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AS & A Level · Practice questions · Power and driving force

Driving force is P over v.

Six original Cambridge-style questions on power, driving force, maximum speed and acceleration on level roads and hills, with full worked solutions.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
The method

F = P/v, then F = ma.

01
Recall
[2 marks]

Define power. A car's engine exerts a driving force of 400 N while the car moves at \(15\ \text{m s}^{-1}\). Find the power.

  • Power is the rate at which work is done. ✓
  • \(P = Fv = 400 \times 15 = 6000\) W \(= 6\) kW. ✓
02
Calculation
[3 marks]

A motor raises a load of mass 50 kg vertically through 12 m at constant speed in 8 s. Find the work done and the power of the motor.

  • Work \(= 50 \times 10 \times 12 = 6000\) J. ✓
  • Power \(= 6000 \div 8 = 750\) W. ✓
03
Calculation
[4 marks]

A car of mass 1200 kg has an engine working at 30 kW. It moves on a level road against a constant resistance of 1200 N. Find the maximum speed and the acceleration when the speed is \(15\ \text{m s}^{-1}\).

  • Top speed: \(30\,000 \div v = 1200\), so \(v = 25\ \text{m s}^{-1}\). ✓
  • At 15: \(F = 2000\) N, \(2000 - 1200 = 1200a\). ✓
  • \(a = 0.667\ \text{m s}^{-2}\). ✓
04
Analysis
[5 marks]

A car of mass 1000 kg moves on a level road with its engine working at a constant power \(P\) W against a constant resistance \(R\) N. Its maximum speed is \(40\ \text{m s}^{-1}\). When its speed is \(20\ \text{m s}^{-1}\) its acceleration is \(0.5\ \text{m s}^{-2}\). Find \(P\) and \(R\).

  • Top speed: \(\dfrac{P}{40} = R\). ✓
  • At 20: \(\dfrac{P}{20} - R = 1000 \times 0.5 = 500\). ✓
  • Substitute: \(\dfrac{P}{20} - \dfrac{P}{40} = \dfrac{P}{40} = 500\). ✓
  • \(P = 20\,000\) (20 kW) and \(R = 500\). ✓
05
Analysis
[5 marks]

A car of mass 1200 kg drives up a straight hill inclined at \(\alpha\) to the horizontal, where \(\sin\alpha = 0.05\). The engine works at 30 kW and the resistance is 1200 N. Find the maximum speed up the hill and the acceleration at the instant the speed is \(10\ \text{m s}^{-1}\).

  • Weight component \(= 1200 \times 10 \times 0.05 = 600\) N down the hill. ✓
  • Top speed: \(\dfrac{30\,000}{v} = 1800\), so \(v = 16.7\ \text{m s}^{-1}\). ✓
  • At 10: \(3000 - 1800 = 1200a\), so \(a = 1\ \text{m s}^{-2}\). ✓
06
Analysis
[6 marks]

A lorry of mass 8000 kg travels down a straight hill inclined at \(\alpha\), where \(\sin\alpha = 0.02\). The resistance to motion is a constant 4000 N and the engine works at 120 kW.

(a) Find the maximum speed of the lorry down the hill.

(b) Find its acceleration at the instant its speed is \(30\ \text{m s}^{-1}\).

  • Weight component down the hill \(= 80\,000 \times 0.02 = 1600\) N. ✓
  • (a) \(\dfrac{120\,000}{v} + 1600 = 4000\), so \(\dfrac{120\,000}{v} = 2400\) and \(v = 50\ \text{m s}^{-1}\). ✓
  • (b) At 30: driving force \(= 4000\) N. ✓
  • \(4000 + 1600 - 4000 = 8000a\), so \(a = 0.2\ \text{m s}^{-2}\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.