PHYS 1111 - Summer 007 - Professor Caillault Homework Solutions Chapter 8
5. Picture the Problem The physical situation is depicted at right. for the Strategy Use W sp = 1 k x i x f work done by the spring. That way the work will always be negative if you start out at x i = 0 because the spring force will always be in the opposite direction from the stretch or compression. The work done by kinetic friction is W fr = µ k mgd, where d is the distance the box is pushed irregardless of direction, because the friction force always acts in a direction opposite the motion. Solution 1. (a) Sum the work done by the spring for each segment of path. Sum the work done by friction for each segment of path 3. (b) Sum the work done by the spring for the direct path from A to B 4. Sum the work done by friction for the direct path from A to B + x ( 4 x 3 ) W sp = 1 k x 1 x 4 = 1 ( 480 N/m ) 0 0.00 m W sp = 0.096 J 0.00 m + 0.00 m + ( 0 J) = 0.096 J (.7 kg) 9.81 m/s = 0.5 J W fr = µ k mg d 1 + d = 0.16 W sp = 1 k x A x B = 1 ( 480 N/m ) 0 0.00 m 0.00 + 0.040 m = 0.096 J W fr = µ k mgd = ( 0.16) (.7 kg) ( 9.81 m/s ) ( 0.00 m) = 0.085 J Insight The work done by friction is always negative, and increases in magnitude with the distance traveled. 11. Picture the Problem The spring in the soap dispenser is compressed by the applied force. Strategy Use equation 8-5 to find the spring constant using the given energy and compression distance data. Solve the same equation for x in order to answer part (b). Solution 1. (a) Solve equation 8-5 for k k = U x = 0.005 J = 00 N/m = 0.0 kn/m 0.0050 m. (b) Solve equation 8-5 for x x = U k = ( 0.0084 J ) = 0.9 cm 00 N/m Insight To compress the spring of this dispenser 0.50 cm requires 1.0 N or about lb of force. 13. Picture the Problem A graph of the potential energy vs. stretch distance is depicted at right.
Strategy The work that you must do to stretch a spring is equal to minus the work done by the spring because the force you exert is in the opposite direction from the force the spring exerts. Use equations 8-1 and 8-5 together to find the spring constant and the required work to stretch the spring the specified distance. Solution 1. (a) Because the stored potential energy in a spring is proportional to the stretch distance squared, the work required to stretch the spring from 5.00 cm to 6.00 cm will be greater than the work required to stretch it from 4.00 cm to 5.00 cm.. (b) Use equations 8-1 and 8-5 to find k 3. Use k and equations 8-1 and 8-5 to find the new W req W req = W spring = ΔU = U 5 U 4 = 1 kx 5 1 kx 4 = 1 k x 5 x 4 k = W req x 5 x 4 = ( 30.5 J) ( 0.0500 m) ( 0.0400 m) = 6.78 104 N/m W req = 1 k x ( x 1 ) = 1 6.78 ( 104 N/m) 0.0600 m ( 0.0500 m) = 37.3 J Insight Using the same procedure we discover that it would take 44.1 J to stretch the spring from 6.00 cm to 7.00 cm. 1. Picture the Problem The trajectory of the rock is depicted at right. Strategy The rock starts at height h, rises to y max, comes briefly to rest, then falls down to the base of the cliff at y = 0. Set the mechanical energy at the point of release equal to the mechanical energy at the base of the cliff and at the maximum height y max in order to find v i and y max. Solution 1. (a) Set E i = E f and solve for v i. (b) Now set E ymax = E f and solve for y max K i + U i = K f + U f 1 mvi + mgh = 1 mv + 0 f v i = v f gh 3 m = ( 9 m/s) 9.81 m/s v i = 15 m/s K ymax + U ymax = K f + U f 0 + mgy max = 1 mv f + 0 y max = v f g = ( 9 m/s) = 43 m 9.81 m/s Insight In part (a) the initial energy is a combination of potential and kinetic, but becomes all kinetic just before impact with the ground. In part (b) the rock at the peak of its flight has zero kinetic energy;
all of its energy is potential energy.
5. Picture the Problem The pendulum bob swings from point B to point A and gains altitude and thus gravitational potential energy. See the figure at right. Strategy Use equation 7-6 to find the kinetic energy of the bob at point B. Use the geometry of the problem to find the maximum change in altitude Δy max of the pendulum bob, and then use equation 8-3 to find its maximum change in gravitational potential energy. Apply conservation of energy between points B and the endpoint of its travel to find the maximum angle θ max the string makes with the vertical. Solution 1. (a) Use equation 7-6 to find K B K B = 1 mv B = 1 ( 0.33 kg).4 m/s = 0.95 J. (b) Since there is no friction, mechanical energy is conserved. If we take the potential energy at point B to be zero, we can say that all of the bob s kinetic energy will become potential energy when the bob reaches its maximum height and comes momentarily to rest. Therefore the change in potential energy between point B and the point where the bob comes to rest is 0.95 J. 3. (c) Find the height change Δy max of the pendulum bob 4. Use equation 8-3 and the result of part (b) to solve for θ max Δy max = L Lcosθ max = L( 1 cosθ max ) ΔU = mgδy max = mgl( 1 cosθ max ) θ max = cos 1 1 ΔU mgl = cos 1 1 0.33 kg 0.95 J 1. m 9.81 m/s = 41 Insight The pendulum bob cannot swing any further than 41 because there is not enough energy available to raise the mass to a higher elevation.
