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841166678 Name_ _____________________ Page 1 of 16 Exam 4 Wednesday, 21 May 2008 Room BLDG 7 – 1350……Webb Auditorium TIME: 2:45 TO 4:45 PM ******************************************************************** NOTE: THIS STUDY GUIDE WILL NOT BE COMPLETELY FINISHED UNTIL SUNDAY, MAY 18, 2008, 5 PM. ********************************************************** ******BE SURE TO SCROLL ALL THE WAY TO END!!! Formula sheet will be provided … see below… Bring ruler with millimeter markings Bring a protractor (unlikely, but just in case) Bring a fully charged calculator ******BE SURE TO SCROLL ALL THE WAY TO END!!! 6/17/2017 841166678 Name_ _____________________ Page 2 of 16 Descriptions of some of the topics covered Unit conversion (Conversion relations will be given) Scientific notation Significant figures and rules for arithmetic operations Measurement uncertainty Proper form 6/17/2017 CONTENT OF EXAM: About 50% MULTIPLE CHOICE ON THE WHOLE COURSE About 50% MOMENTUM AND ENERGY CONSERVATION 841166678 Name_ _____________________ Page 3 of 16 Preliminary Formula sheet (preliminary) Unit Systems Length Mass Time Force SI / MKS m kg s N CGS cm g s dyne BE ft slug s lb Significant figures In Dr. Entenberg’s class, trailing zeroes will be meaningful unless the number has quotes around it, i.e. “75,000” nails. Use a decimal point when you know the correct number of significant figures. Whole numbers have no decimal point; e.g., 3 = 3.0000000……. If unsure, use scientific notation to determine the number of significant figures. Proper form: [(A + A) 10exponent units] where A must be written to 1 significant figure Note that A and A must have 1. Common exponent outside the parentheses 2. Common units outside the parentheses 3. Same number of decimal places inside the parentheses Scientific notation example: 3769. m = 3.769 X 103 m Uncertainty propagation examples: f=X*Y f = fmax – f where fmax = (X + X) * (Y + Y) f=X/Y f = fmax – f where fmax = (X + X) / (Y – Y) Constants Geometry Trigonometry g = 9.80 m/s2 g = 980. cm/s2 g = 32.2 ft/s2 G = 6.67 x 10-11 N m2 / kg2 2 R R2 4 R2 (4/3) R3 R2 H sine = opp / hyp cosine = adj / hyp tangent = opp / adj = sine/cosine Vectors in 2-D Vectors in 2-D Vectors in 2-D Symbol F = |F| = magnitude ( + ONLY) Fx = component along x direction ( + ) Fx= + F cos (or Fx= + F sin ’) Force and Vector NOTATION: Fnet F F1 F2 F3 F4 = = angle wrt a given direction ( + ) Fy = component along y direction ( + ) Fy= + F sin (or Fy= + F cos ’) Fnet = F = F1 + F2 + F3 + F4…. Newton’s Second Law Notation Fnet = F = F1 + F2 + F3 + …. = M a = 0 (Vector notation form) Fnet, x = Fx = F1x + F2x + F3x + ….. = M ax = 0 (Component form, more practical) Fnet, y = Fy = F1y + F2y + F3y + ….. = M ay = 0 (Component form, more practical) Force: Newton’s Third Law All forces occur in “action-reaction” pairs! There is no such thing as an isolated force! 