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# Energy Conservation Practice Problems for AP Physics B & C

based on 2 ratings
By — McGraw-Hill Professional
Updated on Feb 10, 2011

Review the following concepts if necessary:

### Multiple Choice:

Questions 1 and 2

A block of weight mg = 100 N slides a distance of 5.0 m down a 30-degree incline, as shown above.

1. How much work is done on the block by gravity?
1. 500 J
2. 430 J
3. 100 J
4. 50 J
5. 250 J
2. If the block experiences a constant friction force of 10 N, how much work is done by the friction force?
1. –43 J
2. –25 J
3. –500 J
4. –100 J
5. –50 J
3. This problem is for AP Physics C students only:

4. A mass experiences a potential energy U that varies with distance x as shown in the graph above. The mass is released from position x = 0 with 10 J of kinetic energy. Which of the following describes the long-term motion of the mass?
1. The mass eventually comes to rest at x = 0.
2. The mass slows down with constant acceleration, stopping at x = 5 cm.
3. The mass speeds up with constant acceleration.
4. The mass oscillates, never getting farther than 5 cm from x = 0.
5. The mass oscillates, never getting farther than 10 cm from x = 0.
5. Two identical balls of mass m = 1.0 kg are moving towards each other, as shown above. What is the initial kinetic energy of the system consisting of the two balls?
1. 0 joules
2. 1 joules
3. 12 joules
4. 18 joules
5. 36 joules

### Free Response:

1. A 1500-kg car moves north according to the velocity–time graph shown.
1. Determine the change in the car's kinetic energy during the first 7 s.
2. To determine how far the car traveled in these 7 s, the three basic kinematics equations can not be used. Explain why not.
3. Use the velocity–time graph to estimate the distance the car traveled in 7 s.
4. What was the net work done on the car in these 7 s?
5. Determine the average power necessary for the car to perform this motion.

### Solutions

1. E—The force of gravity is straight down and equal to 100 N. The displacement parallel to this force is the vertical displacement, 2.5 m. Work equals force times parallel displacement, 250 J.
2. E—The force of friction acts up the plane, and displacement is down the plane, so just multiply force times distance to get 50 J. The negative sign indicates that force is opposite displacement.
3. D—Think of Chris on a skateboard—on this graph, he will oscillate back and forth about x = 0. Because he starts with a KE of 10 J, he can, at most, have a potential energy of 10 J, which corresponds on the graph to a maximum displacement of 5 cm. (The mass cannot have constant acceleration because constant acceleration only occurs for a constant force; a constant force produces an energy graph that is linear. The mass will not come to rest because we are assuming a conservative force, for which KE can be converted to and from PE freely.)
4. E—Kinetic energy is a scalar, so even though the balls move in opposite directions, the KEs cannot cancel. Instead, kinetic energy ½(1 kg)(6 m/s)2 attributable to different objects adds together algebraically, giving 36 J total.
5.
1. The car started from rest, or zero KE. The car ended up with ½(1500 kg)(40 m/s)2 = 1.2 × 106 J of kinetic energy. So its change in KE is 1.2 × 106 J.
2. The acceleration is not constant. We know that because the velocity–time graph is not linear.
3. The distance traveled is found from the area under the graph. It is easiest to approximate by counting boxes, where one of the big boxes is 10 m. There are, give-or-take, 19 boxes underneath the curve, so the car went 190 m.
4. We cannot use work = force × distance here, because the net force is continually changing (because acceleration is changing). But Wnet = ΔKE is always valid. In part (a) the car's change in KE was found to be 1.2 × 106 J; so the net work done on the car is also 1.2 × 106 J.
5. Power is work divided by time, or 1.2 × 106 J/7 s = 170 kW. This can be compared to the power of a car, 220 horsepower.

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