AS Level Physics chapter 5: Work, energy and power MCQs with answers

By Team Megalecture · Updated 4 Nov 2026

50 exam-style questions on Work, energy and power, each with the answer and a short explanation. Ten of them are below to try. The chapter test draws ten fresh questions each time and is marked the moment you finish, with your score by topic.

50 questions · 2 topics · 10 per test

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What chapter 5 covers

Ten Work, energy and power questions with answers

Try each one, then open the answer. These ten stay the same; the test draws new ones every time.

1. A ball of mass 0.20 kg is thrown vertically upwards at 15 m s−1 and rises 9.0 m before stopping. What is the average force of air resistance on the ball as it rises? (g = 9.81 m s−2)

  1. A
    0.54 N
  2. B
    2.0 N
  3. C
    2.5 N
  4. D
    4.8 N
Show answer

Answer: A. KE lost = ½ × 0.20 × 152 = 22.5 J; GPE gained = 0.20 × 9.81 × 9.0 = 17.7 J. Work against drag = 4.8 J over 9.0 m, so F = 4.8 ÷ 9.0 = 0.54 N. 4.8 is the energy, not the force.

2. A car of mass 1200 kg travels at a constant 15 m s−1 up a straight road that rises 1.0 m for every 20 m travelled along the road. The total resistive force is 600 N. What is the useful output power of the engine? (g = 9.81 m s−2)

  1. A
    8.8 kW
  2. B
    9.0 kW
  3. C
    18 kW
  4. D
    190 kW
Show answer

Answer: C. Component of weight down the slope = 1200 × 9.81 × (1/20) = 589 N. Driving force = 600 + 589 = 1189 N, so P = Fv = 1189 × 15 = 18 kW. Ignoring the slope gives 9.0 kW.

3. The maximum useful output power of a car engine is 45 kW. At its top speed the total resistive force on the car is 1.5 kN. What is the top speed?

  1. A
    0.033 m s−1
  2. B
    3.0 m s−1
  3. C
    30 m s−1
  4. D
    300 m s−1
Show answer

Answer: C. At top speed the driving force equals the resistive force, and P = Fv, so v = 45 000 ÷ 1500 = 30 m s−1.

4. A book of mass 750 g is lifted from the floor onto a shelf 1.8 m above the floor. What is its gain in gravitational potential energy? (g = 9.81 m s−2)

  1. A
    1.4 J
  2. B
    6.6 J
  3. C
    13 J
  4. D
    1.3 × 104 J
Show answer

Answer: C. ΔEp = mgΔh = 0.750 × 9.81 × 1.8 = 13 J. Leaving the mass in grams gives 1.3 × 104 J; leaving out g gives 1.4 J.

5. Power is the work done per unit time. What is the watt expressed in SI base units?

  1. A
    kg m2 s−3
  2. B
    kg m2 s−2
  3. C
    kg m s−2
  4. D
    kg m s−3
Show answer

Answer: A. 1 W = 1 J s−1 and 1 J = 1 N m = kg m2 s−2, so 1 W = kg m2 s−3. kg m2 s−2 is the joule.

6. A climber of mass 65 kg climbs from an altitude of 1200 m to an altitude of 1850 m. What is her gain in gravitational potential energy? (g = 9.81 m s−2)

  1. A
    6.4 × 102 J
  2. B
    4.2 × 104 J
  3. C
    4.1 × 105 J
  4. D
    1.2 × 106 J
Show answer

Answer: C. ΔEp = mgΔh = 65 × 9.81 × (1850 − 1200) = 65 × 9.81 × 650 = 4.1 × 105 J.

7. A student of mass 55 kg runs up a staircase of 20 steps, each 0.18 m high, in 4.5 s. What is the useful power she develops in raising her body? (g = 9.81 m s−2)

  1. A
    44 W
  2. B
    430 W
  3. C
    1.9 kW
  4. D
    8.7 kW
Show answer

Answer: B. Height = 20 × 0.18 = 3.6 m; work = 55 × 9.81 × 3.6 = 1940 J; P = 1940 ÷ 4.5 = 430 W. Using mass for weight gives 44 W; 1.9 kW is the work done, not divided by the time.

8. A pump raises 1500 litres of water from a well into a tank 12 m higher. The density of water is 1000 kg m−3 and 1 litre = 1.0 × 10−3 m3. What is the gain in gravitational potential energy of the water? (g = 9.81 m s−2)

  1. A
    1.8 × 102 J
  2. B
    1.8 × 104 J
  3. C
    1.8 × 105 J
  4. D
    1.8 × 108 J
Show answer

Answer: C. Mass = 1000 × 1500 × 10−3 = 1500 kg, so ΔEp = 1500 × 9.81 × 12 = 1.8 × 105 J. Using the volume 1.5 m3 as the mass gives 1.8 × 102 J.

9. An object starts from rest and is accelerated by a constant resultant force along a straight line. Which describes the graph of its kinetic energy (y-axis) against the distance it has travelled (x-axis)?

  1. A
    a straight line through the origin whose gradient equals the resultant force
  2. B
    a straight line through the origin whose gradient equals the acceleration
  3. C
    a curve through the origin whose gradient increases with distance
  4. D
    a curve through the origin whose gradient decreases with distance
Show answer

Answer: A. Gain in KE = work done = Fs, so Ek ∝ s and the gradient Ek/s is the force F (in J m−1 = N). Ek against time would curve upwards.

10. A box of weight 200 N is pushed 5.0 m up a smooth slope inclined at 30° to the horizontal. How much work is done against gravity? (sin 30° = 0.50, cos 30° = 0.87)

  1. A
    500 J
  2. B
    870 J
  3. C
    1000 J
  4. D
    2000 J
Show answer

Answer: A. Work against gravity = weight × vertical rise = 200 × 5.0 sin 30° = 200 × 2.5 = 500 J. Using cos 30° gives 870 J; ignoring the angle gives 1000 J.

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Ways to practise this chapter

Work, energy and power: questions students ask

How many Work, energy and power MCQs are there?

50, each with the answer and a short explanation, across 2 topics: Energy conservation, Gravitational potential energy and kinetic energy. Every chapter test draws ten of them, so you can take it many times.

Is the AS Level Physics chapter 5 test free?

Yes. The test, the answers and the explanations are free and need no sign-up. Sign in only if you want your scores kept across devices.

How is the chapter test marked?

The moment you press Finish: your score, the right answer and an explanation for every question, and the ones you got wrong saved in My mistakes to retry.

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