Specification: AQA A-level Physics (7408) 3.4.1.7 | Paper: 1 (and 2 as assumed knowledge) | Data and formulae booklet: given in the exam
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📌 KEY DEFINITIONS – learn them word for word
Work done: the product of the force and the distance moved in the direction of the force. Work done is equal to the energy transferred.
Area under a force–displacement graph: the area under a force–displacement graph is equal to the work done, even when the force varies.
The joule: 1 J is the work done when a force of 1 N moves its point of application 1 m in the direction of the force, so 1 J = 1 N m.
The equation
work done = force × displacement × cos θ W = Fs cos θ
| Quantity | Symbol | Unit |
|---|---|---|
| work done | W | joules (J), equivalent to N m |
| force | F | newtons (N) |
| displacement | s | metres (m) |
| angle between the force and the displacement | θ | degrees (°) |
Rearranged: F = W ÷ (s cos θ) | s = W ÷ (F cos θ) | cos θ = W ÷ (Fs) | θ is the angle between the force and the direction of motion: when θ = 0, cos θ = 1; when θ = 90°, cos θ = 0 and no work is done. Check your calculator is in degrees.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) A horizontal force of 25 N pushes a box 4.0 m along a horizontal floor in the direction of the force. Calculate the work done by the force.
Show solution
F: W = Fs cos θ
I: W = 25 × 4.0 × cos 0°
F: W is already the subject (cos 0° = 1)
A: 100 J
Q2 (Core) A child pulls a sledge 15 m across flat snow using a rope at 30° to the horizontal. The tension in the rope is 40 N. Calculate the work done by the tension, to 2 significant figures.
Show solution
F: W = Fs cos θ
I: W = 40 × 15 × cos 30°
F: W is already the subject
A: 600 × 0.866… = 519.6… J = 520 J (2 s.f.)
Watch out: only the horizontal component of the tension (F cos θ) does work, because the sledge moves horizontally.
Q3 (Core) A student lifts a 500 g book vertically from the floor onto a shelf 180 cm high at constant speed. Calculate the work done on the book, to 2 significant figures. (g = 9.81 N kg⁻¹)
Show solution
Convert first: 500 g = 0.500 kg; 180 cm = 1.80 m
F: W = Fs cos θ, with F = mg (constant speed) and θ = 0°
I: W = 0.500 × 9.81 × 1.80 × 1
F: W is already the subject
A: 8.829 J = 8.8 J (2 s.f.)
Q4 (Core) A traveller pulls a wheeled suitcase along a level platform with a force of 80 N at 40° to the horizontal. The force does 1.5 kJ of work. Calculate the distance travelled, to 2 significant figures.
Show solution
Convert first: 1.5 kJ = 1500 J
F: W = Fs cos θ
I: 1500 = 80 × s × cos 40°
F: s = 1500 ÷ (80 × cos 40°) = 1500 ÷ 61.28…
A: 24.47… m = 24 m (2 s.f.)
Q5 (Exam) A student lifts a 12 kg bag vertically through 0.80 m at constant speed, then carries it 20 m along a level corridor at the same height and constant speed. Calculate the total work done by the upward force of the student on the bag. (g = 9.81 N kg⁻¹)
Show solution
Step 1 – lifting
F: W = Fs cos θ
I: W = 12 × 9.81 × 0.80 × cos 0°
A: 94.176 J
Step 2 – carrying
the upward force is at θ = 90° to the horizontal displacement, so W = 12 × 9.81 × 20 × cos 90° = 0
A: total = 94 J (2 s.f.)
Watch out: a force perpendicular to the motion does no work – the same reason the centripetal force on an orbiting satellite does no work.
