answer(6) - Answer, Key Homework 5 Rubin H Landau 1 The...

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Unformatted text preview: Answer, Key Homework 5 Rubin H Landau 1 The work is found through the expression, This print-out should have 8 questions. Check that it is complete before leaving the printer. Zf Also, multiple-choice questions may continue ~s W = F  d~ on the next column or page: nd all choices i before making your selection. Note that only a few usually 4 of the For the force given, problems will have their scores kept for a Z 7:2 m grade. You may make multiple tries to get W= 4 x ^ + 4:74 y ^  dx ^ | a problem right, although it's worth less each 0 time. Worked solutions to a number of these problems even some of the scored ones may Evaluation of the integral yields, be found on the Ph211 home page. 2 ?7:2 m W = 4 N=m  x ?0 2? Dragging a Block 07:01, calculus, numeric, 1 min. Algorithm  001 a = 4:74 1::2 1 4 A 15:5 kg block is dragged over a rough, hor28 2 b = 7:2 m 8 izontal surface by a constant force of 106 N c = 4:0 N=m 3 acting at an angle of angle 26:8 above the 2:0 horizontal. The block is displaced 75:3 m, W = c2b:0 4 and the coe cient of kinetic friction is 0:2. Find the work done by the force 106 N. 2:0 = h4i h27::02i Correct answer: 7124:44 J. Explanation: = 103:68 J Consider the force diagram 2:0 N units hJi = hN=mi hmi hi F µN mg θ ~s Work is W = F  ~ , where ~ is the distance s traveled. In this problem ~ = 5 ^ is only in the s x direction. When a spring is stretched near its elastic limit, the spring force satis es the equation, 005 F = ,k x + x3 If k = 9:23 N=m and = 149 N=m3, calculate the work done by this force when the spring is stretched 0:0514 m from its equilibrium position. Correct answer: ,0:0119326 J.  WA = Fx sx = F cos sx = 106 N cos26:8  75:3 m = 7124:44 J : Explanation: A force F = 4.0 x ^ + 4:74 y ^ N acts on an | object as the object moves in the ^ direction from the origin to b = 7:2 m. Find the work done on the object by the force. Correct answer: 103:68 J. 004 The work done is found by integrating the force of the spring over the distance it is compressed, Z xf F dx W= For this problem, xi Explanation: W= Z xf xi ,k x + x3  dx Answer, Key Homework 5 Rubin H Landau 2 Evaluating the integral where xi = 0 we 1. kinetic energy of the object increases, nd, mechanical energy increases. x4 W = 2 + 4f ,9:23 N=m 0:0514 m2 = 2 149 N=m3 0:0514 m4 + 4 = ,0:0119326 J ,k x2 f mechanical energy decreases. mechanical energy decreases. 1 2 3 4 2. None of these. 3. kinetic energy of the object decreases, 4. kinetic energy of the object increases, Algorithm 4 k = 9:23 N=m 16  40 = 149 N=m3 160 0 xf = 0:0514 m !::04 0 16  2:0 kx W1 = , f 5. kinetic energy of the object decreases, mechanical energy increases. correct Explanation: Therefore, W = ,70 + 50 = ,20J: 2:0 h9:23i h0:0514i2:0 =, 2:0 = ,0:0121926 J 2:0 hJi = , hN=mi hmi hi units W = EK 0; and EK decreases. Since Wnon,conserve = 50 = EMechanical 0; the total mechanical energy increases. f W 2 = 4:0 5 4:0 = h149i h0::0514i 40 = 0:000260003 J 3i 4:0 hJi = hN=m hi hmi units W = W1 + W2 6 = h,0:0121926i + h0:000260003i = ,0:0119326 J hJi = hJi + hJi units x4:0 Pushing a Crate 03 07:03, calculus, numeric, 1 min. 008 You push a 36:9 kg crate at a constant speed of 2:08 m=s across a horizontal oor with coe cient of kinetic friction 0:13. At what rate is work being done on the crate by you? Correct answer: 97:782 W. Explanation: Energies 07:03, calculus, numeric, 1 min. 006 As an object moves from point A to point B only two forces act on it: one force is conservative and does ,70 J of work, the other force is nonconservative and does +50 J of work. Between A and B, In order to keep the crate moving at a constant speed, the force you apply needs to exactly cancel the frictional force. Therefore the force that you need to apply is F = k m g The work you need to apply per unit time power is given by P =Fv Therefore, the rate at which you do work on the crate is P = k m g v = 0:13 36:9 kg 9:8 m=s2 2:08 m=s = 97:782 W Answer, Key Homework 5 Rubin H Landau 3 We can use conservation of energy to relate the initial kinetic energy of the bullet to the Pushing a Crate 03 07:04, trigonometry, multiple choice, 1 min. work done by the frictional force. 009 1 m v2 = f  s At what rate is energy being dissipated by the 2 frictional force? Correct answer: ,97:782 W. Solving for the frictional force, f , Explanation: 2 To keep the crate moving at a constant speed, f = mv the work per unit time that you apply must 2s 2 exactly cancel the energy per unit time that is = 0:00462 kg 891 m=s dissipated by the frictional force. Therefore, 2 0:0206 m the rate at which energy is being dissipated = 89022:6 N : by friction is P = ,k m g v = ,97:782 W Algorithm m = 36:9 kg 16 64  v = 2:08 m=s 0::56  2 24 k = 0:13 0::1 04 g = 9:8 m=s2 f = k m g = h0:13i h36:9i h9:8i = 47:0106 N hNi = hi hkgi hm=s2i P =fv = h47:0106i h2:08i = 97:782 W hWi = hNi hm=si Pf = ,P  = ,h97:782i = ,97:782 W hWi = ,hWi  Tree Stops a Bullet 07:04, calculus, numeric, 1 min. 011 1 2 3 4 5 units 6 units 7 units Assuming that the frictional force is constant, determine how much time elapsed between the moment the bullet entered the tree and the moment it stopped. Correct answer: 4:62402  10,5 s. Explanation: If we assume the frictional force is constant, by Newton's second law we can assume constant deceleration, which implies vavg = vinitial : 2 The time of deceleration is then t = vs avg 0:0206 m = 891 m=s 2 = 4:62402  10,5 s : 07:04, calculus, numeric, 1 min. A 4:62 g bullet moving at 891 m=s penetrates a tree to a depth of 2:06 cm. Use energy considerations to nd the average frictional force that stops the bullet. Correct answer: 89022:6 N. Tree Stops a Bullet 010 Algorithm hkgi = 0:001 kg=g g m i = 0:01 m=cm hcm  Explanation: m = 4:62 g 2 8 v = 891 m=s 240 1960 d = 2:06 cm 6::6 4 mu = m hkgi g = h4:62i h0:001i = 0:00462 kg 1 2 3 4 5 6 Answer, Key Homework 5 Rubin H Landau hkgi = hgi hkg=gi units mi du = d hcm 7 = h2:06i h0:01i = 0:0206 m hmi = hcmi hm=cmi units 2:0 f = m:u0vd 8 2u h0:00462i h891i2:0 = 2:0 h0:0206i = 89022:6 N s 2:0 hNi = hkgi hm=i i units hi hm 9 t = 2:0vdu = 2:0 hh0:0206i 891i = 4:62402  10,5 s h hsi = himmsii h= units 4 ...
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This note was uploaded on 05/06/2011 for the course PHYS 103N taught by Professor Chiu during the Spring '11 term at University of Texas at Austin.

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