The Physics Question Bank

Course: Combined Science + Separate Physics  |  Tier: Foundation + Higher  |  Equation sheet: given in the exam

The equation

work done = force × distance (along the line of action of the force)    W = F s

Quantity	Symbol	Unit
work done	W	joules (J)
force	F	newtons (N)
distance	s	metres (m)

Rearranged: F = W ÷ s   |   s = W ÷ F

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.

Foundation: aim for Q1–8.   Higher: aim for Q4–10. Most questions need rearranging.

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Questions

Q1 (F) A shopper pushes a trolley with a force of 40 N for 25 m. Calculate the work done.

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F: W = F s
I: W = 40 × 25
F: W is already the subject – no rearranging needed
A: 1000 J

Q2 (F) A child does 600 J of work pulling a sledge 15 m across snow. Calculate the pulling force.

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F: W = F s
I: 600 = F × 15
F: F = 600 ÷ 15
A: 40 N

Q3 (F) A lift motor exerts a force of 800 N and does 12 000 J of work. How far does the lift move?

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F: W = F s
I: 12 000 = 800 × s
F: s = 12 000 ÷ 800
A: 15 m

Q4 (F) A cyclist pushes with a driving force of 150 N and does 4.5 kJ of work. How far does she travel?

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Convert first: 4.5 kJ = 4500 J
F: W = F s
I: 4500 = 150 × s
F: s = 4500 ÷ 150
A: 30 m

Q5 (F) A tractor does 1.2 MJ of work pulling a plough 480 m across a field. Calculate the pulling force in kN.

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Convert first: 1.2 MJ = 1 200 000 J
F: W = F s
I: 1 200 000 = F × 480
F: F = 1 200 000 ÷ 480 = 2500 N
A: 2500 ÷ 1000 = 2.5 kN

Q6 (F/H) A car’s brakes do 360 kJ of work to stop the car in 45 m. Calculate the braking force.

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Convert first: 360 kJ = 360 000 J
F: W = F s
I: 360 000 = F × 45
F: F = 360 000 ÷ 45
A: 8000 N

Q7 (F/H) A ferry’s engine provides a driving force of 24 kN and does 4.8 MJ of work. How far does the ferry travel? Give your answer in km.

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Convert first: 24 kN = 24 000 N; 4.8 MJ = 4 800 000 J
F: W = F s
I: 4 800 000 = 24 000 × s
F: s = 4 800 000 ÷ 24 000 = 200 m
A: 200 ÷ 1000 = 0.2 km

Q8 (F/H) A freight train does 3.0 × 10⁹ J of work over a 15 km journey. Calculate the average driving force. Give your answer in standard form.

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Convert first: 15 km = 15 000 m = 1.5 × 10⁴ m
F: W = F s
I: 3.0 × 10⁹ = F × 1.5 × 10⁴
F: F = 3.0 × 10⁹ ÷ 1.5 × 10⁴
A: 2.0 × 10⁵ N

Q9 (H) A weightlifter does 2352 J of work lifting a barbell 2.0 m straight up at a steady speed. Calculate the mass of the barbell. (g = 9.8 N/kg)

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Step 1 – find the lifting force
F: W = F s
I: 2352 = F × 2.0
F: F = 2352 ÷ 2.0 = 1176 N
Step 2 – the lifting force equals the weight
F: W = m g
I: 1176 = m × 9.8
F: m = 1176 ÷ 9.8
A: 120 kg

Q10 (H) A car’s engine does 7.5 MJ of work during a 12.6 km journey. Calculate the average driving force. Give your answer to 2 significant figures.

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Convert first: 7.5 MJ = 7 500 000 J; 12.6 km = 12 600 m
F: W = F s
I: 7 500 000 = F × 12 600
F: F = 7 500 000 ÷ 12 600
A: 595.2… = 600 N (2 s.f.)