when it reaches the top. The speed at the top is zero if you hit it just hard enough to get it there. Ki + Ugi = Kf + Ugf 1 2 mv i + 0 = 0 + mg(2L) 2 vi = *8.28 4gL = 4(9.80)(0.770) = 5.49 m/s We shall take the zero level of gravitational potential energy to be at the lowest level reached by the diver under the water, and consider the energy change from when the diver started to fall until he came to rest. ∆E = 1 1 2 2 mv f – mv i + mgyf – mgyi = fk d cos 180− 2 2 0 – 0 – mg(yi – yf) = –fk d fk = mg(y i – y f) (70.0 kg)(9.80 m/s2)(10.0 m + 5.00 m) = = 2.06 kN d 5.00 m © 2000 by Harcourt College Publishers. All rights reserved. Chapter 8 Solutions *8.29 17 x 1 ⌠ Fx dx = area under the Fx vs x curve. mv2 = ⌡ 2 0 2.00 ⌠ ⌡0 for x = 2.00 m 2(10.0) 5.00 ∴ v&x = 2.00 m = Fx dx = 10.0 N · m = 2.00 m/s Similarly, *8.30 v&x = 4.00 m = 2(19.5) = 2.79 m/s 5.00 v&x = 6.00 m = 2(25.5) = 3.19 m/s 5.00 The distance traveled by the ball from the top of the arc to the bottom is s = πr. The work done by the non-conservative force, the force exerted by the pitcher, is ∆E = Fs cos 0° = F(πR). We shall choose the gravitational potential energy to be zero at the bottom of the arc. Then ∆E = 1 1 2 2 mv f – mv i + mgyf – mgyi becomes 2 2 1 1 2 2 mv f = mv i + mgyi + F(πR) 2 2 or vf = 2 v i + 2gy i + 2F(π R) = m (15.0)2 + 2(9.80)(1.20) + 2(30.)π (0.600) 0.250 vf = 26.5 m/s 8.31 Ui + Ki + ∆E = Uf + Kf m 2gh – fh = 1 1 m v2 + m 2v 2 2 1 2 f = µn = µm1 g m 2gh – µm 1gh = v2 = v= 1 (m + m2) v2 2 1 2(m 2 – µm 1 )(hg) m1 + m2 2(9.80 m/s2)(1.50 m)[5.00 kg – 0.400(3.00 kg)] = 3.74 m/s 8.00 kg © 2000 by Harcourt College Publishers. All rights reserved. 18 Chapter 8 Solutions Goal Solution G: Assuming that the block does not reach the pulley within the 1.50 m distance, a reasonable speed for the ball might be somewhere between 1 and 10 m/s based on common experience. O: We could solve this problem by using ΣF = ma to give a pair of simultaneous equations in the unknown acceleration and tension; then we would have to solve a motion problem to find the final speed. We may find it easier to solve using the work-energy theorem. A: For objects A (block) and B (ball), the work-energy theorem is (KA + KB + UA + UB)i + Wapp – fkd = (KA + KB + UA + UB)f Choose the initial point before release and the final point after each block has moved 1.50 m. For the 3.00-kg block, choose U g = 0 at the tabletop. For the 5.00-kg ball, take the zero level of gravitational energy at the final position. So KAi = KBi = UAi = UAf = UBf = 0. Also, since the only external forces are gravity and friction, W app = 0. We now have 0 + 0 + 0 + mBgyBi – f1d = 1 1 2 2 mAv f + mBv f + 0 + 0 2 2 where the frictional force is f1 = µ1n = µ1mAg and does negative work since the force opposes the motion. Since all of the variables are known except for vf, we can substitute and solve for the final speed. (5.00 kg)(9.80 m/s2)(1.50 m) – (0.400)(3.00 kg)(9.80 m/s2)(1.50 m) 1 1 2 2 = (3.00 kg) v f + (5.00 kg) v f 2 2 1 2 73.5 J – 17.6 J = (8.00 kg) v f 2 or vf = 2(55.9 J) = 3.74 m/s 8.00 kg L: The final speed seems reasonable based on our expectation. This speed must also be less than if the rope were cut and the ball simply fell, in which case its final speed would be v f' = 2gy = 2(9.80 m/s2)(1.50 m) = 5.42 m/s © 2000 by Harcourt College Publishers. All rights reserved. Chapter 8 Solutions *8.32 The initial vertical height of the car above the zero reference level at the base of the hill is h = (5.00 m) sin 20.0° = 1.71 m The energy lost through friction is ∆E = –fs = –(4000 N)(5.00 m) = –2.00 ∞ 104 J We now use, ∆E = 1 1 2 2 mv f – mv i 2 2 + mgyf – mgyi 1 –2.00 × 104 J = (2000 kg) v2 – 0 + 0 – (2000 kg)g(1.71 m) 2 and v = 3.68 m/s 8.33 1 1 2 2 m(v2 – v i ) = – mv i = –160 J 2 2 (a) ∆K = (b) ∆U = mg(3.00 m) sin 30.0° = 73.5 J (c) The energy lost to friction is 86.5 J f= (d) f = µkn = µKmg cos 30.0° = 28.8 N µ= *8.34 86.5 J = 28.8 N 3.00 m 28.8 N = 0.679 (5.00 kg)(9.80 m/s2) cos 30.0° ∆E = (Kf – Ki) + (Ugf – Ugi) But ∆E = Wapp + fs cos 180° where Wapp is the work the boy did pushing forward on the wheels. Thus, Wapp = (Kf – Ki) + (Ugf – Ugi) + fs, or Wapp = 1 2 2 m(v f – v i ) + mg(–h) + fs 2 Wapp = 1 (47.0) [(6.20)2 – (1.40)2] – (47.0)(9.80)(2.60) + (41.0)(12.4) 2 Wapp = 168 J © 2000 by Harcourt College Publishers. All rights reserved. 19 20 8.35 Chapter 8 Solutions 1 2 mv f 2 ∆E = mghi – = (50.0)(9.80)(1000) – 1 (50.0)(5.00) 2 2 ∆E = 489 kJ 8.36 Consider the whole motion: Ki + Ui + ∆E = Kf + Uf (a) 0 + mgyi + f1d1 cos 180° + f2d2 cos 180° = 1 2 mv f + 0 2 (80.0 kg)(9.80 m/s2)1000 m – (50.0 N)(800 m) – (3600 N)(200 m) = 1 2 (80.0 kg) v f 2 1 2 784,000 J – 40,000 J – 720,000 J = (80.0 kg) v f 2 vf = (b) (c) Yes 2(24,000 J) = 24.5 m/s 80.0 kg , this is too fast for safety. Now in the same work-energy equation d2 is unknown, and d1 = 1000 m – d2: 784,000 J – (50.0 N)(1000 m – d2) – (3600 N)d2 = 1 (80.0 kg)(5.00 m/s) 2 2 784,000 J – 50,000 J – (3550 N)d2 = 1000 J d2 = *8.37 733,000 J = 206 m 3550 N (d) Really the air drag will depend on the skydiver's speed. It will be larger than her 784 N weight only after the chute is opened. It will be nearly equal to 784 N before she opens the chute and again before she touches down, whenever she moves near terminal speed. (a) (K + U)i + ∆E = (K + U)f 0+ 1 1 kx2 – fd = mv2 + 0 2 2 2 v= – (3.20 × 10–2 N)(0.150 m) = = 1.40 m/s © 2000 by Harcourt College Publishers. All rights reserved. v2 Chapter 8 Solutions (b) When the spring force just equals the friction force, the ball will stop speeding up. Here Fs = kx, the spring is compressed by = 0.400 cm and the ball has moved 5.00 cm – 0.400 cm = 4.60 cm from the start. (c) 21 Between start and maximum speed points, 1 1 1 2 2 kx i – fx = mv2 + kx f 2 2 2 1 2 8.00(5.00 × 10–2)2 – (3.20 × 10–2)(4.60 × 10–2) v2 + = 1 2 8.00(4.00 × 10–3)2 v = 1.79 m/s © 2000 by Harcourt College Publishers. All rights reserved. 