Work-energy theorem worksheet answers for AP Physics 1 (free PDF download)

Problem 1 Sign of the work

1. Positive: gravity points down and the apple moves down.
2. Negative: friction points against the motion.
3. Zero: the normal force is perpendicular to the horizontal displacement.
4. Zero: the tension is perpendicular to the velocity at every point.
Positive, negative, zero, zero.

Problem 2 Market cart

1. W = F d = 120 × 18 = 2.2 × 10³ J (2160 J).
2. Then we substitute: d = W ÷ F = 4500 ÷ 150 = 30 m.

Problem 3 Tennis serve

1. K = ½ m v² = ½ × 0.057 × 45² = 58 J.
2. Then we substitute: K ∝ v², so doubling the speed gives 4 × 57.7 = 2.3 × 10² J.

Problem 4 Units of energy

1. 1 J = 1 N m = 1 (kg m/s²) × m = 1 kg m²/s².
2. Then we substitute: 2.4 kJ = 2400 J.
3. Finally, we calculate: 72 km/h = 72 ÷ 3.6 = 20 m/s, so K = ½ × 1000 × 20² = 2.0 × 10⁵ J.

Problem 5 Angle matters

W = F d cosθ with F d = 200 J.
1. 200 × 1 = 200 J.
2. Then we substitute: 200 × 0.5 = 100 J.
3. Then we calculate: 200 × 0 = 0 J.
4. Finally, we calculate: 200 × (−1) = −200 J.

Problem 6 From work to speed

1. By the theorem, ΔK = W_net = 150 J.
2. Then we substitute: ½ m v² = 150, so v = √(2 × 150 ÷ 3.0) = 10 m/s.

Problem 7 Force along a rail

1. Area of the triangle = ½ × 4.0 × 30 = 60 J.
2. Then add the rectangle: 6.0 × 30 = 180 J, so the total is 60 + 180 = 240 J.
3. Finally, we calculate: ½ m v² = 240, so v = √(2 × 240 ÷ 5.0) = 9.8 m/s.

Problem 8 Loaded trolley

1. W = 30 × 12 × cos 35° = 2.9 × 10² J (295 J).
2. Then we substitute: W = −14 × 12 = −168 J.
3. Finally, we calculate: W_net = 295 − 168 = 127 J. Then v = √(2 × 127 ÷ 15) = 4.1 m/s. Gravity and the normal force do zero work.

Problem 9 Frictionless ramp

1. The height lost is 2.5 × sin 30° = 1.25 m, so W = m g h = 0.30 × 9.81 × 1.25 = 3.7 J.
2. Zero, since the normal force is perpendicular to the motion.
3. Finally, we calculate: ½ m v² = 3.68, so v = √(2 × 3.68 ÷ 0.30) = 5.0 m/s.

Problem 10 Skateboard rolling to rest

1. K = ½ × 60 × 5.0² = 750 J.
2. Then we substitute: F d = ΔK, so F = 750 ÷ 12 = 62.5 N (about 63 N).
3. Finally, we calculate: μ = F ÷ (m g) = 62.5 ÷ (60 × 9.81) = 0.11.

Problem 11 Curling stone

1. K = ½ × 19 × 2.4² = 55 J (54.7 J).
2. Then we substitute: f = K ÷ d = 54.7 ÷ 33 = 1.7 N.
3. Finally, we calculate: μ = 1.66 ÷ (19 × 9.81) = 0.0089, very small because ice is slippery.

Problem 12 Kinetic energy against distance

1. Gradient = 100 J ÷ 5.0 m = 20 N.
2. Then we substitute: ½ m v² = 100 J, so m = 2 × 100 ÷ 6.0² = 5.6 kg.
3. Finally, we calculate: a = F ÷ m = 20 ÷ 5.56 = 3.6 m/s².

Problem 13 Doubling the speed

1. F d = ½ m v², so d = ½ × 1200 × 15² ÷ 6000 = 22.5 m.
2. Then we substitute: d = ½ × 1200 × 30² ÷ 6000 = 90 m.
3. Finally, we calculate: K depends on v², so doubling the speed quadruples the kinetic energy and the distance (4 × 22.5 = 90 m), not doubles it.

Problem 14 Stretching a spring

1. A straight line through the origin; the area under it is the work done by the stretching force.
2. Then we substitute: W = ½ k x² = ½ × 150 × 0.20² = 3.0 J.
3. Finally, we calculate: W = ½ × 150 × (0.40² − 0.20²) = ½ × 150 × 0.12 = 9.0 J. The second half costs three times as much, because the force is larger there.

Problem 15 Ranking kinetic energies

1. K = ½ × 0.15 × 30² = 67.5 J.
2. Then we substitute: K = ½ × 65 × 6.0² = 1170 J.
3. Finally, we calculate: K = ½ × 0.0040 × 400² = 320 J.
Runner (1.2 × 10³ J) > bullet (3.2 × 10² J) > ball (68 J).

Problem 16 Pushing a crate up a ramp

1. W = −m g sin 20° × d = −25 × 9.81 × 0.342 × 4.0 = −3.4 × 10² J (−336 J).
2. Then we substitute: f = μ m g cos 20° = 0.25 × 245 × 0.940 = 57.6 N, so W = −2.3 × 10² J (−230 J).
3. Equilibrium along the ramp: F = 83.9 + 57.6 = 1.4 × 10² N.
4. Finally, we calculate: W = 141.5 × 4.0 = 5.7 × 10² J. Sum: 566 − 336 − 230 = 0, because the speed does not change (ΔK = 0).

Problem 17 Force that changes sign

1. Triangle: ½ × 4.0 × 12 = 24 J.
2. Triangle below the axis: ½ × 2.0 × (−6.0) = −6.0 J.
3. Then we calculate: K_i = ½ × 2.0 × 3.0² = 9.0 J; K_f = 9.0 + 24 − 6.0 = 27 J; v = √(2 × 27 ÷ 2.0) = 5.2 m/s.
4. Finally, we calculate: At x = 4.0 m, where the force changes sign and the net work is largest: K = 9 + 24 = 33 J, so v = √33 = 5.7 m/s.

Problem 18 Constant velocity paradox

1. W = 40 × 6.0 = 240 J.
2. Friction equals 40 N and opposes the motion, so W = −240 J.
3. The claim is wrong. Bilal does 240 J of positive work, but friction does −240 J, so the net work is zero. The theorem then gives ΔK = 0, consistent with constant speed. The 240 J does not vanish: it is transferred to thermal energy of the box and the floor.

Problem 19 Design: work and kinetic energy

1. Measure the distance d travelled from rest with a metre rule and the speed v at the end with a light gate or video; measure the masses on a balance.
2. Tilt the track slightly until the cart rolls at constant speed when pushed, so the weight component along the track cancels friction.
3. Next, plot ΔK = ½ (m + M) v² against W = M g d for the system. The gradient should be 1.0.
4. Residual friction or pulley friction removes some work, so ΔK is less than M g d and the gradient comes out below 1.

Problem 20 Falling through the air

1. W_g = 75 × 9.81 × 120 = 8.8 × 10⁴ J (88 290 J); ΔK = ½ × 75 × 30² = 3.4 × 10⁴ J (33 750 J).
2. Then we substitute: W_air = ΔK − W_g = 33 750 − 88 290 = −5.5 × 10⁴ J; average drag = 54 540 ÷ 120 = 4.5 × 10² N.
3. Finally, we calculate: ΔK = 0, so W_air = −W_g = −75 × 9.81 × 300 = −2.2 × 10⁵ J.