The questions are on the work energy theorem test. Take it under real conditions before reading the mark scheme.
Quick answer: Work done is the energy a force transfers, W = F s cos θ, measured in joules. This mark scheme gives the model answers and the marks for the 8 questions of the Physics 11-12 test on work energy theorem, from definitions to heavy and light balls, with the equivalent answers to accept.
The model answers show the full method and the final answer in bold. Marks are given for the method as well as the result, so a wrong answer with correct working still earns most of the marks. Check your units too.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB A.3
Question 1 Definitions (2 marks)
- Work done is the energy transferred by a force, equal to force × displacement in the direction of the force.
- The net work done on an object equals its change in kinetic energy.
Marking: 1 mark each; accept W = F s cos θ with symbols defined for (a).
Question 2 Rope on a sack (3 marks)
- W = F s cos θ = 90 × 8.0 × cos 25° = 720 × 0.906 = 653 J.
- Friction does −40 × 8.0 = −320 J, so Wnet = 653 − 320 = 333 J.
Marking: (a) 1 mark for the cosine term, 1 for the result; (b) 1 mark.
Question 3 Carrying a box (2 marks)
The lifting force acts upwards, while the box moves horizontally. The force is at 90° to the displacement, so cos 90° = 0 and the work done is zero.Marking: 1 mark for force perpendicular to displacement; 1 mark for W = F s cos 90° = 0.
Question 4 Smooth and rough ramp (3 marks)
- m g h = ½ m v², so v = √(2 × 9.81 × 0.45) = √8.829 = 3.0 m/s.
- KE = 2.0 × 9.81 × 0.45 − 1.5 = 7.329 J, so v = √(2 × 7.329 ÷ 2.0) = 2.7 m/s.
Marking: (a) 1 for energy equation, 1 for result; (b) 1 for result.
Question 5 Hoist efficiency (3 marks)
- P = F v = m g v = 120 × 9.81 × 0.50 = 589 W.
- Efficiency = 589 ÷ 900 × 100 % = 65 %.
Marking: (a) 1 for F = m g, 1 for result; (b) 1 mark.
Question 6 Force-displacement graph (3 marks)
- Area = 30 × 4 = 120 J.
- Area of trapezium = ½ × (30 + 10) × 4 = 80 J.
- Total work = 200 J = ½ × 5.0 × v², so v = √80 = 8.9 m/s.
Marking: 1 mark for each part.
Question 7 Doubling the speed (2 marks)
The braking work F d equals the kinetic energy, so F d = ½ m v². With the same F, d is proportional to v². Doubling v multiplies v² by 4, so the distance is four times as large.Marking: 1 mark for F d = ½ m v²; 1 mark for d ∝ v² and the factor 4.
Question 8 Heavy and light balls (2 marks)
m g h = ½ m v², so v = √(2 g h) and the mass cancels. Both reach the ground with the same speed √(2 × 9.81 × 4.0) = 8.9 m/s. The claim is wrong.Marking: 1 mark for the energy equation with m cancelling; 1 mark for the conclusion.
