Homework #3: Conservation of Energy

1. (I) A spring has a spring stiffness constant, k, of How much must this spring be stretched to store 25 J of potential energy?

2. (I) A 7.0-kg monkey swings from one branch to another 1.2 m higher. What is the change in potential energy?

3. (II) A 1200-kg car rolling on a horizontal surface has speed when it strikes a horizontal coiled spring and is brought to rest in a distance of 2.2 m. What is the spring stiffness constant of the spring?

4. (II) A 1.60-m tall person lifts a 2.10-kg book from the ground so it is 2.20 m above the ground. What is the potential energy of the book relative to (a) the ground, and (b) the top of the person’s head? (c) How is the work done by the person related to the answers in parts (a) and (b)?

5. (II) A spring with hangs vertically next to a ruler. The end of the spring is next to the 15-cm mark on the ruler. If a 2.5-kg mass is now attached to the end of the spring, where will the end of the spring line up with the ruler marks?

6. Find the height from which you would have to drop a ball so that it would have a speed of 9.0 m/s just before it hits the ground.

7. (I) Jane, looking for Tarzan, is running at top speed and grabs a vine hanging vertically from a tall tree in the jungle. How high can she swing upward? Does the length of the vine affect your answer?

8. (I) A novice skier, starting from rest, slides down a frictionless 35.0º incline whose vertical height is 185 m. How fast is she going when she reaches the bottom?

9. (I) A sled is initially given a shove up a frictionless 28.0º incline. It reaches a maximum vertical height 1.35 m higher than where it started. What was its initial speed?

10. (II) In the high jump, Fran’s kinetic energy is transformed into gravitational potential energy without the aid of a pole. With what minimum speed must Fran leave the ground in order to lift her center of mass 2.10 m and cross the bar with a speed of

11. (II) A 65-kg trampoline artist jumps vertically upward from the top of a platform with a speed of (a) How fast is he going as he lands on the trampoline, 3.0 m below (Fig. 6–38)? (b) If the trampoline behaves like a spring with spring stiffness constant how far does he depress it?

12. A child and a sled with a combined mass of 50.0 kg slide down a frictionless slope. If the sled starts from rest and has a speed of 3.00 m/s at the bottom, what is the height of the hill?

13. A 0.250-kg block is placed on a light vertical spring (k = 5.00 × 103 N/m) and pushed downwards, compressing the spring 0.100 m. After the block is released, it leaves the spring and continues to travel upwards. What height above the point of release will the block reach if air resistance is negligible?

14. (II) A 62-kg bungee jumper jumps from a bridge. She is tied to a bungee cord whose unstretched length is 12 m, and falls a total of 31 m. (a) Calculate the spring stiffness constant k of the bungee cord, assuming Hooke’s law applies. (b) Calculate the maximum acceleration she experiences.

15. A 70-kg diver steps off a 10-m tower and drops from rest straight down into the water. If he comes to rest 5.0 m beneath the surface, determine the average resistive force exerted on him by the water.

16. An airplane of mass 1.5 × 104 kg is moving at 60 m/s. The pilot then revs up the engine so that the forward thrust by the air around the propeller becomes 7.5 × 104 N. If the force exerted by air resistance on the body of the airplane has a magnitude of 4.0 × 104 N, find the speed of the airplane after it has traveled 500 m. Assume that the airplane is in level flight throughout this motion.

17. A 2.1 × 103-kg car starts from rest at the top of a 5.0-m-long driveway that is inclined at 20° with the horizontal. If an average friction force of 4.0 × 103 N impedes the motion, find the speed of the car at the bottom of the driveway.

18. A 25.0-kg child on a 2.00-m-long swing is released from rest when the ropes of the swing make an angle of 30.0° with the vertical. (a) Neglecting friction, find the child’s speed at the lowest position. (b) If the actual speed of the child at the lowest position is 2.00 m/s, what is the mechanical energy lost due to friction?

19. (I) Two railroad cars, each of mass 7650 kg and traveling in opposite directions, collide head-on and come to rest. How much thermal energy is produced in this collision?

20. (II) A 21.7-kg child descends a slide 3.5 m high and reaches the bottom with a speed of How much thermal energy due to friction was generated in this process?

21. (II) You drop a ball from a height of 2.0 m, and it bounces back to a height of 1.5 m. (a) What fraction of its initial energy is lost during the bounce? (b) What is the ball’s speed just as it leaves the ground after the bounce? (c) Where did the energy go?

22. (II) A 110-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 350 N. For the first 15 m the floor is frictionless, and for the next 15 m the coefficient of friction is 0.30. What is the final speed of the crate?

Solutions #3: Conservation of Energy

26. The elastic PE is given by where x is the distance of stretching or compressing of the spring from its natural length.

27. Subtract the initial gravitational PE from the final gravitational PE.

29. Assume that all of the kinetic energy of the car becomes PE of the compressed spring.

30. (a) Relative to the ground, the PE is given by

(b) Relative to the top of the person’s head, the PE is given by

as the answer to part (a), 45.3 J. In part (a), the PE is calculated relative to the starting location of the application of the force on the book. The work done by the person is not related to the answer to part (b).

32. The spring will stretch enough to hold up the mass. The force exerted by the spring will be equal to the weight of the mass.

Thus the ruler reading will be .

