CHAPTER 9 SOLUTIONS TO PROBLEMS AND QUESTIONS

Citation: Jacqueline D. Spears and Dean Zollman, Instructor's Guide for The Fascination of Physics, (The Benjamin /Cummings Publishing Company, Inc., Menlo Park, CA 1985). Permission granted by the publisher and authors.

A. Review of Chapter Material

A1. The answers to the "A1" questions are found directly in the text. Look for each term printed in boldface type.

A2. Energy and work are measured in units of joules.

A3. Physicists relate the concept of work to the transfer of energy. Motion in the direction in which the force acts provides us evidence that energy has been transferred.

A4. The gravitational potential energy of an object depends on the object's mass, its height above some convenient reference point, and the acceleration due to gravity on the planet on which the object is located.

A5. An object's location, or height, depends upon the reference frame chosen for measurement. Consequently, gravitational potential energy is a relative concept.

A6. The kinetic energy of an object depends on its mass and its speed.

A7. The kinetic energy of an object depends directly on its mass and on the square of its speed. Doubling an object's speed will cause a greater increase in its kinetic energy.

A8. In physics "to conserve" means to keep a quantity the same throughout an interaction. In everyday life "to conserve" means to save.

A9. New forms of energy are frequently discovered by looking at situations in which energy seems not to be conserved.

A10. The various forms of energy can be classified into two general types- kinetic energy and potential energy.

A11. The energy equivalent of an object is determined by multiplying the object's mass by the speed of light squared. The law of conservation of energy must include the object's rest mass as part of the total energy of the system.

B. Using the Chapter Material

B1. Work = force x distance = 9800 N x 100 m = 980,000 joules. Gravitational potential energy is transferred to the concrete.

B2. Conservation of energy allows us to predict that the concrete slab's kinetic energy just before it hits the ground will equal its gravitational potential energy just before it began to fall. The gravitational potential energy can be calculated two ways. (1) It is the same as the work done to lift the slab. In problem B1 that was calculated as 980,000 joules. (2) Gravitational potential energy = mass x acceleration due to gravity x height 100 kg x 9.8 (m/s)/s X 100 m = 980,000 joules.

B3. When friction is included, the system includes both the concrete slab and the air surrounding it. The gravitational potential energy that the slab has while on the scaffolding must equal the kinetic energy of the slab immediately before it strikes the ground plus the thermal energy transferred to the surrounding air. Since some gravitational potential energy is lost to thermal energy, the kinetic energy of the slab will be less than the value calculated in problem B2.

B4. In (a) and (c) there is no change in the energy of the objects. Therefore, no work was done on the objects.

B5. Gravitational potential energy = mass x acceleration due to gravity x height = 70 kg x 3.7 (m/s)/s x 3 m = 777 J.

B6. Gravitational potential energy = mass x acceleration due to gravity x height = 500 kg x 9.8 (m/s)/s x 10 m = 49,000 J. The elevator is 20 meters above the basement (10 m + 10 m). Gravitational potential energy = mass x acceleration due to gravity x height = 500 kg x 9.8 (m/s)/s x 20 m = 98,000 J.

B7. Kinetic energy = (1/2) x (mass) x (speed)2

Speed Kinetic Energy

1 m/s 25 Joules

2 m/s 100 Joules

3 m/s 225 Joules

4 m/s 400 Joules

The kinetic energy increases as the square of the speed. Thus, when the speed of an object doubles, its kinetic energy increases by a factor of four; when its speed triples, its kinetic energy increases by a factor of nine; etc.

B9. When the gymnast is directly above the bar, her gravitational potential energy is at its greatest value. When she is directly below the bar, her gravitational potential energy is at its lowest value. Conservation of energy predicts that her kinetic energy, and thus her speed, must be greatest when her gravitational potential energy is the least. Her speed will be greatest where- her gravitational potential energy is least -at the bottom of the swing.

B10. The kinetic energy needed will be equal to the gravitational potential energy which the rock will have when it reaches the top of the window. Gravitational potential energy = mass x acceleration due to gravity x height = 0.3 kg x 9.8 (m/s)/s x 10 m = 29.4 J. So, Romeo must give the rock 29.4 J of kinetic energy.

B11. Energy = mass x (speed of light)2 = 70 kg x (3 x 108 m/s)2 = 6.3 x 1018 J. The actual value will depend on your mass, but should be somewhere near to this value.

C. Extensions to New Situations

C1. We apply conservation of energy to this situation. The total energy at the top of the hill must be equal to the kinetic energy of the sled at the bottom of the hill plus the energy which is transformed into thermal energy. Thus, the energy transformed into thermal energy is the difference between the gravitational potential energy at the top of the hill and the kinetic energy at the bottom. Energy transformed = 800 J - 200 J = 600J.