8.38 (a) The mass moves down distance 1.20 m + x. Choose y = 0 at its lower point. Ki + Ugi + Usi + ∆E = Kf + Ugf + Usf 0 + mgyi + 0 + 0 = 0 + 0 + 1 2 kx 2 (1.50 kg)9.80 m/s2 (1.20 m + x) = 1 (320 N/m) x2 2 0 = (160 N/m)x2 – (14.7 N)x – 17.6 J x= 14.7 N ± (–14.7 N)2 – 4(160 N/m)(–17.6 N · m) 2(160 N/m) x= 14.7 N ± 107 N 320 N/m The negative root tells how high the mass will rebound if it is instantly glued to the spring. We want x = 0.381 m (b) From the same equation, (1.50 kg)1.63 m/s2 (1.20 m + x) = 1 (320 N/m) x2 2 0 = 160x2 – 2.44x – 2.93 The positive root is x = 0.143 m (c) The full work-energy theorem has one more term: mgyi + fyi cos 180° = 1 2 kx 2 (1.50 kg) 9.80 m/s2 (1.20 m + x) – 0.700 N(1.20 m + x) = 1 (320 N/m) x2 2 17.6 J + 14.7 N x – 0.840 J – 0.700 N x = 160 N/m x2 160x2 – 14.0x – 16.8 = 0 x= 14.0 ± (14.0)2 – 4(160)(–16.8) 320 x = 0.371 m 8.39 Choose Ug = 0 at the level of the horizontal surface. Then ∆E = (Kf – Ki) + (Ugf – Ugi) becomes: m = 3.00 kg –f1s – f2x = (0 – 0) + (0 – mgh) or h √ – (µ mg)x = –(µkmg cos 30.0°) sin 30.0ϒ↵ k h = 60.0 cm θ = 30.0° –mgh Thus, the distance the block slides across the horizontal surface before stopping is: x= or h – h cot 30.0° = h µk x = 1.96 m 1 k 1 cot 30.0ϒ√ = (0.600 m) cot 30.0ϒ√ √ 0 . 200 ↵ ↵ *8.40 The total mechanical energy of the diver is Emech = K + Ug = 1 mv2 + mgh. Since the diver 2 has constant speed, dEmech dv dh = mv + mg = 0 + mg(–v) = –mgv dt dt dt The rate he is losing mechanical energy is then dEmech = mgv = (75.0 kg)(9.80 m/s2)(60.0 m/s) = 44.1 kW dt 8.41 U(r) = Fr = – 8.42 Fx = – A r ∂U d A A =– = 2 ∂r dr r r ƒU ƒ(3x 3 y − 7 x =– = –(9x2y – 7) = 7 – 9x2y ƒx ƒx ƒ(3x 3 y − 7 x ƒU Fy = – =– = –(3x3 – 0) = –3x3 ƒy ƒy Thus, the force acting at the point (x, y) is F = Fx i + Fy j = (7 – 9x2y)i – 3x3 j *8.43 (a) There is an equilibrium point wherever the graph of potential energy is horizontal: At r = 1.5 mm and 3.2 mm, the equilibrium is stable. At r = 2.3 mm, the equilibrium is unstable. A particle moving out toward r → ∞ approaches neutral equilibrium. (b) The particle energy cannot be less than –5.6 J. The particle is bound if –5.6 J ≤ E < 1 J . (c) If the particle energy is –3 J, its potential energy must be less than or equal to –3 J. Thus, its position is limited to 0.6 mm ≤ r ≤ 3.6 mm . (d) K + U = E. Thus, Kmax = E – Umin = –3.0 J – (–5.6 J) = 2.6 J (e) Kinetic energy is a maximum when the potential energy is a minimum, at r = 1.5 mm . (f) –3 J + W = 1 J. Hence, the binding energy is W = 4 J . *8.44 stable 8.45 neutral unstable (a) Fx is zero at points A, C and E; Fx is positive at point B and negative at point D. (b) A and E are unstable, and C is stable. (c) Fx B E C A x (m) D 8.46 (a) As the pipe is rotated, the CM rises, so this is stable equilibrium. (b) As the pipe is rotated, the CM moves horizontally, so this is neutral equilibrium. (c) As the pipe is rotated, the CM falls, so this is unstable equilibrium. CM O CM a 8.47 (a) O CM b O c When the mass moves distance x, the length of each spring changes from L to x2 + L2 , so each exerts force k( x2 + L2 – L) toward its fixed end. The ycomponents cancel out and the x-components add to: x 2kLx = –2kx + x2 + L2 x2 + L2 Fx = –2k( x2 + L2 – L) Choose U = 0 at x = 0. Then at any point ⌡ ⌡ Fxdx = – ⌠ U(x) = – ⌠ 2kLx ⌡xxdx – 2kL ⌠ ⌡x x dx –2kx + dx = 2k ⌠ 0 0 0 x2 + L2 x2 + L2 U(x) = kx2 + 2kL(L – x2 + L2) x x 0 (b) U(x) = 40.0x2 + 96.0(1.20 – x2 + 1.44 ) U(x)(J) 6.00 5.00 4.00 3.00 x, m U, J 1 0.8 0.6 0 0.2 0.4 −0.4 0 0 −0.2 −0.6 −1 −0.8 2.00 1.00 x(m) 0.200 0.400 0.600 0.800 1.00 1.50 0.011 0.168 0.802 2.35 5.24 20.8 2.00 2.50 51.3 99.0 For negative x, U(x) has the same value as for positive x. The only equilibrium point (i.e., where Fx = 0) is x = 0 . (c) Ki + Ui + ∆E = Kf + Uf 0 + 0.400 J + 0 = vf = 8.48 8.49 1 2 mv f + 0 2 0.800 J m (a) E = mc2 = (9.11 χ 10–31 kg)(2.998 × 108 m/s)2 = 8.19 × 10-14J (b) 3.60 × 10-8J (c) 1.80 × 1014J (d) 5.38 × 1041J (a) Rest energy = mc2 = (1.673 ∞ 10–27 kg)(2.998 ∞ 108 m/s)2 = 1.50 × 10-10J (b) E = γmc2 = (c) K = γmc2 – mc2 = 1.07 × 10–9 J – 1.50 × 10–10 J = 9.15 × 10-10J mc2 1 – (v/c)2 = 1.50 × 10–10 J 1 – (.990)2 = 1.07 × 10-9J 8.50 The potential energy of the block is mgh. An amount of energy µkmgs cos θ is lost to friction on the incline. Therefore the final height ymax is found from mgymax = mgh – µkmgs cos θ where s= ymax sin θ ∴ mgymax = mgh – µkmgymax cot θ h Solving, ymax h ymax = 1 + µk cot θ *8.51 θ m = mass of pumpkin R = radius of silo top vi ≈ 0 n θt initially ∑Fr = mar ⇒ n – mg cos θ = –m v R mg later v2 R When the pumpkin is ready to lose contact with the surface, n = 0. Thus, at the point where it leaves the surface: v2 = Rg cos θ. Choose Ug = 0 in the θ = 90.0°plane. Then applying conservation of energy from the starting point to the point where the pumpkin leaves the surface gives Kf + Ugf = Ki + Ugi 1 mv2 + mgR cos θ = 0 + mgR 2 Using the result from the force analysis, this becomes 1 mRg cos θ + mgR cos θ = mgR, which reduces to 2 cos θ = 2 , and gives θ = cos–1 (2/3) = 3 48.2ϒ as the angle at which the pumpkin will lose contact with the surface. 