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OCR A Physics · Module 3: Forces and motion

Work, energy & power: practice questions with answers

6 exam-style questions on work, energy & power. Try each one on paper first, then open the answer to see the full working and the common slips that cost marks.

  1. Resolving vectors2 marks

    A student pulls a suitcase 12 m along a level floor using a handle at 40° above the horizontal. The pull is 45 N. Calculate the work done by the pull.

    W = J

    Show the answer and working

    Answer W = 414 J

    W = Fs cos θ = 45 × 12 × cos 40° = 414 J.

    • Got 347 J? You used sin 40°. Work done = Fs cos θ, using the component along the direction of motion: 45 × 12 × cos 40° = 414 J.
    • Got 540 J? You ignored the angle. Only the horizontal component does work: W = 45 × 12 × cos 40° = 414 J.
  2. Rearranging equations3 marks

    A 0.15 kg ball is dropped from a height of 6.0 m. Air resistance transfers 15% of its initial gravitational potential energy to thermal energy. Calculate its speed just before it hits the ground.

    v = m s⁻¹

    Show the answer and working

    Answer v = 10 m s⁻¹

    ½mv² = 0.85 × mgh, so v = √(2 × 0.85 × 9.81 × 6.0) = 10 m s⁻¹. The mass cancels.

    • Got 10.8 m s⁻¹? You ignored the 15% loss. Only 85% of mgh becomes kinetic energy: ½v² = 0.85 × 9.81 × 6.0, so v = 10 m s⁻¹.
    • Got 100 m s⁻¹? You stopped before taking the square root. ½mv² = 0.85mgh gives v² = 100, so v = 10 m s⁻¹.
  3. Using the right equation3 marks

    A lift motor raises a 650 kg load through 18 m in 24 s at constant speed. The motor is 80% efficient. Calculate the electrical power input to the motor.

    P = kW

    Show the answer and working

    Answer P = 5.98 kW

    Useful power = mgh ÷ t = 650 × 9.81 × 18 ÷ 24 = 4.78e+03 W. Input = 4.78e+03 ÷ 0.80 = 5.98 kW.

    • Got 3.83 kW? You multiplied by the efficiency instead of dividing. Input = useful output ÷ efficiency = 4.78e+03 W ÷ 0.80 = 5.98 kW.
    • Got 4.78 kW? This is the useful output power only. Divide by the efficiency: 4.78 ÷ 0.80 = 5.98 kW.
  4. Unit prefixes2 marks

    A car moves at a constant 30 m s⁻¹ against a total resistive force of 480 N. Calculate the useful power delivered to the wheels, in kW.

    P = kW

    Show the answer and working

    Answer P = 14.4 kW

    At constant speed, driving force = resistive force, so P = Fv = 480 × 30 = 14 400 W = 14.4 kW.

    • Got 14400 kW? That is the power in watts. Divide by 1000 to get kW: 14 400 W = 14.4 kW.
    • Got 0.016 kW? You divided force by speed. Power = Fv = 480 × 30 = 14 400 W = 14.4 kW.
  5. Reading graphs1 mark

    What does the area under a force-displacement graph represent?

    • A Power
    • B Work done
    • C Momentum
    Show the answer and working

    Answer B: Work done

    Force × displacement is work done, so the area under the graph is the work.

    • Why not A? Power is work done per unit time, which needs a time axis.
    • Why not C? Momentum relates to force against time (impulse), not displacement.
  6. Using the right equation1 mark

    A box of mass m is raised through height h, once by lifting it vertically and once by pushing it up a smooth ramp to the same height. Which statement is correct?

    • A More energy is transferred when it is lifted vertically.
    • B The gain in gravitational potential energy is greater on the ramp because the path is longer.
    • C The gain in gravitational potential energy is mgh in both cases.
    Show the answer and working

    Answer C: The gain in gravitational potential energy is mgh in both cases.

    ΔGPE = mgΔh depends on the vertical change in height only.

    • Why not A? The gain in GPE depends only on the final height, not on the route.
    • Why not B? GPE depends on vertical height, not the length of the path.