5.19 With only a conservative force acting on the falling ball,
or
Applying this to the motion of the ball gives
or

33. The only forces acting on Jane are gravity and the vine tension. The tension pulls in a centripetal direction, and so can do no work – the tension force is perpendicular at all times to her motion. So Jane’s mechanical energy is conserved. Subscript 1 represents Jane at the point where she grabs the vine, and subscript 2 represents Jane at the highest point of her swing. The ground is the zero location for PE . We have , , and (top of swing). Solve for y2, the height of her swing.

No, the length of the vine does not enter into the calculation, unless the vine is less than 0.7 m long. If that were the case, she could not rise 1.4 m high. Instead she would wrap the vine around the tree branch.

34. The forces on the skier are gravity and the normal force. The normal force is perpendicular to the direction of motion, and so does no work. Thus the skier’s mechanical energy is conserved. Subscript 1 represents the skier at the top of the hill, and subscript 2 represents the skier at the bottom of the hill. The ground is the zero location for PE . We have , , and (bottom of the hill). Solve for v2, the speed at the bottom.

35. The forces on the sled are gravity and the normal force. The normal force is perpendicular to the direction of motion, and so does no work. Thus the sled’s mechanical energy is conserved. Subscript 1 represents the sled at the bottom of the hill, and subscript 2 represents the sled at the top of the hill. The ground is the zero location for PE . We have , , and . Solve for v1, the speed at the bottom.

Notice that the angle is not used in the calculation.

36. We assume that all the forces on the jumper are conservative, so that the mechanical energy of the jumper is conserved. Subscript 1 represents the jumper at the bottom of the jump, and subscript 2 represents the jumper at the top of the jump. Call the ground the zero location for PE . We have , , and . Solve for v1, the speed at the bottom.

5.27 Since no non-conservative forces do work, we use conservation of mechanical energy, with the zero of potential energy selected at the level of the base of the hill. Then, with yields
Note that this result is independent of the mass of the child and sled.

5.35 Choose at the level of the release point and use conservation of mechanical energy from release until the block reaches maximum height. Then,
and we have , or
which yields

37. (a) Since there are no dissipative forces present, the mechanical energy of the person – trampoline –

Earth combination will be conserved. The level of the unstretched trampoline is the zero level for both the elastic and gravitational PE. Call up the positive direction. Subscript 1 represents the jumper at the top of the jump, and subscript 2 represents the jumper upon arriving at the trampoline. There is no elastic PE involved in this part of the problem. We have , , and . Solve for v2, the speed upon arriving at the trampoline.

The speed is the absolute value of .

(b) Now let subscript 3 represent the jumper at the maximum stretch of the trampoline. We have , , , , and . There is no elastic energy at position 2, but there is elastic energy at position 3. Also, the gravitational PE at position 3 is negative, and so . A quadratic relationship results from the conservation of energy condition.

Since .

The first term under the quadratic is about 1000 times smaller than the second term, indicating that the problem could have been approximated by not even including gravitational PE for the final position. If that approximation would have been made, the result would have been found by taking the negative result from the following solution.

5.39 We shall take at the lowest level reached by the diver under the water. The diver falls a total of 15 m, but the non-conservative force due to water resistance acts only during the last 5.0 m of fall. The work-energy theorem then gives
or
This gives the average resistance force as

5.40 Since the plane is in level flight, and the work-energy theorem reduces to , or
This gives

5.41 Choose at the level of the bottom of the driveway.
Then becomes
.
Solving for the final speed gives , or
or

5.42 (a) Choose at the level of the bottom of the arc. The child’s initial vertical displacement from this level is
In the absence of friction, we use conservation of mechanical energy as
, or , which gives

(b) With a non-conservative force present, we use
, or
Thus, of energy is spent overcoming friction.

42. Consider this diagram for the jumper’s fall.

(a) The mechanical energy of the jumper is conserved. Use y

for the distance from the 0 of gravitational PE and x for the amount of bungee cord “stretch” from its unstretched length. Subscript 1 represents the jumper at the start of the fall, and subscript 2 represents the jumper at the lowest point of the fall. The bottom of the fall is the zero location for gravitational PE , and the location where the bungee cord just starts to be stretched is the zero location for elastic PE . We have , , , , , and . Apply conservation of energy.

(b) The maximum acceleration occurs at the location of the maximum force, which

occurs when the bungee cord has its maximum stretch, at the bottom of the fall. Write Newton’s 2nd law for the force on the jumper, with upward as positive.

47. Use conservation of energy, where all of the kinetic energy is transformed to thermal energy.

48. Apply the conservation of energy to the child, considering work done by gravity and work changed into thermal energy. Subscript 1 represents the child at the top of the slide, and subscript 2 represents the child at the bottom of the slide. The ground is the zero location for PE . We have , , , and . Solve for the work changed into thermal energy.

51. (a) Calculate the energy of the ball at the two maximum heights, and subtract to find the amount of

energy “lost”. The energy at the two heights is all gravitational PE, since the ball has no KE at those maximum heights.

(b) Due to energy conservation, the KE of the ball just as it leaves the ground is equal to its final PE.

(c) The energy “lost” was changed primarily into heat energy – the temperature of the ball and the ground would have increased slightly after the bounce. Some of the energy may have been changed into acoustic energy (sound waves). Some may have been lost due to non-elastic deformation of the ball or ground.

52. Since the crate moves along the floor, there is no change in gravitational

PE, so use the work-energy theorem: . There are two forces doing work: , the pulling force, and , the frictional force. since the crate starts from rest. Note that the two forces doing work do work over different distances.