6/17/2017 841166678 Name_ _____________________ Page 4 of 16 Variables, definitions, and equations Mass Linear variables X and Y motions “linear mass” m or M Angular variables Tangential variables Rigid body rotation Point P on rigid body “moment of inertia” angular mass I = mrperp2 (see below for more info) Position Displacement Average velocity Average acceleration Linear variables X X = Xfinal – Xinitial vxavg = x / t axavg = vx / t Angular variables = final – initial avg = t avg = t Constant acceleration equations Velocity time equation Position time equation Velocity position equation aX = (aX)avg = constant Vxf = Vxi + ax t Xf = Vxi t + ½ ax t2 Vxf 2 = Vxi 2 + 2 ax X = avg = constant f t fi t + ½ t2 f 2 = i 2 + 2 Force: Newton’s 2nd Law Centripetal acceleration Fx = M ax ac = V2 / r Torque: Newton’s 2nd Law …………………………………………………………… = I Work Energy conservation Momentum Momentum conservation 6/17/2017 W = |F| cos |s| Ef = Ei Etotal = K + U (no friction) Etotal = K + U + Ethermal (friction) K = ½ M V2 Ugravity = M g Y Uspring = ½ k X2 px = M vx M1V1i + M2V2i = M1V1i + M2V2i = + r F +CCL , -CL = + r F sin() K = ½ I 2 L=I Isystem i system i = Isystem f system f I1i1i + I2i2i = I1f1f + I2f2f Tangential variables Stan = r Vtan = r atan = r 841166678 Name_ _____________________ Various Force Magnitudes Gravitational force (general) Gravity force at earth’s surface Friction force magnitude Normal force magnitude Tension force magnitude Spring force magnitude Page 5 of 16 F = G M1 M2 / r2 G = 6.67 x 10-11 N m2 / kg2 Wt = M g where g = [G ME / RE2] = 9.80 m/s2 fs < fsmax fsmax =s Fn fkin = k Fn Fn or n Ft or T |F| = k |x| Two dimensional circular motion: Centripetal (radial) acceleration = ar = V2 / r = 2 r Center of mass (center of gravity) {Newton’s Universal Law of Gravitation} Tangential acceleration = at = r Xcg = (M1 X1 + M2 X2 + … + MN XN) / (M1 + M2 + … + MN) Moment of inertia: point mass (m r2), hoop (M R2), solid disk (½ M R2), sphere [(2/5) M R2 ], rod, at end [(1/3) M L2 ] 6/17/2017 841166678 Name_ _____________________ Page 6 of 16 You are given the Earth with its mass of 5.98 x 1024 kg and its radius of 6.37 x 106 m. You are also given that g = 9.80 m /s2 and that G = 6.67 X 10-11 N m2 / kg2 (a) Use the standard method to calculate the weight of a 70.0 kg person standing on the Earth’s surface. Use SI units and 3 significant figures. (686. N) (b) Use Newton’s Law of Universal Gravitation to calculate the weight of the same 70.0 kg person. Start by making a cartoon sketch of the person standing on the surface of the earth; sketch and label the radius of the Earth on your drawing. Use SI units and 3 significant figures. (686. N) 6/17/2017 841166678 Name_ _____________________ Page 7 of 16 IN CLASS EXAMPLE y axis T x axis 60 A man pulls on a 70.0 kg crate with a rope. The rope makes an angle of 60.0 with respect to the vertical direction. The tension in the rope is 450. N. As the crate slides, its motion is opposed by a kinetic friction force of 140. N. The crate is displaced to right a distance of 8.00 m. (a) Sketch and label all the forces that act on the crate. Place the tail of the vector at the approximate point where the force acts. (b) Calculate the work by the man on the crate. 3.12 x 103 J (c) Calculate the work by the force of friction on the crate. – 1.12 x 103 J (d) Calculate the work done by any other forces on the crate. Normal 0, Gravity 0 (e) Calculate the kinetic energy of the crate after it has moved the distance of 8.00 m. 2.00 x 103 J 6/17/2017 841166678 Name_ _____________________ Page 8 of 16 EOC Problem 50 modified… You are given a child on a swing. If you wish, you can assume that the mass of the child is 35.0 kg. She is at the top of her motion when the angle is 45.0°. (a) Let the zero for gravitational potential energy be at the bottom of her swing. Show