8.52 (a) UA = mgR = (0.200 kg)(9.80 m/s2)(0.300 m) = 0.588 J (b) KA + UA = KB + UB A KB = KA + UA – UB = mgR= 0.588 J 2KB = m C R B 2R/3 2(0.588 J) 0.200 kg (c) vB = = 2.42 m/s (d) UC = mghC = (0.200 kg)(9.80 m/s2)(0.200 m) = 0.392 J KC = KA + UA – UC = mg(hA – hC) KC = (0.200 kg)(9.80 m/s2)(0.300 – 0.200) m = 0.196 J 8.53 1 1 2 mvB = (0.200 kg)(1.50 m/s) 2 = 0.225 J 2 2 (a) KB = (b) ∆E = ∆K + ∆U = KB – KA + UB – UA = KB + mg(hB – hA) = 0.225 J + (0.200 kg)(9.80 m/s2)(0 – 0.300 m) = 0.225 J – 0.588 J = –0.363 J (c) *8.54 It's possible to find an effective coefficient of friction, but not the actual value of µ since n and f vary with position. v = 100 km/h = 27.8 m/s The retarding force due to air resistance is R= 1 1 DρAv2 = (0.330)(1.20 kg/m3)(2.50 m2)(27.8 m/s) 2 = 382 N 2 2 Comparing the energy of the car at two points along the hill, Ki + Ugi + ∆E = Kf + Ugf or Ki + Ugi + ∆We – R(∆s) = Kf + Ugf where ∆We is the work input from the engine. Thus, ∆We = R(∆s) + (Kf – Ki) + (Ugf – Ugi) Recognizing that Kf = Ki and dividing by the travel time ∆t gives the required power input from the engine as ∆We ∆s ∆y = R + mg = Rv + mgv sin θ P= ∆t ∆t ∆t P = (382 N)(27.8 m/s) + (1500 kg)(9.80 m/s2)(27.8 m/s)sin 3.20° P = 33.4 kW = 44.8 hp *8.55 At a pace I could keep up for a half-hour exercise period, I climb two stories up, forty steps each 18 cm high, in 20 s. My output work becomes my final gravitational energy, mgy = 85 kg(9.80 m/s2)(40 × 0.18 m) = 6000 J making my sustainable power 6000 J = ~102 W 20 s 8.56 k = 2.50 × 104 N/m (a) m = 25.0 kg E = KA + UgA + UsA = 0 + mgxA + xA = –0.100 m Ugx = 0 = Usx = 0 = 0 1 2 kx 2 A E = (25.0 kg)(9.80 m/s2)(–0.100 m) + 1 (2.50 ↔10 4 Ν/µ )(0.100µ ) 2 2 E = –24.5 J + 125 J = 100 J (b) Since only conservative forces are involved, the total energy at point C is the same as that at point A. KC + UgC + UsC = KA + UgA + UsA 0 + (25.0 kg)(9.80 m/s2)xC + 0 = 0 + –24.5 J + 125 J ⇒ xC = 0.410 J (c) KB + UgB + UsB = KA + UgA + UsA 1 2 (25.0 kg) vB + 0 + 0 = 0 + –24.5 J + 125 J ⇒ vB = 2.84 m/s 2 (d) K and v are at a maximum when a = ΣF = 0 (i.e., when the magnitude of the m upward spring force equals the magnitude of the downward gravitational force). This occurs at x < 0 where k x = mg or x = (25.0kg )(9.80m / s 2 ) = 9.80 × 10–3 m 2.50 ↔10 4 Ν/m Thus, K = Kmax at x = –9.80 mm (e) Kmax = KA + (UgA – Ugx = –9.80 mm) + (UsA – Usx = –9.80 mm), or 1 2 (25.0 kg) vmax = (25.0 kg)(9.80 m/s2)[(–0.100 m) – (–0.0098 m)] 2 + 1 (2.50 × 104 N/m) [(–0.100 m)2 – (–0.0098 m)2] 2 yielding vmax = 2.85 m/s 8.57 ∆E = Wf Ef – Ei = – f · dBC 1 k ∆x2 – mgh = – µmgd 2 1 µ= mgh – 2 k · ∆x2 mgd = 0.328 A 3.00 m 6.00 m B C Goal Solution G: We should expect the coefficient of friction to be somewhere between 0 and 1 since this is the range of typical µk values. It is possible that µk could be greater than 1, but it can never be less than 0. O: The easiest way to solve this problem is by considering the energy conversion for each part of the motion: gravitational potential to kinetic energy from A to B, loss of kinetic energy due to friction from B to C, and kinetic to elastic potential energy as the block compresses the spring. Choose the gravitational energy to be zero along the flat portion of the track. A: Putting the energy equation into symbols: UgA – W expanding into specific variables: mgyA – f1dBC = = Usf 1 2 kx s where f1 = µ1mg 2 solving for the unknown variable: µ1mgd = mgy – substituting: µ1 = BC 1 2 kx 2 or µ1 = y kx2 – d 2mgd 3.00 m (2250 N/m)(0.300 m)2 – = 0.328 6.00 m 2(10.0 kg)(9.80 m/s2)(6.00 m) L: Our calculated value seems reasonable based on the friction data in Table 5.2. The most important aspect to solving these energy problems is considering how the energy is transferred from the initial to final energy states and remembering to subtract the energy resulting from any non-conservative forces (like friction). 8.58 The nonconservative work (due to friction) must equal the change in the kinetic energy plus the change in the potential energy. Therefore, –µkmgx cos θ = ∆K + 1 2 kx – mgx sin θ 2 and since vi = vf = 0, ∆K = 0. Thus, –µk(2.00)(9.80)(cos 37.0°)(0.200) = (100)(0.200)2 – (2.00)(9.80)(sin 37.0°)(0.200) 2 and we find µk = 0.115 . Note that in the above we had a gain in elastic potential energy for the spring and a loss in gravitational potential energy. The net loss in mechanical energy is equal to the energy lost due to friction. 