33.Picture the Problem The initial and final states of the system are depicted at right. The values given in the example are m 1 =.40 kg, m = 1.80 kg, d = 0.500 m, and µ k = 0.450. Strategy The nonconservative work done by friction changes the mechanical energy of the system. Use equation 8-9 to find ΔE and set it equal to the work done by friction. Solve the resulting expression for the final velocity of the system. Solution 1. Write equation 8-9 to find W nc. The nonconservative work is done by friction 3. Substitute the expression from step into step 1 and solve for v f v i ( K i + U i ) W nc = E f E i = K f + U f = 1 ( m 1 + m )v f + m 1 gh + m gy f 1 ( m 1 + m )v i + m 1 gh + m gy i v f v i W nc = 1 m 1 + m W nc = f k d = µ k m 1 gd + m g y f y i µ k m 1 gd = 1 m 1 + m v f v i gd ( m µ k m 1 ) = 1 ( m 1 + m ) v f v i ( m 1 + m ) gd m µ k m 1 + m g 0 d 0.500 m 9.81 m/s = v f v i = = 1.68 m / s 4. Finally, solve for v f v f = v + 1.68 m /s = 1.3 m/s i + 1.68 m /s = 1.8 m/s 1.80 kg ( 0.450) (.40 kg) (.40 + 1.80 kg) Insight If the blocks are not given the initial speed of 1.3 m/s, their final speed is 1.30 m/s, as in example 8-10. 37. Picture the Problem The skater travels up a hill (we know this for reasons given below), changing his kinetic and gravitational potential energies, while both his muscles and friction do nonconservative work on him. Strategy The total nonconservative work done on the skater changes his mechanical energy according to equation 8-9. This nonconservative work includes the positive work W nc1 done by his muscles and the negative work W nc done by the friction. Use this relationship and the known change in potential energy to find Δy. Solution 1. (a) The skater has gone uphill because the work done by the skater is larger than that done by friction, so the skater has gained mechanical energy. However, the final speed of the skater is less than the initial speed, so he has lost kinetic energy. Therefore he must have gained potential energy, and has gone uphill.
. (b) Set the nonconservative work equal to the change in mechanical energy and solve for Δy W nc = W nc1 + W nc = ΔE = E f E i = 1 m v ( f v i ) + mgδy W nc1 + W nc = ( K f + U f ) K i + U i Δy = W nc1 + W nc 1 m v f v i mg ( 340 J) + ( 715 J) 1 81.0 kg = 81.0 kg ( 1. m/s) (.50 m/s) 9.81 m/s = 3.65 m Insight Verify for yourself that if the skates had been frictionless but the skater s muscles did the same amount of work, the skater s final speed would have been 4.37 m/s. He would have sped up if it weren t for friction! 41. Picture the Problem The U vs. x plot is depicted at right. Strategy Describe the motion of the object, keeping in mind that objects tend to move to the minimum potential energy, and when they do their kinetic energy is maximum. The turning points on the potential energy plot are points A and E. Solution At point A, the object is at rest. As the object travels from point A to point B, some of its potential energy is converted into kinetic energy and the object s speed increases. As the object travels from point B to point C, some of its kinetic energy is converted back into potential energy and its speed decreases. From point C to point D, the speed increases again, and from point D to point E, the speed decreases. Insight The object momentarily comes to rest at point E, but then turns around and accelerates back towards D and retraces its path all the way to point A, at which time the cycle begins again. 61. Picture the Problem The child slides from rest at point A and lands at point B as indicated in the figure at right. Strategy Use the conservation of mechanical energy to find the horizontal speed of the child at the bottom of the slide in terms of h. Then use equation 4-9, the landing site of a projectile launched horizontally, to find the speed the child should have in order to land.50 m down range. Set the speeds equal to each other and solve for h.
Solution 1. Use equation 8-3 and let E A = E bottom to find v bottom K A + U A = K bottom + U bottom 1 mv A + mgy A = 1 mv bottom + mgy bottom = 1 mv bottom 0 + mg h + 1.50 m + mg 1.50 m gh = v bottom. Use equation 4-9 to find the appropriate v bottom for the child to land.50 m down range x = v bottom y bottom g v bottom = x g y bottom 3. Set the two velocities equal to each other and solve for h gh = x g y bottom gh = x g y bottom x h = =.50 m 4y bottom 4 1.50 m = 1.04 m Insight These are the sort of calculations an engineer might make to determine how high to build a slide so that a child will land in a certain place. However, an engineer should account for the nonzero friction when making his design. 71. Picture the Problem The mass oscillates back and forth on a frictionless horizontal surface. Strategy As the mass oscillates back and forth, its kinetic energy is converted into spring potential energy and back again. Use equation 8-6 and the conservation of mechanical energy to find the turning points of the motion. Solution 1. (a) Set E i = E f, where the initial state corresponds to maximum speed and the final state is at the turning point. U i + K i = U f + K f 0 + 1 mv i = 1 kx + 0. Solve the expression from step 1 for x x = ±v i m k 3. (b) Write a ratio to find the multiplicative factor for x when k is doubled x new = ±v i m k x old v i m k = ± ( 1.7 m/s) = 1. kg 560 N/m = ±0.11 m Insight The maximum compression distance is reduced because the stiffer spring stores more energy with less compression.