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Next exam on November 14 – 15, 2026:
All concepts

Work and Power

功和功率

The defining equation for work done by a constant force and the sign convention, total work done by several forces, and average versus instantaneous power.

11 real-exam questions · 4 sittings21 in mock papers13 practice

Formulas

  • Work done by a constant force

    Definition
    W=Fscos⁡θW=Fs\cos\theta

    Applies when：θ is the angle between force and displacement; work is negative when θ>90°

  • Definition of power

    Definition
    P=WtP=\frac{W}{t}

    Applies when：Gives the average power over that interval

  • Power in terms of force and velocity

    Formula
    P=Fvcos⁡θP=Fv\cos\theta

    Applies when：Instantaneous v gives instantaneous power; average v gives average power

How the real exam tests it

Questions from real CSCA papers that test this concept. Work each one out before you open the answer.

  1. CSCA Physics December 2025 Past Paper
    A horizontal pulling force F of 20 N is applied to an object of mass 10 kg . The object moves 5 m on a rough horizontal surface with a friction force f=5 Nf = 5 \mathrm {~N}. The work done by the pulling force FF is ( ).
    • A．100 J
    • B．75 J
    • C．- 25 J
    • D．25 J
    Show answer & solution

    Answer: A

    1 Work by the pulling force: W=Fscos⁡0=20 N×5 m=100 JW = F s \cos 0 = 20 \mathrm {~N} \times 5 \mathrm {~m} = 100 \mathrm {~J}.
    See it in the full CSCA Physics December 2025 Past Paper
  2. CSCA Physics March 2026 Past Paper
    A horizontal force F pulls an object 5 m , doing 50 J of work. F equals
    • A．5 N
    • B．10 N
    • C．15 N
    • D．20 N
    Show answer & solution

    Answer: B

    W=FsW = F s (same direction, cos⁡0∘=1)\left. \cos 0 ^ { \circ } = 1 \right) . F=50/5=10 NF = 50 / 5 = 10 \mathrm {~N}. Answer: BB.
    Take the full CSCA Physics March 2026 Past Paper — timed & graded
  3. CSCA Physics April 2026 Past Paper
    As shown in the figure, a tower crane uses a steel cable to lift a load of mass mm vertically with acceleration aa through a height hh. The gravitational acceleration is gg. Which statement about this process is correct? Question 45 diagram
    • A．Work done by the cable tension is m(a+g)hm ( a + g ) h
    • B．Work done by gravity on the load is - mah
    • C．Work done by the net force on the load is mghm g h
    • D．The load's gravitational potential energy decreases by mghm g h
    Show answer & solution

    Answer: A

    Newton's second law in the vertical direction: T−mg=ma⇒T=m(g+a)T - m g = m a \Rightarrow T = m ( g + a ). (A) WT=T⋅h=m(g+a)hW _ { T } = T \cdot h = m ( g + a ) h - correct ✓\checkmark. (B) Gravity does negative work: WG=−mghW _ { G } = - m g h, not - mah - wrong. (C) Net-force work: Wnet =(T−mg)h=mahW _ { \text {net } } = ( T - m g ) h = m a h, not mghm g h - wrong. (D) The load rises by hh, so gravitational PE increases by mghm g h (not decreases) - wrong.
    Take the full CSCA Physics April 2026 Past Paper — timed & graded

Real-exam evidence

Recorded past-paper appearances are listed below; online verification is labelled separately.

11 questions across 4 sittings1 of 4 sittings available to verify free online

21 questions across 6 mock papers — our own papers, so a coverage signal rather than real-exam evidence

13 related practice questions availablePractise free: Work and Energy

Sitting evidence comes from the CSCA past papers on record; questions whose original paper is missing are excluded, and sprint papers are deliberately not counted.

Related concepts

Kinetic Energy and the Work–Energy TheoremConservation of Mechanical Energy