8.59 (a) Since no nonconservative work is done, ∆E = 0 Also ∆K = 0 therefore, Ui = Uf where Ui = (mg sin θ)x and Uf = 1 2 kx 2 Substituting values yields (2.00)(9.80) sin 37.0° = (100) x and solving we find 2 x = 0.236 m (b) ∑F = ma. Only gravity and the spring force act on the block, so –kx + mg sin θ = ma For x = 0.236 m, a = –5.90 m/s2 The negative sign indicates a is up the incline. The acceleration depends on position . (c) U(gravity) decreases monotonically as the height decreases. U(spring) increases monotonically as the spring is stretched. K initially increases, but then goes back to zero. F=– d (–x3 + 2x2 + 3x)i = (3x2 – 4x – 3)i dx (b) F = 0 when x = 1.87 and –0.535 (c) The stable point is at x = –0.535 point of minimum U(x) 5 F(x) 2 The unstable point is at x = 1.87 maximum in U(x) 1 (a) −1 *8.60 U(x) −5 8.61 (K + U)i = (K + U)f 0 + (30.0 kg)(9.80 m/s2)(0.200 m) + = 1 (250 N/m)(0.200 m) 2 2 1 (50.0 kg) v2 + (20.0 kg)(9.80 m/s2)(0.200 m) sin 40.0° 2 58.8 J + 5.00 J = (25.0 kg)v2 + 25.2 J v = 1.24 m/s 30.0 kg 20.0 kg 20.0 cm 40° 8.62 (a) Between the second and the third picture, ∆E = ∆K + ∆U – µmgd = – k m 1 1 2 mv i + kd2 2 2 vi 1 (50.0 N/m) d2 + 0.250(1.00 kg)(9.80 m/s2)d 2 1 – (1.00 kg)(3.00 m/s2) = 0 2 d= (b) ∆U d [–2.45 ± 21.35] N = 0.378 m 50.0 N/m v Between picture two and picture four, ∆E = ∆K + – f(2d) = – 1 1 2 mv2 + mv i 2 2 v= 0 D v= (3.00 m/s)2 – = 2.30 m/s 2 (2.45 N)(2)(0.378 m) (1.00 kg) vf = 0 (c) For the motion from picture two to picture five, ∆E = ∆K + ∆U 1 (1.00 kg)(3.00 m/s) 2 2 –f(D + 2d) = – D= 8.63 9.00 J – 2(0.378 m) = 1.08 m 2(0.250)(1.00 kg)(9.80 m/s2) (a) T vT R vB ∆x k m B Initial compression of spring: 1 2 1 kx = mv2 2 2 1 1 (450 N/m)(∆x) 2 = (0.500 kg)(12.0 m/s) 2 2 2 ∴ ∆x = 0.400 m (b) Speed of block at top of track: ∆E = Wf 1 1 2 2 mghT + 2 mvT – mghB + 2 mvB = – f(πR) 1 1 2 (0.500 kg)(9.80 m/s2)(2.00 m) + (0.500 kg) vT – (0.500 kg)(12.0 m/s) 2 2 2 = – (7.00 N)(π)(1.00 m) 2 0.250vT = 4.21 (c) ∴ vT = 4.10 m/s Does block fall off at or before top of track? Block falls if ar < g 2 ar = vT (4.10)2 = = 16.8 m/s2 R 1.00 therefore ar > g and the block stays on the track . 8.64 Let λ represent the mass of each one meter of the chain and T represent the tension in the chain at the table edge. We imagine the edge to act like a frictionless pulley. (a) 3λg n f T P T 5λg T T For the five meters on the table with motion impending, ∑Fy = 0 + n – 5λg = 0 n = 5λ g fs ≤ µs n = 0.6(5λg) = 3λg ∑Fx = 0 + T – fs = 0 T = fs T≤ 3λg The maximum value is barely enough to support the hanging segment according to + T – 3λ g = 0 ∑Fy = 0 T = 3λg so it is at this point that the chain starts to slide. (b) Let x represent the variable distance the chain has slipped since the start. Then length (5 – x) remains on the table, with now + n – (5 – x)λg = 0 ∑Fy = 0 n = (5 – x)λg fk = µk n = 0.4(5 – x)λg = 2λg – 0.4xλg Consider energies at the initial moment when the chain starts to slip, and a final moment when x = 5, when the last link goes over the brink. Measure heights above the final position of the leading end of the chain. At the moment the final link slips off, the center of the chain is at yf = 4 meters. Originally, 5 meters of chain is at height 8 m and the middle of the dangling 3 = 6.5 m. segment is at height 8 – 2 Ki + Ui + ∆E = Kf + Uf 1 ⌠f 0 + (m1gy1 + m2gy2)i + ⌡ fk dx cos θ = mv2 + mgy 2 f i ⌡ (2λg – 0.4xλg) dx cos 180° (5λg)8 + (3λg)6.5 + ⌠ 5 0 1 = 2 (8λ) v2 + (8λg)4 ⌡ 40.0 g + 19.5 g – 2.00 g ⌠ 5 0 ⌡ dx + 0.400 g ⌠ 5 0 x dx = 4.00v2 + 32.0 g 5 x 2 5 27.5 g – 2.00 gx 0 + 0.400 g 0 = 4.00v2 2 27.5 g – 2.00 g(5.00) + 0.400 g(12.5) = 4.00v2 22.5 g = 4.00v2 v= 8.65 (a) (22.5 m)(9.80 m/s2) 4.00 = 7.42 m/s On the upward swing of the mass: vi Ki + Ui + ∆E = Kf + Uf L 1 2 mv i + 0 + 0 = 0 + mgL(1 – cos 2 θ) θ m vi = 2gL(1 – cos θ) (b) (a) (b) vi = 2(9.80 m/s2)(1.20 m)(1 – cos 35.0°) vi = 2.06 m/s 8.66 Launch speed is found from 4 1 mg h = mv2 5 2 h 4 2g h 5 v= vy = v sin θ The height y above the water (by conservation of energy) is found from mgy = h 1 1 2 2 mvy + mg since mvx is constant in projectile motion 5 2 2 y= 1 2 h 1 2 h v + = v sin2 θ + 5 2g y 5 2g y= 1 4 h 4 h 2g h sin2 θ + = h sin2 θ + 2g 5 5 5 5 θ y h/5 8.67 (a) Take the original point where the ball is released and the final point where its upward swing stops at height H and horizontal displacement x = L2 – (L – H)2 = Pivot Pivot F F L L 2LH – H2 m H m Since the wind force is purely horizontal, it does work (b) (a) Wwind = ∨ F ⋅ ds = F ∨ dx = F 2LH – H2 [The wind force potential energy change would be –F 2LH – H2 ] The work-energy theorem can be written: Ki + Ugi + Wwind = Kf + Ugf, or 0 + 0 + F 2LH – H2 = 0 + mg H giving F22LH – F2H2 = m2g2H2 Here H = 0 represents the lower turning point of the ball's oscillation, and the upper limit is at F2(2L) = (F2 + m2g2)H. Solving for H yields H= 2L 2LF2 = F2 + m2g2 1 + (mg/F)2 As F → 0, H → 0 as is reasonable. As F → ∞, H → 2L, which is unreasonable. 