with a drawing that the Y coordinate for the potential energy function M g Y is Y = L ( 1 – cos( ) ) where is the angle between the vertical and the rope of the swing and L is the length of the rope. (b) Starting with Efinal = Einitial, find her speed at the bottom of the motion. (4.15 m/s) 6/17/2017 841166678 Name_ _____________________ Total Kinetic Energy Page 9 of 16 Ktotal = Ktranslational + Krotational Calculate the total kinetic energy of a bowling ball of mass M, radius R, and speed V. Assume that the bowling ball is a sphere with a uniform distribution of mass; ignore the holes. The ball rolls without slipping on a smooth level surface. Be sure to show the formulas and the numerical setup. Let M = 7.00 kg, R = 11.0 cm, and V = 3.00 m/s. Ktrans = ½ M V2 = ½ (7.00 kg) (3.00 m/s)2 = 31.5 Joules Krot Krot = ½ I 2 = ½ ( 2/5 M R2 ) ( V / R )2 = ½ I 2 = ½ ( 2/5 7.00 kg (0.110 m)2 ) ( 3.00 m/s / 0.110 m )2 = 12.6 Joules Ktotal = 44.1 Joules 6/17/2017 841166678 Name_ _____________________ Page 10 of 16 Use energy conservation to show that a circular object of mass M and moment of inertia I , that rolls down a ramp of height h without slipping has the following final speed. Vfinal = [ 2 g h / ( 1 + ( I / M R2) ] ½ Explain which object (hoop, cylindrical disk, or sphere) with mass M and radius R will reach the bottom first? 6/17/2017 841166678 Name_ _____________________ Page 11 of 16 Let the mass of the solid circular disk be M = 7.00 kg and let m = 3.50 kg. The radius of the mass-less pulley is Rp = 0.150 meter and Rdisk = 0.300 meter. Let the system start from rest. There is no friction. Find the following quantities after the mass m has gone down by 20.0 m. RMS with aluminum disk attached (a) Find the speed V of the mass m Efinal = Einitial ½ m Vf2 + ½ I f2 + m g Yf = ½ m Vi2 + ½ I i2 + m g Yi ½ 3.50 Vf2 + ½ ( ½ 7.00 Rd2 ) (Vf / Rp)2 + 0 = 0 + 0 + 3.50 9.80 20.0 2 2 ½ 3.50 Vf + ½ ( 14.0 Vf ) + 0 = 0 + 0 + 686. Vf = 8.85 m/s (b) Find the angular velocity of the disk f = Vf / Rp = 59.0 radians/s (c) Find the total energy of the system 686. Joules 6/17/2017 R m Figure 1 Mass/pulley system 841166678 Name_ _____________________ Page 12 of 16 Example 10.8 Thermal energy created sledding down a hill. (a) George jumps onto his sled and starts from rest at the top of a 5.00 m high hill. His speed at the bottom is 8.00 m/s. The mass of George and the sled is 55.0 kg. How much thermal energy was produced in the process? E = K + U + Ethermal Note: System = (sled and rider) + surface of slope + earth Ei = Ef Ki + Ui + Eth,i = Kf + Uf + Eth,f Ki + Ui = Kf + Uf + (Eth,f – Eth,i) Ki + Ui = Kf + Uf + Ethermal Note: Ethermal = (Eth,f – Eth,i) ½ M Vi2 + M g Yi = ½ M Vf2 + M g Yf + Ethermal 0 + 55.0 · 9.80 · 5.00 = ½ 55.0 · 8.002 + 0 + Ethermal Ethermal = 935. J (b) Suppose that the length of the hill is 12.0 m long. What is the average kinetic friction force on the sled as it moves down the hill? (77.9 N) Explain your reasoning for this calculation. 