Work energy theorem worksheet answers for A Level physics (free PDF download)

Problem 1 Key relationships

1. W = F s cos θ.
2. KE = ½ m v² and ΔE_p = m g Δh.
3. P = W ÷ t and P = F v.
4. Efficiency = useful output ÷ total input × 100 %.
W = F s cos θ, KE = ½ m v², ΔE_p = m g Δh, P = W/t = F v.

Problem 2 Wheelbarrow of bricks

1. W = F s = 140 × 25 = 3500 J.
2. 3500 J ÷ 1000 = 3.5 kJ.
W = 3500 J = 3.5 kJ.

Problem 3 Convert, then calculate

1. 72 km/h = 72 000 m ÷ 3600 s = 20 m/s.
2. 3.6 MJ ÷ 3.6 MJ per kWh = 1.0 kWh.
3. E = P t = 0.450 kW × 2.0 h = 0.90 kWh, and 0.90 × 3.6 = 3.24 MJ.
20 m/s; 1.0 kWh; 0.90 kWh = 3.2 MJ.

Problem 4 A footballer’s kick

1. KE = ½ m v² = ½ × 0.43 × 24² = 0.215 × 576 = 123.8 J.
2. KE is proportional to v², so doubling the speed multiplies it by 4: 4 × 123.84 = 495 J. Check: ½ × 0.43 × 48² = 0.215 × 2304 = 495.4 J.
KE = 124 J; at 48 m/s, KE = 495 J.

Problem 5 Rice on the shelf

1. ΔE_p = m g Δh = 15 × 9.81 × 1.4 = 206 J.
2. The lifting force transfers this energy to the sack, so the work done is 206 J.
3. Doubling Δh doubles the gain: 15 × 9.81 × 2.8 = 412 J.
206 J; 206 J; 412 J.

Problem 6 Lift motor

1. P = W ÷ t = 48 000 ÷ 12 = 4000 W = 4.0 kW.
2. Efficiency = 4.0 ÷ 6.0 × 100 % = 66.7 %.
4.0 kW; efficiency = 67 %.

Problem 7 Dragging a sled

1. F cos θ = 65 × cos 40° = 65 × 0.766 = 49.8 N.
2. W = F s cos θ = 49.8 × 18 = 896 J.
3. A smaller angle makes cos θ larger, so a bigger part of the force acts along the motion. At 0° the full 65 N is used.
W = 896 J.

Problem 8 Positive, negative or zero

1. Negative: friction points opposite to the motion.
2. Zero: gravity is perpendicular to the displacement.
3. Zero: tension is always perpendicular to the velocity.
4. Positive: gravity points in the direction of motion.
5. W = F s cos 180° = 35 × 6.0 × (−1) = −210 J.
−210 J.

Problem 9 Hockey stick strike

1. W_net = ΔKE = ½ × 0.17 × 30² − 0 = 76.5 J.
2. F = W ÷ s = 76.5 ÷ 0.12 = 638 N.
W = 76.5 J; average F = 6.4 × 10² N.

Problem 10 Force on a runway

Work equals the area under the graph.
1. Triangle: ½ × 2 × 40 = 40 J.
2. Rectangle: 3 × 40 = 120 J.
3. Final triangle: ½ × 2 × 40 = 40 J, so the total is 40 + 120 + 40 = 200 J.
4. ½ m v² = 200, so v = √(2 × 200 ÷ 4.0) = √100 = 10 m/s.
200 J; 10 m/s.

Problem 11 Swinging bob

1. m g h = ½ m v², so v = √(2 g h) = √(2 × 9.81 × 0.35) = √6.867 = 2.6 m/s.
2. KE = m g h = 0.50 × 9.81 × 0.35 = 1.7 J.
3. The speed is unchanged because m cancels from both sides; only the kinetic energy doubles to 3.4 J.
v = 2.6 m/s; KE = 1.7 J.