2(2.00 m) = 1.44 m 1 + [(2.00 kg)(9.80 m/s2)/14.7 N]2 (b) H= (c) Call θ the equilibrium angle with the vertical. ΣFx = 0 ⇒ T sin θ = F, and ΣϖFy = 0 ⇒ T cos θ = mg Dividing: tan θ = F 14.7 N = = 0.750, or θ = 36.9° mg 19.6 N Therefore, Heq = L(1 – cos θ) = (2.00 m)(1 – cos 36.9°) = 0.400 m (d) As F → ∞, tan θ → ∞, θ → 90.0° and Heq →ϖϖ L A very strong wind pulls the string out horizontal, parallel to the ground. Thus, (Heq)max = L 8.68 Call φ = 180° – θ the angle between the upward vertical and the radius to the release point. Call vr the speed here. By conservation of energy Ki + Ui + ∆E = Kr + Ur vi = 1 1 2 2 mv i + mgR + 0 = mv r + mgR cos φ 2 2 The path after string is cut m 2 gR + 2 gR = v r + 2 gR cos φ R The components of velocity at release are vx = vr cos φ and vy = vr sin φ so for the projectile motion we have x = vx t R sin φ = vr cos φ t y = vy t – 1 2 gt 2 – R cos φ = vr sin φ t – 1 2 gt 2 By substitution –R cos φ = vr sin φ R sin φ g R2 sin2 φ – vr cos φ 2 v2 cos2 φ r with sin2 φ + cos2 φ = 1, 2 gR sin2 φ = 2v r cos φ = 2 cos φ(3 gR – 2 gR cos φ) sin2 φ = 6 cos φ – 4 cos2 φ = 1 – cos2 φ 3 cos2 φ – 6 cos φ + 1 = 0 cos φ = 6 ± 36 – 12 6 Only the – sign gives a value for cos φ that is less than one: cos φ = 0.1835 φ = 79.43° C θ vr = 3 gR – 2 gR cos φ so θ = 100.6° Rg 8.69 Applying Newton's second law at the bottom (b) and top (t) of the circle gives 2 mvb Tb – mg = R 2 mvt and –Tt – mg = – R mg 2 Adding these gives vt Tb = Tt + 2mg + Tt 2 m(v b – v t ) R Tb Also, energy must be conserved and ∆U + ∆K = 0 2 So, 2 m(v b – v t ) + (0 – 2mgR) = 0 and 2 2 2 m(v b – v t ) = 4mg R vb mg Substituting into the above equation gives Tb = Tt + 6mg 8.70 (a) Energy is conserved in the swing of the pendulum, and the stationary peg does no work. So the ball's speed does not change when the string hits or leaves the peg, and the ball swings equally high on both sides. (b) Relative to the point of suspension, Ui = 0, Uf = –mg[d – (L – d)] From this we find that –mg(2d – L) + 1 mv2 = 0 2 Also for centripetal motion, mg = mv2 where R = L – d. R Upon solving, we get d = 3L 5 L θ d Peg 8.71 (a) The potential energy associated with the wind force is +Fx, where x is the horizontal distance traveled, with x positive when swinging into the wind and negative when swinging in the direction the wind is blowing. The initial energy of Jane is, (using the pivot point of the swing as the point of zero gravitational energy), Ei = (K + Ug + Uwind)i = 1 2 mv i – mgL cos θ – FL sin θ 2 where m is her mass. At the end of her swing, her energy is Ef = (K + Ug + Uwind)f = 0 – mgL cos φ + FL sin φ so conservation of energy (Ei = Ef) gives 1 2 mv i – mgL cos θ – FL sin θ = –mgL cos φ + FL sin φ 2 2gL(cos θ – cos φ) + 2 This leads to vi = But D = L sin φ + L sin θ, so that sin φ = FL (sin θ + sin φ) m D 50.0 – sin θ = – sin 50.0° = 0.484 L 40.0 which gives φ = 28.9°. Using this, we have vi = 6.15 m/s . (b) Here (again using conservation of energy) we have, –MgL cos φ + FL sin φ + 1 Mv2 = –MgL cos θ – FL sin θ 2 where M is the combined mass of Jane and Tarzan. Therefore, v = 2gL(cos φ – cos θ) – 2 FL (sin φ + sin θ) which gives v = 9.87 m/s M as the minimum speed needed. 8.72 Find the velocity at the point where the child leaves the slide, height h: (U + K)i = (U + K)f mgH + 0 = mgh + 1 mv2 2 v = 2g(H – h) H Use Newton's laws to compare h and H. R θ (Recall the normal force will be zero): ∑ Fr = mar = mv2 R mg sin θ – n = mg sin θ = mv2 R m(2 g)(H – h) R Put θ in terms of R: sin θ = h R h 2 mg(H – h) mg = R R h= 2 H 3 2 3 R, the assumption that the child will leave the slide at a height H is no 3 2 longer valid. Then the velocity will be too large for the centripetal force to keep the child 3 R, the child will leave the track at h = R. on the slide. Thus if H ≥ 2 Notice if H ≥ 8.73 Case I: Surface is frictionless 1 1 2 mv2 = kx 2 2 k= mv2 (5.00 kg)(1.20 m/s)2 = = 7.20 ∞ 102 N/m x2 10–2 m2 Case II: Surface is rough, µk = 0.300 1 1 2 mv2 = kx – µkmgx 2 2 5.00 kg 2 v = 2 1 (7.20 ↔10 2 Ν/m )(10 − 1 m ) 2 – (0.300)(5.00 kg)(9.80 m/s2)(10–1 m) 2 v = 0.923 m/s 8.74 ΣFy = n – mg cos 37.0° = 0, ∴ n = mg cos 37.0° = 400 N f = µN = (0.250)(400) = 100 N Wf = ∆E (–100)(20.0) = ∆UA + ∆UB + ∆KA + ∆KB ∆UA = mAg(hf – hi) = (50.0)(9.80)(20.0 sin 37.0°) = 5.90 ∞ 103 ∆UB = mBg(hf – hi) = (100)(9.80)(–20.0) = – 1.96 ∞ 104 ∆KA = ∆KB 1 2 2 m (v – v i ) 2 A f mB 1 2 2 m (v – v i ) = ∆KA = 2∆KA mA 2 B f Adding and solving, ∆KA = 3.92 kJ Chapter 9 Even Answers 2. 4. 6. 8. 10. 12. 14. 16. 18. 20. 22. 24. 26. 28. 30. 32. 34. 36. 38. 40. 42. 44. 46. 48. 50. 52. 54. 56. 58. 60. 62. 64. 66. 68. 70. 72. 74. (a) 0 (b) 1.06 j kg ⋅ m/s 31.0 m/s (a) 6.00 m/s toward the left (b) 8.40 J 364 kg·m/s, 438 N (a) 5.40 N·s in direction of vf (b) –27.0 J (a) (9.05i + 6.12j) N ⋅ s (b) (377i + 255j) N ~103 N – F = 3750 N, no broken bones 4M gl m 15.6 m/s (a) 2.50 m/s (b) ∆K = –37.5 kJ 2.66 m/s 0.556 m 7.94 cm vgreen = 7.07 m/s, vblue = 5.89 m/s (a) v i/ 2 (b) ±45.0° (a) 1.07 m/s at –29.7˚ (b) ∆K/Ki = –0.318 vorange = 3.99 m/s, vyellow = 3.01 m/s (45.4i + 80.6j) m/s, or 92.5 m/s at 60.6˚ rcm = (0i + 1.00j) m 4.67 × 106 m See Instructor’s Manual (b) (–2.00i – 1.00j) m (c) (3.00i – 1.00j) m/s (d) (15.0i – 5.00j) kg ⋅ m/s (a) (–0.189i + 0.566j) m/s (b) 0.596 m/s at 108˚ (c) rCM = (–0.189i + 0.566j)t m (a) v1f = –0.780 m/s, v2f = 1.12 m/s (b) 0.360i m/s (a) 8000 kg/s (b) 6.91 km/s (a) 430 kg (b) 14.3 s 291 N M + m gd 2 m 2h (a) –0.667 m/s (b) 0.952 m 3.20 × 104 N, 7.13 MW (a) 3.54 m/s (b) 1.77 m (c) 3.54 × 104 N (d) No, the normal force of the rail contributes upward momentum to the system m 1v 1 + m 2v 2 m 1m 2 (b) xm = (v1 – v2) m1 + m2 k(m 1 + m 2 ) m – m 2m 2m 1 m2 – m1 1 2 2 (c) v1f = v1 + v2, v2f = v1 + v m 1 + m 2 m 1 + m 2 m 1 + m 2 m 1 + m 2 2 See Instructor’s Manual (a) 6.30 m/s (b) 6.17 m/s 2vi and 0 (a) (20.0i + 7.00j) m/s (b) (4.00 i) m/s2 (c) (4.00 i) m/s2 (d) (50.0i + 35.0j) m (e) 600 J (f) 674 J (g) 674 J (a) v = © 2000 by Harcourt College Publishers. All rights reserved. 