Q6 (Exam) A 1200 kg car rolls 400 m down a road inclined at 5.0° to the horizontal. Calculate the work done on the car by gravity, to 2 significant figures. (g = 9.81 N kg⁻¹)
Show solution
F: W = Fs cos θ, where θ is the angle between the weight (vertical) and the displacement (along the slope), θ = 90° − 5.0° = 85°
I: W = 1200 × 9.81 × 400 × cos 85°
F: W is already the subject (equivalently, W = mg × vertical drop = 11 772 × 400 sin 5.0° = 11 772 × 34.86…)
A: 4.10… × 105 J = 4.1 × 105 J (2 s.f.)
Watch out: using 400 m as the height. Gravity only does work over the vertical distance, 400 sin 5.0° = 34.9 m.
Q7 (Exam) A catapult is pulled back 5.0 cm. The force–displacement graph is a straight line from 0 N at 0 cm to 300 N at 5.0 cm. All the stored energy is transferred to a 30 g stone. Calculate the launch speed of the stone, to 2 significant figures.
Show solution
Convert first: 5.0 cm = 0.050 m; 30 g = 0.030 kg
Step 1 – work done = area under the graph
W = ½ × 0.050 × 300 = 7.5 J
Step 2 – speed
F: Eₖ = ½ m v²
I: 7.5 = ½ × 0.030 × v²
F: v = √(2 × 7.5 ÷ 0.030) = √500 = 22.36… m s⁻¹
A: 22 m s⁻¹ (2 s.f.)
Watch out: using F × s = 300 × 0.050 = 15 J doubles the energy. The force is not constant, so you must use the area of the triangle.
Q8 (Exam) A tractor drags a log 400 m along level ground with a chain. The tension in the chain is 900 N and it does 2.7 × 105 J of work. Calculate the angle between the chain and the ground, to 2 significant figures.
Show solution
F: W = Fs cos θ
I: 2.7 × 105 = 900 × 400 × cos θ
F: cos θ = 2.7 × 105 ÷ 3.6 × 105 = 0.75, so θ = cos⁻¹ 0.75
A: 41.4…° = 41° (2 s.f.)
Q9 (Stretch) A 40 kg crate is pulled 5.0 m up a ramp inclined at 15° to the horizontal, at constant speed. The rope pulls with a force of 300 N at 25° above the surface of the ramp. Calculate the work done against friction and the average friction force, to 2 significant figures. (g = 9.81 N kg⁻¹)
Show solution
Step 1 – work done by the rope
W = Fs cos θ = 300 × 5.0 × cos 25° = 1359.4… J
Step 2 – work done against gravity
vertical height = 5.0 × sin 15° = 1.294… m; ΔEₚ = mgΔh = 40 × 9.81 × 1.294… = 507.8… J
Step 3 – work done against friction
the speed is constant, so no Eₖ is gained: 1359.4… − 507.8… = 851.6… J
A: work against friction = 850 J (2 s.f.); friction force = 851.6… ÷ 5.0 = 170.3… N = 170 N (2 s.f.)
Watch out: the angle in W = Fs cos θ is measured from the slope (25°), not from the horizontal (40°).
Q10 (Stretch) A 0.80 kg trolley starts from rest on a level frictionless track. It is pulled by a string at 30° above the horizontal. The graph of tension against horizontal displacement rises steadily from 0 N to 2.0 N over the first 1.0 m and then stays at 2.0 N for a further 1.5 m. Calculate the speed of the trolley after 2.5 m, to 2 significant figures.
Show solution
Step 1 – area under the graph
area = triangle + rectangle = ½ × 1.0 × 2.0 + 2.0 × 1.5 = 1.0 + 3.0 = 4.0 N m
Step 2 – work done in the direction of motion
the graph shows the tension, not its horizontal component, so W = 4.0 × cos 30° = 3.464… J
Step 3 – speed
F: W = ΔEₖ = ½ m v²
I: 3.464… = ½ × 0.80 × v²
F: v = √(2 × 3.464… ÷ 0.80) = 2.942… m s⁻¹
A: 2.9 m s⁻¹ (2 s.f.)
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