The problems are on the work energy theorem worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Work done is the energy a force transfers, W = F s cos θ, measured in joules. These worked answers correct the 20 problems of the Physics 11-12 worksheet on work energy theorem, from key relationships to wind turbine power: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the formula, the substitution and the result with its unit. Compare your method first, because a correct method with a small arithmetic slip earns most of the credit in an exam. Round only at the end.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB A.3
Problem 1 Key relationships
1. W = F s cos θ.2. KE = ½ m v² and ΔEp = 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, ΔEp = 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. ΔEp = 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. Wnet = 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. ΔEp = m g h = 8.0 × 9.81 × 2.5 = 196.2 J.
3. Wf = 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. ΔEp = 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.