2 Chapter 9 Even Answers © 2000 by Harcourt College Publishers. All rights reserved. Chapter 9 Solutions 9.1 m = 3.00 kg, v = (3.00i – 4.00j) m/s (a) p = mv = (9.00i – 12.0j) kg ⋅ m/s Thus, px = 9.00 kg ⋅ m/s and py = –12.0 kg ⋅ m/s (b) p= 2 2 p x + py = (9.00)2 + (–12.0)2 = 15.0 kg ⋅ m/s θ = tan–1 (py/px) = tan–1 (–1.33) = 307° *9.2 (a) At maximum height v = 0, so p = 0 (b) Its original kinetic energy is its constant total energy, Ki = 1 1 2 mv i = (0.100) kg(15.0 m/s)2 = 11.2 J 2 2 At the top all of this energy is gravitational. Halfway up, one-half of it is gravitational and the other half is kinetic: K = 5.62 J = v= 1 (0.100 kg) v2 2 2 × 5.62 J = 10.6 m/s 0.100 kg Then p = mv = (0.100 kg)(10.6 m/s)j p = 1.06 kg · m/s j *9.3 The initial momentum = 0. Therefore, the final momentum, pf , must also be zero. We have, (taking eastward as the positive direction), pf = (40.0 kg)(vc) + (0.500 kg)(5.00 m/s) = 0 vc = – 6.25 × 10–2 m/s (The child recoils westward.) *9.4 pbaseball = pbullet (0.145 kg)v = (3.00 × 10–3 kg)(1500 m/s) = 4.50 kg · m/s © 2000 by Harcourt College Publishers. All rights reserved. v= 4.50 kg · m/s = 31.0 m/s 0.145 kg © 2000 by Harcourt College Publishers. All rights reserved. Chapter 9 Solutions *9.5 I have mass 85.0 kg and can jump to raise my center of gravity 25.0 cm. I leave the ground with speed given by 2 2 v f – v i = 2a(x – xi) 2 0 – vi = 2(–9.80 m/s2)(0.250 m) vi = 2.20 m/s Total momentum is conserved as I push the earth down and myself up: 0 = –(5.98 × 1024 kg)ve + (85.0 kg)(2.20 m/s) v e ~10–23 m/s 9.6 (a) For the system of two blocks ∆p = 0, 3M M or pi = pf Before Therefore, (a) 0 = Mvm + (3 M)(2.00 m/s) 2.00 m/s v Solving gives vm = – 6.00 m/s (motion toward the left) 3M M After *9.7 (b) 1 2 1 1 2 2 kx = MvM + (3 M) v3M = 8.40 J 2 2 2 (a) The momentum is p = mv, so v = p/m and the kinetic energy is K = (b) K= 1 mv2 implies v = 2 p = mv = m 2K/m = 9.8 1 1 p 2 p2 mv2 = m = 2 2 m 2m 2K/m , so 2mK I = ∆p = m ∆v = (70.0 kg)(5.20 m/s) = 364 kg · m/s F= ∆p 364 = = 438 N ∆ t 0.832 © 2000 by Harcourt College Publishers. All rights reserved. (b) 3 Chapter 9 Solutions 4 I = ∫ Fdt = area under curve (a) 9.9 F (N) 20,000 1 = (1.50 × 10–3 s)(18000 N) = 13.5 N · s 2 9.10 (b) 13.5 N · s F= = 9.00 kN 1.50 × 10–3 s (c) From the graph, we see that Fmax = 18.0 kN F = 18,000 N 15,000 10,000 5,000 t (ms) 0 0 1 2 3 Assume the initial direction of the ball in the –x direction. (a) Impulse, I = ∆p = pf – pi = (0.0600)(40.0)i – (0.0600)(50.0)(–i) = 5.40i N · s (b) Work = Kf – Ki = 1 (0.0600) [(40.0)2 – (50.0)2] = 2 –27.0 J 9.11 y ∆p = F ∆t ∆py = m(vfy – viy) = m(v cos 60.0°) – mv cos 60.0° = 0 ∆px = m(–v sin 60.0° – v sin 60.0°) = –2mv sin 60.0° 60° = –2(3.00 kg)(10.0 m/s)(0.866) = –52.0 kg · m/s Fave = 60° ∆px –52.0 kg · m/s = = –260 N ∆t 0.200 s Goal Solution G: If we think about the angle as a variable and consider the limiting cases, then the force should be zero when the angle is 0 (no contact between the ball and the wall). When the angle is 90° the force will be its maximum and can be found from the momentum-impulse equation, so that F < 300N, and the force on the ball must be directed to the left. O: Use the momentum-impulse equation to find the force, and carefully consider the direction of the velocity vectors by defining up and to the right as positive. © 2000 by Harcourt College Publishers. All rights reserved. Chapter 9 Solutions 5 A: ∆p = F ∆t ∆py = mvyf – mvyx = m(v cos 60.0° – v cos 60.0°) = 0 So the wall does not exert a force on the ball in the y direction. ∆px = mvxf – mvx = m(–v sin 60.0° – v sin 60.0°) = –2mv sin 60.0° ∆px = –2(3.00 kg)(10.0 m/s)(0.866) = –52.0 kg ⋅ m/s – ∆p ∆pxi –52.0 i kg ⋅ m/s F = = = = –260 i N ∆t ∆t 0.200 s L: The force is to the left and has a magnitude less than 300 N as expected. 9.12 Take x-axis toward the pitcher (a) pix + Ix = pfx 0.200 kg(15.0 m/s)(–cos 45.0°) + Ix = 0.200 kg(40.0 m/s) cos 30.0° Ix = 9.05 N · s piy + Iy = pfy 0.200 kg(15.0 m/s)(–sin 45.0°) + Iy = 0.200 kg(40.0 m/s) sin 30.0° I = (9.05i + 6.12j) N · s (b) 1 1 I = (0 + Fm)(4.00 ms) + Fm (20.0 ms) + Fm (4.00 ms) 2 2 Fm × 24.0 × 10–3 s = (9.05i + 6.12j) N · s Fm = (377i + 255j) N 9.13 The force exerted on the water by the hose is F= ∆pwater mvf – mvi (0.600 kg)(25.0 m/s) – 0 = = = 15.0 N ∆t ∆t 1.00 s According to Newton's 3rd law, the water exerts a force of equal magnitude back on the hose. Thus, the holder must apply a 15.0 N force (in the direction of the velocity of the exiting water steam) to hold the hose stationary. © 2000 by Harcourt College Publishers. All rights reserved. 6 Chapter 9 Solutions *9.14 If the diver starts from rest and drops vertically into the water, the velocity just before impact is found from Kf + Ugf = Ki + Ugi 1 2 mvimpact + 0 = 0 + mgh ⇒ vimpact = 2 2gh With the diver at rest after an impact time of ∆t, the average force during impact is given by – m(0 – vimpact) –m 2g h F = = ∆t ∆t or – m 2g h F= ∆t (directed upward) Assuming a mass of 55 kg and an impact time of ≈1.0 s, the magnitude of this average force is – (55 kg) F = *9.15 2(9.8 m/s2)(10 m) = 770 N, or ~ 103 N 1.0 s (200 g)(55.0 m/s) = (46.0 g)v + (200 g)(40.0 m/s) v = 65.2 m/s 9.16 For each skater, – m ∆v (75.0 kg)(5.00 m/s) F = = = 3750 N ∆t (0.100 s) – Since F < 4500 N, there are no broken bones. 