Energy, work and power

AS Level Mathematics · Mechanics. A short explanation of the idea, the rules to remember, the mistake to avoid, a worked example and practice questions with answers.

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The idea

Work is done when a force moves its point of application. Only the component of the force in the direction of motion does work. A force at right angles to the motion, such as the normal reaction, does no work.

A moving body has kinetic energy, and a body raised through a height gains gravitational potential energy. If no resistance or driving force acts, the total of these two is conserved. If such forces act, use the work–energy principle: work done by the driving force minus work done against resistance equals the gain in total mechanical energy.

Power is the rate of doing work. For a vehicle, power equals driving force multiplied by speed.

Rules to remember

Common mistake

Using the distance along the slope in mgh. Potential energy depends on the vertical height only, while work against friction uses the distance moved along the slope.

Worked example

A particle of mass 2 kg is projected up a rough slope at 6 m s−1. It comes to rest after travelling 3 m along the slope, having risen a vertical height of 1.2 m. Find the frictional force, assumed constant. (g = 10 m s−2)

  1. Loss of KE = (1/2) × 2 × 62 = 36 J.
  2. Gain in PE = 2 × 10 × 1.2 = 24 J.
  3. Work done against friction = 36 − 24 = 12 J.
  4. Friction × 3 = 12, so friction = 12 ÷ 3 = 4.

Answer: 4 N

Practice questions

Try each one, then open the answer.

1. Find the kinetic energy of a car of mass 800 kg moving at 15 m s−1.

  1. A
    180 kJ
  2. B
    90 kJ
  3. C
    6 kJ
  4. D
    12 kJ
Show answer

Answer: B. KE = (1/2)mv2 = (1/2) × 800 × 225 = 90 000 J = 90 kJ. Leaving out the 1/2 gives 180 kJ.

2. A block of mass 4 kg is pulled at constant speed 5 m up a rough plane inclined at 30° to the horizontal by a force parallel to the plane. The frictional force is 8 N. Find the work done by the pulling force. (g = 10 m s−2)

  1. A
    100 J
  2. B
    240 J
  3. C
    200 J
  4. D
    140 J
Show answer

Answer: D. No change in KE, so work done = gain in PE + work against friction = 40 × (5 sin 30°) + 8 × 5 = 100 + 40 = 140 J. The height gained is 2.5 m, not 5 m.

3. A car travels at a constant speed of 25 m s−1 along a level road against a total resistance of 800 N. Find the power output of the engine.

  1. A
    20 kW
  2. B
    32 W
  3. C
    2 kW
  4. D
    800 W
Show answer

Answer: A. At constant speed the driving force equals the resistance, 800 N. Power = Fv = 800 × 25 = 20 000 W = 20 kW.

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