Problem 12 Motorbike thrust

1. P = F v, so F = P ÷ v = 18 000 ÷ 25 = 720 N.
2. W = P t = 18 000 × 300 = 5.4 × 10⁶ J.
F = 720 N; W = 5.4 MJ.

Problem 13 Solar water pump

1. W = m g h = 540 × 9.81 × 6.0 = 31 784 J.
2. P = 31 784 ÷ 60 = 530 W.
3. Efficiency = 530 ÷ 750 × 100 % = 70.6 %.
4. Friction in the pump and thermal energy in the motor and wires warm the surroundings.
3.18 × 10⁴ J; 530 W; efficiency = 71 %.

Problem 14 Braking distance

1. KE = ½ × 950 × 15² = 475 × 225 = 106 875 J.
2. The work done by the brakes equals the kinetic energy: F d = KE, so d = 106 875 ÷ 4750 = 22.5 m.
3. KE at 30 m/s = ½ × 950 × 900 = 427 500 J, so d = 427 500 ÷ 4750 = 90 m, four times as far.
1.07 × 10⁵ J; 22.5 m; 90 m.

Problem 15 Find the errors

1. Error one: the force needed is the weight m g, not the mass, so g is missing. Error two: power is work divided by time, not multiplied.
2. W = m g h = 20 × 9.81 × 3.0 = 588.6 J, and P = W ÷ t = 588.6 ÷ 4.0 = 147 W.
W = 589 J; P = 1.5 × 10² W.

Problem 16 Sliding down a slope

1. h = s sin 30° = 5.0 × 0.50 = 2.5 m.
2. ΔE_p = m g h = 8.0 × 9.81 × 2.5 = 196.2 J.
3. W_f = 12 × 5.0 = 60 J.
4. KE = 196.2 − 60 = 136.2 J, so v = √(2 × 136.2 ÷ 8.0) = √34.05 = 5.8 m/s.
196 J lost; 60 J against friction; v = 5.8 m/s.

Problem 17 A bouncing ball

1. v = √(2 g h) = √(2 × 9.81 × 1.5) = √29.43 = 5.4 m/s.
2. v = √(2 × 9.81 × 1.1) = √21.58 = 4.6 m/s.
3. Energy lost = m g (1.5 − 1.1) = 0.060 × 9.81 × 0.40 = 0.235 J.
4. Fraction lost = 0.40 ÷ 1.5 = 26.7 %.
5.4 m/s; 4.6 m/s; 0.24 J lost; 27 %.

Problem 18 Lorry on a hill

1. m g sin θ = 12 000 × 9.81 × 0.050 = 5886 N.
2. Constant speed means zero net force: driving force = 6000 + 5886 = 11 886 N.
3. P = F v = 11 886 × 15 = 178 290 W = 178 kW.
4. Efficiency = 178.3 ÷ 450 × 100 % = 39.6 %.
1.19 × 10⁴ N; 178 kW; efficiency = 40 %.

Problem 19 Lift with a load

1. ΔE_p = 1500 × 9.81 × 30 = 441 450 J.
2. ΔKE = ½ × 1500 × 2.0² = 3000 J.
3. W = 441 450 + 3000 = 444 450 J.
4. P = 444 450 ÷ 18 = 24 692 W = 24.7 kW.
5. Input = 24.69 ÷ 0.85 = 29.0 kW.
W = 4.44 × 10⁵ J; useful P = 24.7 kW; input = 29.0 kW.

Problem 20 Wind turbine power

1. In one second a column of air of length v and volume A v passes, with mass ρ A v. Its kinetic energy is ½ (ρ A v) v² = ½ ρ A v³, so P = ½ ρ A v³.
2. A = π × 20² = 1257 m². P = ½ × 1.2 × 1257 × 8.0³ = 0.6 × 1257 × 512 = 3.86 × 10⁵ W.
3. Efficiency = 150 ÷ 386 × 100 % = 38.9 %.
4. P is proportional to v³, so doubling the speed multiplies it by 8: 8 × 386 = 3088 kW.
386 kW; efficiency = 39 %; 3.09 MW.