9.17 Momentum is conserved (10.0 × 10–3 kg)v = (5.01 kg)(0.600 m/s) v = 301 m/s © 2000 by Harcourt College Publishers. All rights reserved. Chapter 9 Solutions Goal Solution G: A reasonable speed of a bullet should be somewhere between 100 and 1000 m/s. O: We can find the initial speed of the bullet from conservation of momentum. We are told that the block of wood was originally stationary. A: Since there is no external force on the block and bullet system, the total momentum of the system is constant so that ∆p = 0 p1i + p2i = p1f + p2f (0.0100 kg)v1i + 0 = (0.0100 kg)(0.600 m/s)i + (5.00 kg)(0.600 m/s) i v1i = (5.01 kg)(0.600 m/s)i = 301 i m/s 0.0100 kg L: The speed seems reasonable, and is in fact just under the speed of sound in air (343 m/s at 20°C). © 2000 by Harcourt College Publishers. All rights reserved. 7 8 9.18 Chapter 9 Solutions Energy is conserved for the bob between bottom and top of swing: Ki + Ui = Kf + Uf 1 2 Mvb + 0 = 0 + Mg 2l 2 2 vb = g 4l l vb = 2 gl m Momentum is conserved in the collision: M v v/2 v mv = m + M · 2 gl 2 v= 9.19 4M m gl (a) and (b) Let vg and vp be the velocity of the girl and the plank relative to the ice surface. Then we may say that vg – vp is the velocity of the girl relative to the plank, so that vg – vp = 1.50 (1) But also we must have mgvg + mpvp = 0, since total momentum of the girl-plank system is zero relative to the ice surface. Therefore 45.0vg + 150vp = 0, or vg = – 3.33 vp initial Putting this into the equation (1) above gives vg – 3.33 vp – vp = 1.50, or vp = – 0.346 m/s Then vg = – 3.33(–0.346) = 1.15 m/s 9.20 Gayle jumps on the sled: p1i + p2i = p1f + p2f (50.0 kg)(4.00 m/s) = (50.0 kg + 5.00 kg)v2 v2 = 3.64 m/s They glide down 5.00 m: Ki + Ui = Kf + Uf 1 1 2 (55.0 kg)(3.64 m/s) 2 + 55.0 kg(9.8 m/s2)5.00 m = (55.0 kg) v3 2 2 © 2000 by Harcourt College Publishers. All rights reserved. vb final Chapter 9 Solutions v3 = 10.5 m/s © 2000 by Harcourt College Publishers. All rights reserved. 9 10 Chapter 9 Solutions Brother jumps on: 55.0 kg(10.5 m/s) + 0 = (85.0 kg)v4 v4 = 6.82 m/s All slide 10.0 m down: 1 1 2 (85.0 kg)(6.82 m/s) 2 + 85.0 kg (9.80 m/s2)10.0 m = (85.0 kg) v5 2 2 v5 = 15.6 m/s 9.21 pi = pf (a) mc vic + mT viT = mc vfc + mTvfT vfT = 1 [m v + mT viT – mc vfc] mT c ic vfT = 1 [(1200 kg)(25.0 m/s) + (9000 kg)(20.0 m/s) – (1200 kg)(18.0 m/s)] 9000 kg = 20.9 m/s (b) East K lost = K i – K f = 1 1 1 1 2 2 2 m c vic + mT viT – mcvfc – m v 2 2 2 2 T fT = 1 2 2 2 2 [mc (vic – vfc ) + mT(viT – vfT ] 2 = 1 [(1200 kg)(625 – 324)(m2/s2) + (9000 kg)(400 – 438.2)m2/s2] 2 K lost = 8.68 kJ becomes internal energy. (If 20.9 m/s were used to determine the energy lost instead of 20.9333, the answer would be very different. We keep extra significant figures until the problem is complete!) 9.22 (a) mv1i + 3mv2i = 4mvf where m = 2.50 × 104 kg vf = (b) 4.00 + 3(2.00) = 2.50 m/s 4 Kf – Ki = 1 1 1 2 2 2 (4m) v f – m v1i + (3m)v 2i 2 2 2 = 2.50 × 104 [12.5 – 8.00 – 6.00] = –3.75 × 104 J © 2000 by Harcourt College Publishers. All rights reserved. Chapter 9 Solutions *9.23 (a) 11 The internal forces exerted by the actor do not change total momentum. vi m m m m 4.00 m/s 2.00 m/s m m m m (4m)vi = (3m)(2.00 m/s) + m(4.00 m/s) vi = (b) Wactor = Kf – Ki = W actor = (c) *9.24 6.00 m/s + 4.00 m/s = 2.50 m/s 4 1 1 [(3m)(2.00 m/s)2 + m(4.00 m/s)2] – (4m)(2.50 m/s) 2 2 2 (2.50 × 104 kg) [12.0 + 16.0 – 25.0](m/s)2 = 37.5 kJ 2 The explosion considered here is the time reversal of the perfectly inelastic collision in problem 9.22. The same momentum conservation equation describes both processes. We call the initial speed of the bowling ball vi and from momentum conservation, (7.00 kg)(vi) + (2.00 kg)(0) = (7.00 kg)(1.80 m/s) + (2.00 kg)(3.00 m/s) gives vi = 2.66 m/s 9.25 (a) Following Example 9.8, the fraction of total kinetic energy transferred to the moderator is f2 = 4m 1m 2 (m 1 + m 2 ) 2 where m2 is the moderator nucleus and in this case, m2 = 12m1 f2 = 4m 1(12m 1) 48 = = 0.284 or 28.4% (13m 1)2 169 of the neutron energy is transferred to the carbon nucleus. © 2000 by Harcourt College Publishers. All rights reserved. 12 Chapter 9 Solutions (b) KC = (0.284)(1.6 × 10–13 J) = 4.54 × 10–14 J Kn = (0.716)(1.6 × 10–13 J) = 1.15 × 10–13 J 9.26 v1, speed of m1 at B before collision. 1 2 2 m 1v1 = m1gh A m1 5.00 m 2 × 9.80 × 5.00 = 9.90 m/s v1 = m2 B v1f, speed of m1 at B just after collision. v1f = m1 – m2 1 v = – (9.90) m/s = –3.30 m/s m1 + m2 1 3 At the highest point (after collision) m 1ghmax = hmax = 9.27 1 m (–3.30)2 2 1 (–3.30 m/s)2 = 0.556 m 2(9.80 m/s2) At impact momentum is conserved, so: m 1v1 = (m 1 + m 2)v2 After impact the change in kinetic energy is equal to the work done by friction: 1 2 (m + m 2) v2 = ff d = µ(m1 + m2)gd 2 1 1 2 (0.112 kg) v2 = 0.650(0.112 kg)(9.80 m/s2)(7.50 m) 2 2 v 2 = 95.6 m2/s2 v2 = 9.77 m/s (12.0 × 10–3 kg)v1 = (0.112 kg)(9.77 m/s) v1 = 91.2 m/s © 2000 by Harcourt College Publishers. All rights reserved. C Chapter 9 Solutions 9.28 13 We assume equal firing speeds v and equal forces F required for the two bullets to push wood fibers apart. These equal forces act backwards on the