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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;
~ 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
2
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
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
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
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.
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
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
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)
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.
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.
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
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
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 
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
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
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)
=
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.
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
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
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
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
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
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
24.6°
18
Chapter 1 Solutions
h
55°
r
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
19
20
Chapter 1 Solutions
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
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

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
(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
2
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
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
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
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
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
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
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
7
8
Chapter 2 Solutions
*2.20
= 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
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
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.
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!
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
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
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
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
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
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
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
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
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 .
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).
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
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
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
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)
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
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.
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
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 
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 .
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
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
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)
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
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
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
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.
Chapter 2 Solutions
35
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)
2
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
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
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
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
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
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 .
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
Chapter 3 Solutions
y
30.0°
x
100 m
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°
 
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
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)


Chapter 3 Solutions
D = 8.06 at (180° – 82.9°) = 8.06 at 97.2°
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
θA + B = tan–1  –  = –71.6° = 288°
 2
2
θA – 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
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
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.
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
35.0°
x
16
Chapter 3 Solutions
B
= [1902 + (49.7)2]1/2 = 196 cm , θ = tan–1  –
49.7
= –14.7°
 190 
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
x
18
Chapter 3 Solutions
|D| = (4.00)2 + (11.0)2 + (15.0)2 m = 19.0 m
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
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
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
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
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
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.
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
x′
θ
β
x
26
Chapter 3 Solutions
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.
2
64.
66.
68.
70.
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.
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
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
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 
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 
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
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.
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
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
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
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
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
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.
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.
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
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 
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
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.
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°
Chapter 4 Solutions
ar = 29.7 m/s2
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
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
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
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
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
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
.
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
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
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
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
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
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
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
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
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
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.
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
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
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
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
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
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°
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
39
40
Chapter 4 Solutions
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
2
70.
72.
74.
76.
78.
(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

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
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
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
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
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,
F1
b
5
6
Chapter 5 Solutions
ae =
(70.0 kg)(9.80 m/s2)
= ~ 10–22 m/s2
5.98 × 1024 kg
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
(–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
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
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°
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
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.
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
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
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
(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
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
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
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
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
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.
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)
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
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)
+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
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
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/sv 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
mg
Chapter 5 Solutions
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
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
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) .
Chapter 5 Solutions
Part (g):
T2, max – T2, min = Mmaxg – Mming = 6mgµs cos θ
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
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
31
32
Chapter 5 Solutions
(e)
The mass of the block makes no difference.
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 .
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
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.
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
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
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)
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
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
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
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.
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
44
Chapter 5 Solutions
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
2
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
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.
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
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
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
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
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
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
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

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
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)
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°
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.
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
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
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
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
ln2 
1
c=–
20.0
= 3.47 × 10 –2 s–1
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
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
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
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
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
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.
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)
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 )
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
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
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
27
28
Chapter 6 Solutions
(a)
v = k1/2r–1/2
so
v ∝ r –1/2
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
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°
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
32
Chapter 6 Solutions
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.
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
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
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
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 
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
20
Chapter 6 Solutions
39
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.
2
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
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)
mg = 147 N
Chapter 7 Solutions
A · B = A xB x + A yB y + A zB z
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
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
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
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.
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
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
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
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
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).
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
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
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
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
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
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
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
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
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
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.
Chapter 7 Solutions
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.
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
= 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
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
(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
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
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)
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 .
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.
29
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
2
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 =
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⋅⌡
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
dy
3
4
Chapter 8 Solutions
= 25.0 + 20.0 – 1.00 – 4.00 = 40.0 J
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
5
6
Chapter 8 Solutions
viy =
= 19.8 m/s
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
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
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
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
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
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
Chapter 8 Solutions
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°
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
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
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
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
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
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
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
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
Ugx = 0 = Usx = 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 – Ugx = –9.80 mm) + (UsA – Usx = –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 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
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
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
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 =
2
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
v=
4.50 kg · m/s
= 31.0 m/s
0.145 kg
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
(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.
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.
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
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).
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
vb
final
Chapter 9 Solutions
v3 = 10.5 m/s
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
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.
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
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)
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)
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°
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)
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)
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 – m```