C2. (a) In designing roller coasters, engineers must consider the kinetic energy, the gravitational potential energy, and the thermal energy transformed as the roller coaster does work against friction. (b) If the roller coaster has zero speed at the top of the first hill, then all of its energy is the gravitational potential energy due to its height above its lowest point. It would not have enough energy to go any higher. Consequently, the roller coaster must be designed so that the first hill is the highest hill. (c) All hills must be shorter than the first one. However, it is not necessary for each hill to be shorter than the previous one. The roller coaster will have enough energy to climb to greater heights than say the fourth hill as long as the energy which has been transferred to friction is not too great.

C3. (a) In Chapter 5 we learned that in this type of collision the first ball stopped and the second one continued at the same speed that the first one had before the collision. (b) Before the collision the moving ball has a kinetic energy = 1/2 x mass x (speed)2 = 1/2 x 5 kg x (2 m/s)2 = 10 J. The stationary ball has a kinetic energy of zero. After the collision the kinetic energies are Just reversed: the first ball has zero, while the second one has a kinetic energy of 10 J. (c) The total kinetic energy of the system before the collision = 10 J = the total kinetic energy of the system after the collision kinetic energy is conserved. (d) If the balls stick together, then the total mass of the moving object after the collision is 10 kg. Conservation of momentum tells us that 5 kg x 2 m/s = 10 kg m/s = 10 kg x speed after. The speed with which the two balls move off after the collision is 1 m/s. (e) The total kinetic energy before the collision is the same value as we calculated in part (b) above. After the collision:

kinetic energy = 1/2 x mass x (speed)2 = 1/2 x 10 kg x (1 m/s)2 = 5 J. (f) The kinetic energy is not conserved in this sticky collision. Some of the energy must have gone into some other form of energy. (g) Kinetic energy is conserved only in collisions in which the objects do not stick together.

C4. (a) When the ball strikes the floor, some of its energy is transferred to the floor, resulting in a loss of kinetic energy. As the ball rises, its kinetic energy is converted into gravitational potential energy. Since the ball has less kinetic energy after the bounce than before, it cannot rise to the same height from which it was originally dropped. (b) With each bounce, the ball loses some kinetic energy. Each successive bounce will be lower than the previous one.

C5. (a) Gravitational potential energy = mass x acceleration due to gravity x height = 0.5 kg x 9.8 (m/s)/s x 4 m = 19.6 J. (b) Applying conservation of energy we conclude that the kinetic energy Just before it strikes the ground will be the same as its gravitational potential energy at the start of its fall = 19.6 J. (c) Gravitational potential energy = mass x acceleration due to gravity x height = 0.5 kg x 9.8 (m/s)/s X 3.5 m = 17.15 J. (d) Again applying conservation of energy to the situation, the energy just after it left the ground must equal the energy it had at the top of the bounce. Thus, the kinetic energy just after the first bounce will be 17.15 J. (e) The energy transformed into other forms is the difference between the energy the ball had just before the bounce and the energy it had just after the bounce. That energy = 19.6 J - 17.15 J = 2.45 J. (f) We could assume that the ball transfers the same amount of energy to the floor during each bounce. The gravitational potential energy at the top of successive bounces would be 19.6 J, 17.15 J, 14.7 J, 12.25 J and so on. The height for each successive bounce can be determined from the definition for gravitational potential energy (gravitational potential energy = mass x acceleration due to gravity x height.)

C6. (a) Momentum before = (mass 1) x (velocity 1) + (mass 2) x (velocity 2) = 1 kg x 1 m/s, right + 1 kg x 1 m/s, right = 2 kg m/s, right. To test whether momentum is conserved, we calculate the momentum after the collision and compare it with the momentum before. If two balls move outward at a speed of 1 m/s, momentum after = 2 kg m/s, right. Momentum is conserved. If one ball moves outward at a speed of 2 m/s, momentum after = 1 kg x 2 m/s, right = 2 kg m/s, right. Momentum is also conserved in this situation. If four balls move outward at a speed of 0.71 m/s, momentum after = 1 kg x 0.71 m/s, right + 1 kg x 0.71 m/s, right + 1 kg x 0.71 m/s, right + 1 kg x 0.71 m/s, right = 2.84 kg m/s, right. Momentum is not conserved in this situation. (b) Kinetic energy before = 1/2 (mass 1) x (speed 1)2 + 1/2 (mass 2) x (speed 2)2 = (1/2) (1 kg) x (1 m/s)2 + (1/2) (1 kg) x (1 m/s)2 = 1 J. If two balls move outward at a speed of 1 m/s after the collision, kinetic energy will still be 1 J. Kinetic energy is conserved. If one ball moves outward with a speed of 2 m/s, kinetic energy = (1/2) x (1 kg) x (2 m/s)2 = 2 J. Kinetic energy is not conserved in this situation. Four balls at 0.71 m/s: kinetic energy = (1/2) x (1 kg) x (0.71 m/s)2 + (1/2) x (1 kg) x (0.71 m/s)2 + (1/2) x (1 kg) x (0.71 m/s)2 + (1/2) x (1 kg) x (0.71 m/s)2 = 1.0 J. Kinetic energy is conserved in this situation. (c) Momentum is conserved for cases I and II; kinetic energy is conserved for cases I and III. Both momentum and kinetic energy are conserved only for case I. If one ball is released, only one ball will move outward after the collision.

