The problems are on the conservation of energy worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Conservation of energy says the total energy of an isolated system stays constant: energy changes form, such as gravitational or spring potential energy becoming kinetic energy, but is never created or destroyed. These worked answers correct the 20 problems of the AP Physics 1 worksheet on conservation of energy, from name the energy store to spot the error: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the law used, the substitution and the result with its unit in bold. Compare the method as well as the number: in AP free-response items, a correct setup earns marks even when the arithmetic slips.
Level: AP Physics 1 (algebra-based) · Curriculum: AP P1 Unit 3
Problem 1 Name the energy store
1. Elastic (spring) potential energy.
2. Gravitational potential energy (plus a little kinetic energy if it still moves sideways; at the very top of a vertical flight it is mostly potential).
3. Kinetic energy.
4. Gravitational potential energy.
5. The floor lies above the basement reference level, so h > 0 and Ug = mgh is positive; only changes in U are physically meaningful. Answer: potential energy depends on the chosen reference level.
Problem 2 Lifting a water container
1. ΔU = mgh = 12 kg × 9.81 m/s² × 1.5 m = 177 J (176.6 J).
2. The gain is unchanged: 177 J. The container’s U at the shelf becomes zero and its U on the floor becomes −177 J, but the difference is the same.
Problem 3 Energy in a spring
1. Us = ½kx² = ½ × 400 × (0.050)² = 0.50 J.
2. x = 0.100 m gives ½ × 400 × (0.100)² = 2.0 J. 2.0 J, which is four times as much, because Us is proportional to x².
Problem 4 Motor power
1. P = W/t = 4500 J ÷ 30 s = 150 W.
2. t = 120 s, so W = Pt = 150 W × 120 s = 1.8 × 10⁴ J.
Problem 5 Unit conversions
1. 2.5 kW × 1000 = 2500 W.
2. 0.75 × 3.6 × 10⁶ J = 2.7 × 10⁶ J.
3. 8400 J ÷ 60 s = 140 W.
Problem 6 Dropped cricket ball
1. K = mgh = 0.15 × 9.81 × 1.8 = 2.649 J, so K = 2.6 J.
2. v = √(2gh) = √(2 × 9.81 × 1.8) = √35.3 = 5.9 m/s.
Problem 7 Reading a bar chart
1. Six squares of U: 6 × 0.50 J = 3.0 J.
2. Two squares of Eth: 2 × 0.50 J = 1.0 J.
3. U = mgh, so h = 3.0 J ÷ (0.20 kg × 9.81 N/kg) = 3.0 ÷ 1.962 = 1.5 m.
4. K = 4 × 0.50 J = 2.0 J, so v = √(2K/m) = √(2 × 2.0 ÷ 0.20) = √20 = 4.5 m/s.
Problem 8 Spring-launched ball
1. Us = ½ × 200 × (0.060)² = 0.36 J.
2. At the top the speed is zero, so all the spring energy has become gravitational: mgh = 0.36 J. h = 0.36 ÷ (0.050 × 9.81) = 0.36 ÷ 0.4905 = 0.73 m above the release position.
Problem 9 Crate on a rough ramp
1. ΔU = mgh = 25 × 9.81 × 1.2 = 294 J.
2. K = ½mv² = ½ × 25 × (2.0)² = 50 J.
3. Eint = 294.3 − 50 = 244 J.
4. Eint = fkd, so fk = 244.3 ÷ 3.0 = 81 N.
Problem 10 Climbing power
1. ΔU = mgh = 65 × 9.81 × 4.2 = 2.68 × 10³ J (2678 J).
2. P = ΔU/t = 2678 ÷ 9.0 = 298 W.
Problem 11 Motorbike on the highway
1. At constant velocity the net force is zero, so the engine force equals the resistive force: 350 N.
2. P = Fv = 350 × 20 = 7.0 × 10³ W (7.0 kW).
3. t = 900 s, so E = Pt = 7000 × 900 = 6.3 × 10⁶ J, and 6.3 × 10⁶ ÷ 3.6 × 10⁶ = 1.75 kWh.
Problem 12 Kettle energy
1. E = Pt = 1500 W × 180 s = 2.7 × 10⁵ J.
2. 2.7 × 10⁵ ÷ 3.6 × 10⁶ = 0.075 kWh. (Check: 1.5 kW × 0.050 h = 0.075 kWh.)
Problem 13 Reading an energy graph
1. P = slope = 240 J ÷ 4.0 s = 60 W.
2. The line is flat, so 0 W: the motor is switched off or idle.
3. P = (640 − 240) J ÷ 4.0 s = 100 W.
4. Pavg = 640 J ÷ 10 s = 64 W.
Problem 14 Swinging pendulum
1. v = √(2gh) = √(2 × 9.81 × 0.20) = √3.924 = 2.0 m/s (1.98 m/s).
2. K = mgh = 0.30 × 9.81 × 0.20 = 0.59 J.
3. Half the speed means a quarter of the kinetic energy, so ¼ of the energy is kinetic and ¾ is potential. h = ¾ × 0.20 = 0.15 m.
Problem 15 Minimum height for a loop
1. At the top both weight and the normal force point down; at the limit the normal force is zero, so mg = mv²/R, which gives v² = gR.
2. Take the bottom as h = 0. mgH = mg(2R) + ½mv² = mg(2R) + ½mgR = 2.5mgR. So H = 2.5R = 2.5 × 3.0 m = 7.5 m.
Problem 16 Block and spring on rough floor
1. Us = ½ × 500 × (0.20)² = 10 J.
2. fk = μkmg = 0.30 × 2.0 × 9.81 = 5.9 N.
3. All the spring energy ends as thermal energy: 10 J = fkd, so d = 10 ÷ 5.886 = 1.7 m from the release position.
Problem 17 Uphill cycling
1. F = mg sinθ = 80 × 9.81 × sin 5.0° = 784.8 × 0.0872 = 68 N.
2. The rider moves at constant speed, so P = Fv = 68.4 × 4.0 = 2.7 × 10² W (274 W).
3. Pchem = 274 ÷ 0.24 = 1.1 × 10³ W.
Problem 18 Pumping water from a well
1. 1500 kg ÷ 60 s = 25 kg/s.
2. Lifting: mgh per second = 25 × 9.81 × 12 = 2943 W. Kinetic: ½mv² per second = ½ × 25 × (3.0)² = 112.5 W. Total useful power = 3055 W ≈ 3.1 × 10³ W.
3. Pin = 3055 ÷ 0.70 = 4365 W ≈ 4.4 × 10³ W.
Problem 19 Mass on a vertical spring
1. At maximum extension the mass is momentarily at rest. Take the lowest point as the final state: mgx = ½kx², so x = 2mg/k = 2 × 0.50 × 9.81 ÷ 40 = 0.25 m (0.245 m).
2. At x/2 = 0.1226 m: K = mg(x/2) − ½k(x/2)² = 0.6015 − 0.3007 = 0.3007 J. v = √(2 × 0.3007 ÷ 0.50) = 1.1 m/s.
3. The net force is zero at x = mg/k = 0.12 m, but the mass arrives there with kinetic energy, so it overshoots; the spring force then exceeds the weight and decelerates it.
Problem 20 Spot the error
1. He used the length of the slope (5.0 m) as the height. The height is the vertical drop: h = 5.0 sin 30° = 2.5 m.
2. v = √(2gh) = √(2 × 9.81 × 2.5) = √49.05 = 7.0 m/s.
3. No change: mass cancels in mgh = ½mv², so a 4.0 kg cart reaches the same speed (ignoring friction).