two bullets. For the first, K i + ∆E = K f 1 (7.00 × 10–3 kg) v2 + F(8.00 × 10–2 m) cos 180° = 0 2 For the second, pi = pf (7.00 × 10–3 kg)v = (1.014 kg)vf vf = (7.00 × 10–3)v 1.014 Again, Ki + ∆E = Kf 1 1 2 (7.00 × 10–3 kg) v2 + Fd cos 180° = (1.014 kg) v f 2 2 Substituting, 1 1 7.00 × 10 –3 v (7.00 × 10–3 kg) v2 – Fd = (1.014 kg) 2 2 1.014 Fd = 2 1 1 7.00 × 10–3 2 (7.00 × 10–3 kg) v2 – (7.00 × 10–3 kg) v 2 2 1.014 Substituting again, Fd = F(8.00 × 10–2 m) 1 – 7.00 × 10–3 1.014 d = 7.94 cm *9.29 (a) First, we conserve momentum in the x direction (the direction of travel of the fullback). (90.0 kg)(5.00 m/s) + 0 = (185 kg)V cos θ where θ is the angle between the direction of the final velocity V and the x axis. We find V cos θ = 2.43 m/s (1) Now consider conservation of momentum in the y direction (the direction of travel of the opponent). (95.0 kg)(3.00 m/s) + 0 = (185 kg)(V sin θ) which gives, V sin θ = 1.54 m/s (2) © 2000 by Harcourt College Publishers. All rights reserved. 14 Chapter 9 Solutions Divide equation (2) by (1) tan θ = 1.54 = 0.633 2.43 From which θ = 32.3° Then, either (1) or (2) gives V = 2.88 m/s (b) Ki = 1 1 (90.0 kg)(5.00 m/s) 2 + (95.0 kg)(3.00 m/s) 2 = 1.55 × 103 J 2 2 Kf = 1 (185 kg)(2.88 m/s) 2 = 7.67 × 102 J 2 Thus, the kinetic energy lost is 783 J into internal energy. *9.30 The initial momentum of the system is 0. Thus, (1.20m)vBi = m(10.0 m/s) and vBi = 8.33 m/s or Ki = 1 1 1 m(10.0 m/s)2 + (1.20m)(8.33 m/s) 2 = m(183 m2/s2) 2 2 2 Kf = 1 1 1 1 m(vG)2 + (1.20m)(vB) 2 = m(183 m2/s2) 2 2 2 2 2 2 v G + 1.20vB = 91.7 m2/s2 (1) From conservation of momentum, mvG = (1.20m)vB or vG = 1.20vB (2) Solving (1) and (2) simultaneously, we find vG = 7.07 m/s (speed of green puck after collision) and vB = 5.89 m/s (speed of blue puck after collision) © 2000 by Harcourt College Publishers. All rights reserved. Chapter 9 Solutions *9.31 We use conservation of momentum for both northward and eastward components. For the eastward direction: M(13.0 m/s) = 2MVf cos 55.0° For the northward direction: MV = 2MVf sin 55.0° Divide the northward equation by the eastward equation to find: V = (13.0 m/s) tan 55.0° = 18.6 m/s = 41.5 mi/h Thus, the driver of the north bound car was untruthful. *9.32 (a) pi = pf so p xi = p xf and p yi = p yf mvi = mv cos θ + mv cos φ (1) 0 = mv sin θ – mv sin φ (2) From (2), sin θ = sin φ so θ = φ Furthermore, energy conservation requires 1 1 1 2 mv i = mv2 + mv2 2 2 2 so (b) v= vi 2 Hence, (1) gives vi = 2vi cos θ 2 θ = 45.0° φ = 45.0° © 2000 by Harcourt College Publishers. All rights reserved. 15 16 9.33 Chapter 9 Solutions By conservation of momentum (with all masses equal), 5.00 m/s + 0 = (4.33 m/s) cos 30.0° + v2fx v 2 f x = 1.25 m/s 0 = (4.33 m/s) sin 30.0° + v2fy v2fy = –2.16 m/s v = 2.50 m/s at – 60.0° Note that we did not need to use the fact that the collision is perfectly elastic. 9.34 (a) Use Equations 9.24 and 9.25 and refer to the figures below. before v1i b v1f sin θ v1f t θ v1f cos θ after θt φ v2f cos φ t φ −v2f sin φt v2f Let the puck initially at rest be m 2. m1v1i = m1v1f cos θ + m2v2f cos φ 0 = m1v1f sin θ – m2v2f sin φ (0.200 kg)(2.00 m/s) = (0.200 kg)(1.00 m/s) cos 53.0° + (0.300 kg)v2f cos φ 0 = (0.200 kg)(1.00 m/s) sin 53.0° – (0.300 kg)(v2f sin φ) © 2000 by Harcourt College Publishers. All rights reserved. Chapter 9 Solutions 17 From these equations we find tan φ = sin φ 0.160 = = 0.571, cos φ 0.280 φ = 29.7° Then v2f = (b) 9.35 flost = (0.160 kg · m/s) = 1.07 m/s (0.300 kg)(sin 29.7°) ∆K K f – K i = = – 0.318 Ki Ki m 1v1i + m 2v2i = (m 1 + m 2)vf 3.00(5.00)i – 6.00j = 5.00v v = (3.00i – 1.20j) m/s 9.36 p xf = p xi vO mvO cos 37.0° + mvY cos 53.0° = m(5.00 m/s) 0.799vO + 0.602vY = 5.00 m/s (1) O v1 = 5.00 m/s O 37.0° Y 53.0° p yf = p yi mvO sin 37.0° – mvY sin 53.0° = 0 Y vY 0.602vO = 0.799vY (2) before after Solving (1) and (2) simultaneously, vO = 3.99 m/s 9.37 and vY = 3.01 m/s p xf = p xi vO mvO cos θ + mvY cos (90.0° – θ) = mvi vO cos θ + vY sin θ = vi O (1) O v1 θt Y 90.0° − θt p yf = p yi mvO sin θ – mvY sin (90.0° – θ) = 0 vO sin θ = vY cos θ (2) Y before From equation (2), vO = vY (cos θ/sin θ) (3) © 2000 by Harcourt College Publishers. All rights reserved. after vY 18 Chapter 9 Solutions Substituting into equation (1), cos2 θ + vY sin θ = vi sin θ vY so vY(cos2 θ + sin2 θ) = vi sin θ, and v Y = v i sin θ Then, from equation (3), vO = vi cos θ 9.38 The horizontal and vertical components of momentum are conserved: (5.00 g)(250 m/s) cos 20.0˚ – (3.00 g)(280 m/s) cos 15.0˚ = (8.00 g)vfx vfx = 45.4 m/s (5.00 g)(250 m/s) sin 20.0˚ + (3.00 g)(280 m/s) sin 15.0˚ = (8.00 g)vfy vfy = 80.6 m/s v = 45.4 m/s i + 80.6 m/s j 9.39 = 92.5 m/s at 60.6˚ m0 = 17.0 × 10–27 kg vi = 0 (the parent nucleus) m1 = 5.00 × 10–27 kg v1 = 6.00 × 106 j m/s m2 = 8.40 × 10–27 kg v2 = 4.00 × 106 i m/s (a) m 1v 1 + m 2v 2 + m 3v 3 = 0 where m3 = m0 – m1 – m2 = 3.60 × 10–27 kg (5.00 × 10–27)(6.00 × 106 j) + (8.40 × 10–27)(4.00 × 106 i) + (3.60 × 10–27)v3 = 0 v3 = (–9.33 × 106 i – 8.33 × 106 j) m/s (b) E= = 1 1 1 2 2 2 m 1v1 + m 2v2 + m 3v3 2 2 2 1 [(5.00 × 10–27)(6.00 × 106)2 + (8.40 × 10–27)(4.00 × 106)2 2 + (3.60 × 10–27)(12.5 × 106)2] E = 4.39 × 10–13 J © 2000 by Harcourt College Publishers. All rights reserved. Chapter 9 Solutions *9.40