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

Kinetic Energy and the Work–Energy Theorem

动能与动能定理

The expression for kinetic energy, the statement and derivation of the work–energy theorem, and its use for variable forces and multi-stage processes.

6 real-exam questions · 4 sittings16 in mock papers18 practice

Formulas

  • Kinetic energy

    Definition
    Ek=12mv2E_k=\frac{1}{2}mv^2

    Applies when：A scalar, never negative, independent of the direction of motion

  • Work–energy theorem

    Law
    Wnet=ΔEk=12mv2−12mv02W_{\text{net}}=\Delta E_k=\frac{1}{2}mv^2-\frac{1}{2}mv_0^2

    Applies when：Holds for variable forces and curved paths too — its advantage over Newton's second law

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 ball of mass m=3 kgm = 3\,\mathrm{kg} moves on a smooth horizontal surface. An external force changes its speed from v0=2 m/sv_0 = 2\,\mathrm{m/s} to v=6 m/sv = 6\,\mathrm{m/s}. The work done by the external force is ( ).
    • A．54 J
    • B．60 J
    • C．48 J
    • D．6 J
    Show answer & solution

    Answer: C

    1 Work-energy theorem: W=ΔK=(1/2)m(v22−v12)\mathrm { W } = \Delta \mathrm { K } = ( 1 / 2 ) \mathrm { m } \left( \mathrm { v } 2 ^ { 2 } - \mathrm { v } 1 ^ { 2 } \right). 2  m=3 kg,v1=2 m/s,v2=6 m/s\mathrm {~m} = 3 \mathrm {~kg} , \mathrm { v } 1 = 2 \mathrm {~m} / \mathrm { s } , \mathrm { v } 2 = 6 \mathrm {~m} / \mathrm { s }. 3 W=0.5×3×(36−4)=1.5×32=48 JW = 0.5 \times 3 \times ( 36 - 4 ) = 1.5 \times 32 = 48 \mathrm {~J}.
    See it in the full CSCA Physics December 2025 Past Paper
  2. CSCA Physics April 2026 Past Paper
    A 20 N horizontal force moves an object 10 m , with friction 5 N . The increase in the kinetic energy of the object during this process is
    • A．50 J
    • B．200 J
    • C．100 J
    • D．150 J
    Show answer & solution

    Answer: D

    By the work-energy theorem, ΔEk=Wnet \Delta E _ { k } = W _ { \text {net } }. Net force =20−5=15 N= 20 - 5 = 15 \mathrm {~N}, so ΔEk=15×10=150 J\Delta E _ { k } = 15 \times 10 = 150 \mathrm {~J}.
    Take the full CSCA Physics April 2026 Past Paper — timed & graded
  3. CSCA Physics March 2026 Past Paper
    An object is thrown vertically upward from the ground. During the ascent, gravity does −100 J-100 \mathrm {~J} of work on it. Then
    • A．Kinetic energy decreases by 100 J
    • B．Gravitational PE decreases by 100 J
    • C．Mechanical energy decreases by 100 J
    • D．Kinetic energy increases by 100 J
    Show answer & solution

    Answer: A

    By the work-energy theorem, the net work equals the change in kinetic energy. With no air resistance, only gravity does work during the ascent: WG=−100 J=Ek2−Ek1W_G = -100 \mathrm {~J} = E_{k2} - E_{k1}, so the kinetic energy decreases by 100 J (A correct). Gravity does negative work, so the gravitational potential energy increases by 100 J (B wrong: it increases, not decreases). With no air resistance, mechanical energy is conserved (C wrong). Answer: A.
    Take the full CSCA Physics March 2026 Past Paper — timed & graded

Real-exam evidence

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

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

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

18 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.

Glossary

velocity

Related concepts

Work and PowerConservation of Mechanical Energy