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

Forces and Equilibrium

力与物体的平衡

The conditions for and magnitudes of gravity, elastic force and friction, composition and resolution of forces, and the equilibrium condition for concurrent forces.

20 real-exam questions · 4 sittings22 in mock papers11 practice

Formulas

  • Hooke's law

    Law
    F=kxF=kx

    Applies when：Within the elastic limit; x is the deformation and k the spring constant

  • Sliding friction

    Formula
    f=μFNf=\mu F_N

    Applies when：Applies while surfaces slide; Fₙ is the normal force, not necessarily the weight

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 spring with elasticity constant k=100 N/m\mathrm { k } = 100 \mathrm {~N} / \mathrm { m } is stretched from its original length of 10 cm to 15 cm under a force FF. The magnitude of the force FF is ().
    • A．5 N
    • B．25 N
    • C．10 N
    • D．15 N
    Show answer & solution

    Answer: A

    1 Use Hooke's law: F=k x. 2 Extension x=15 cm−10 cm=5 cm=0.05 m\mathrm { x } = 15 \mathrm {~cm} - 10 \mathrm {~cm} = 5 \mathrm {~cm} = 0.05 \mathrm {~m}. 3 F=100 N/m×0.05 m=5 NF = 100 \mathrm {~N} / \mathrm { m } \times 0.05 \mathrm {~m} = 5 \mathrm {~N}.
    See it in the full CSCA Physics December 2025 Past Paper
  2. CSCA Physics March 2026 Past Paper
    A spring has natural length 0.4 m and spring constant k=15 N/m\mathrm { k } = 15 \mathrm {~N} / \mathrm { m }. When compressed by 3 N , the new length is
    • A．0.2 m
    • B．0.4 m
    • C．0.6 m
    • D．0.8 m
    Show answer & solution

    Answer: A

    By Hooke's law: x=F/k=3/15=0.2 mx = F / k = 3 / 15 = 0.2 \mathrm {~m}. Compressed, so new length =0.40.2=0.2 m= 0.40 .2 = 0.2 \mathrm {~m}. If stretched, it would be 0.6 m (option C). Answer: A.
    Take the full CSCA Physics March 2026 Past Paper — timed & graded
  3. CSCA Physics April 2026 Past Paper
    A spring has stiffness k=50 N/mk = 50 \mathrm {~N} / \mathrm { m } and natural length 10 cm . If it is stretched to 14 cm , the applied force is
    • A．6 N
    • B．4 N
    • C．2 N
    • D．8 N
    Show answer & solution

    Answer: C

    Extension Δx=14−10=4 cm=0.04 m\Delta x = 14 - 10 = 4 \mathrm {~cm} = 0.04 \mathrm {~m}. By Hooke's law, F=kΔx=50×0.04=2 NF = k \Delta x = 50 \times 0.04 = 2 \mathrm {~N}.
    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.

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

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

11 related practice questions availablePractise free: Newton's Laws of Motion

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

Newton's Laws of MotionCircular MotionGravitation and Satellites