6/17/2017 841166678 Name_ _____________________ Page 13 of 16 Problem 62 A new event for the Winter Olympics. An athlete will sprint 100. m, starting from rest, then leap onto a 20.0 kg bobsled. The person and bobsled will then slide down a 50.0 m long ice-covered ramp, sloped at 20.0°, and into a spring with a carefully calibrated spring constant of 2000. N/m. Lisa, whose mass is 40.0 kg, can reach a speed of 12.0 m/s in the 100. m dash. Ignore friction on the icecovered ramp! (a) How fast will Lisa and the sled be going when they start to go down the ramp? There will be an inelastic collision between Lisa and the sled. Energy will be lost because the friction between Lisa and the sled. Momentum will be conserved because the sled will slide without friction on the ice. M1 V1i + M2 V2i = M1 V1f + M2 V2f V1f = V2f = Vf = 8.00 m/s (b) How fast will Lisa and the sled be going when they first hit the spring? Energy will be conserved as Lisa and the sled go down the frictionless ramp which is 17.1 m high. Momentum will not be conserved because Lisa and the sled are under the influence of gravity. Vtop = 8.00 m/s Vbottom = 20.0 m/s (c) How far will Lisa and the sled compress the spring? Energy will be conserved as Lisa and the sled compress the spring whose potential energy is U = ½ k X2. Momentum will not be conserved because Lisa and sled are interacting with the spring. ½ M Vf2 + ½ k Xf2 = ½ M Vi2 + ½ k Xi2 M = M1 + M2 = 60.0 kg, Vi = Vbottom = 20.0 m/s, Xi = 0 Xf = 3.46 m 6/17/2017 841166678 Name_ _____________________ Page 14 of 16 Example 10.11 Energy Transformations in a perfectly inelastic collision Show that Ethermal = 1.84 J Only momentum is conserved in this inelastic interaction M1 V1i + M2 V2i = M1 V1f + M2 V2f V1f = V2f = Vf 0.200 3.00 + 0.400 (-2.25) = 0.200 Vf + 0.400 Vf 0.600 + ( – 0.900) = 0.600 Vf Vf = – 0.500 m/s The energy lost goes into thermal energy E = Ksystem + U + Ethermal U = constant = 0 because level stays the same. Ei = Ef Ksystemi + Eth,i = Ksystemf + Eth,f Ksystemi = Ksystemf + Ethermal Note: Ethermal = (Eth,f – Eth,i) ½ 0.200 3.002 + ½ 0.400 (-2.25)2 = ½ 0.200 (-.500)2 + ½ 0.400 (-.500)2 + Ethermal 0.900 + 1.013 = 0.025 + 0.050 + Ethermal Ethermal = 1.84 J 6/17/2017 841166678 Name_ _____________________ Page 15 of 16 Example from class…Velocities in an elastic collision M1 M2 x-axis Given a one dimensional elastic collision between a ball and a block of wood at rest. The mass of the ball is 8.00 kg and the mass of the wood is 16.0 kg. The initial speed of the ball is 45.0 m/s. (a) Using the appropriate equations, determine the final x-component of velocity (V1f) for the ball (M1) and the final x-component of velocity (V2f) for block of wood (M2). (-15.0 m/s, +30.0 m/s) (b) Show with a calculation (to 3 significant figures) that momentum was conserved in this collision. 360. kg m/s = 360 kg m/s (c) Show with a calculation (to 3 significant figures) that this collision was elastic. (8100. Joules = 8100. Joules) 6/17/2017 841166678 Name_ _____________________ Average Power = Average Power = Units Page 16 of 16 Work / Time Fx Vx [ Power ] = [ Work ] / [ Time ] = Joules / Second = Watt Problem 10 – 38 (a) How much work does an elevator motor do to lift a 1000. kg elevator a height of 100. m? 9.80 x 105 Joules (b) How much power must the motor supply to do this in 50.0 s at constant speed? 1.96 x 104 Watts Problem 10 – 44 A 710. kg car drives at a constant speed of 23.0 m/s. It is subject to a drag force of 500. N. What power is required from the car’s engine to the drive the car (a) on level ground? 1.15 x 104 Watts (b) up a hill with a slope of 2.00°. (optional) Example from class How much does it cost to operate a 100. Watt bulb for 1 day? Assume that a “kilowatt hour” (kwh) costs 15.0 cents. $0.36 6/17/2017