Work, energy and power

NUST NET (Engineering) · Physics · 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 part of the force along the motion counts, so a force at right angles to the motion does no work. If the force changes with distance, the work is the area under the force-displacement graph.

Work done on a body changes its energy. The net work equals the change in kinetic energy. When there is no friction, kinetic plus potential energy stays constant, which often gives the answer faster than the equations of motion. With friction, the work done against it is the mechanical energy lost.

Power is the rate of doing work.

Rules to remember

Common mistake

Treating kinetic energy as proportional to speed. It depends on v2, so doubling the speed makes the kinetic energy (and the braking distance) four times as large.

Worked example

A spring of spring constant 200 N m−1 is compressed by 0.10 m and used to fire a 0.020 kg ball along a smooth horizontal track. With what speed does the ball leave the spring?

  1. Energy stored = ½kx2 = ½ × 200 × (0.10)2 = 1.0 J.
  2. All of it becomes kinetic energy: ½mv2 = 1.0 J.
  3. v2 = (2 × 1.0)/0.020 = 100, so v = 10.

Answer: 10 m s−1

Practice questions

Try each one, then open the answer.

1. A 1000 kg car climbs a slope that rises 1 m for every 20 m along the road, at a steady 10 m s−1 against a friction force of 500 N. Taking g = 10 m s−2, the power output of its engine is

  1. A
    5 kW
  2. B
    10 kW
  3. C
    100 kW
  4. D
    105 kW
Show answer

Answer: B. Weight component along the slope = mg sin θ = 10 000 × 1/20 = 500 N. Total force = 500 + 500 = 1000 N, so P = Fv = 1000 × 10 = 10 kW. Ignoring the slope gives 5 kW.

2. A motor lifts a 50 kg load vertically at a steady 2.0 m s−1. Taking g = 10 m s−2, its useful power output is

  1. A
    100 W
  2. B
    250 W
  3. C
    500 W
  4. D
    1000 W
Show answer

Answer: D. At steady speed the lifting force equals the weight, 500 N, so P = Fv = 500 × 2.0 = 1000 W. 100 W leaves out g, and 250 W divides by the speed instead of multiplying.

3. A 5 kg body has a kinetic energy of 40 J. Its momentum is

  1. A
    20 kg m s−1
  2. B
    10 kg m s−1
  3. C
    8 kg m s−1
  4. D
    200 kg m s−1
Show answer

Answer: A. p = √(2mK) = √(2 × 5 × 40) = √400 = 20 kg m s−1. (Or v = √(2K/m) = 4 m s−1, and p = 5 × 4.) 200 is simply m × K, which is not momentum.

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