C7. (a) In both cases you are doing work against the force due to gravity. If you consider only that force, the amount of work done will be the same in both situations. If you include the frictional force when walking, then the work done will be greater for walking. (b) When walking, you do the work. When riding the elevator, the elevator's motor does the work. (c) Change in gravitational potential energy = (mass) x (acceleration due to gravity) x (height) = 75 kg x 9.8 (m/s)/s X 400 m = 294,000 J.

C8. ( a) Both kinetic energy and gravitational potential energy are involved in this situation. When the plane takes off, it must change the kinetic energy of everything inside it. The greater the mass that must be accelerated is, the more energy will be required. The greater the mass that must be elevated is, the more gravitational potential energy will have to be provided. The fuel consumed by the airplane depends on the mass of its contents. (b) The same arguments that apply to the luggage in part (a) can apply to people. The airlines must use more fuel to increase both the gravitational potential energy and the kinetic energy of people whose mass is large than of people whose mass is small. (c) Both gravitational potential energy and kinetic energy increase in direct proportion to the mass of the object. Thus, you would set up a scheme in which the fare was related directly to the mass of the passenger. An average fare could be established for the average mass person. Then, the difference between each person's mass and this average could be added (for heavier persons) or subtracted (for lighter persons) from this fare.

C9. (a) The work done = Change in energy. In this case the change in energy is the change in kinetic energy = (1/2) x (mass) x (speed)2 = (1/2) x (100 kg) x (245 m/s)2 = 3,001,250 J. (b) The work done = Change in energy. The change in kinetic energy = (1/2) x (mass) x (speed)2 = (1/2) x (0.5 kg) x (245 m/s)2 = 15,006 J. (c) Gravitational potential energy = (mass) x (acceleration due to gravity) x (height). For the paint: Gravitational potential energy = (100 kg) x (9.8 (m/s)/s) X (10,000 m) = 9,800,000 J. For the pillow: Gravitational potential energy = (0.5 kg) x (9.8 (m/s)/s) x (10,000 m) = 49,000J. (d) As shown in the calculations in (a), (b), and (c), the amount of energy needed to increase the energy of the pillow and the airplane's paint is rather large. If these items were not present, the airplane would use less fuel and cost the airline less to operate. In TWA's case the airline executives decided that airplanes without paint just did not look right, so they abandoned the experiment.

C10. (a) When the ball starts moving, it gains kinetic energy. Therefore, the spring must have had some form of energy stored in it prior to its release. This is elastic potential energy. (b) As the sticks are rubbed together, they have kinetic energy which produces the fire. Thus, the fire must be a form of energy (thermal energy). (c) Before leaving the floor the player has no gravitational potential energy relative to the floor. In the air he does have gravitational potential energy. Thus, energy must have been stored in him to be available to change his gravitational potential energy. This energy is called chemical potential energy. (d) As a car starts moving, it changes its kinetic energy. Thus, it must have used energy stored in its fuel (chemical potential energy). (e) The gravitational potential energy of the elevators are increased when they are lifted. Thus, the electricity must be providing the energy.

C11. Energy = (mass) x (speed of light)2. In these cases the difference in total mass before and after the interaction is the mass which is converted into energy. (a) Mass difference = (3.918 x 10-25 kg) - (3.915 x 10-25 kg) = 0.003 x 10-25 kg = 3 x 10-22 kg. Energy = (3 x 10-22 kg) x (3 x 108 m/s) = 2.7 x 10-5 J. (b) Total mass before = 4 x (1.673 x 10-27 kg) = 6.692 x 10-27 kg; total mass after = (6.6443 x 10-27 kg) + O.0091 x 10-27 kg) + (0.0091 x 10-27 kg) = 6.6625 x 10-27 kg. Mass difference = (6.692 x 10-27 kg) - (6.6625 x 10-27 kg) = 0.0295 x 10-27 kg = 2.95 x 10-29 kg. Energy = (2.95 x 10-29 kg) x (3 x 108 m/s)2 = 2.655 x 10-12 J. (d) Mass difference = 2 x (0.0091 x 10-27 kg)= 0.0182 x 10-27 kg = 1.82 x 10-29. Energy = (1.82 x 10-29 kg) x (3 x 108 m/s)2 = 1.638 x 10-12 J.