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Hingga ujian berikutnya 14–15 November 2026:
Semua konsep

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 soal ujian asli · 4 kali ujian16 soal tryout18 soal latihan

Ujian CSCA dikerjakan dalam bahasa Mandarin atau Inggris — peserta memilih sendiri. Karena itu konsep dan rumus di sini disajikan dalam kedua bahasa tersebut; halaman ini menampilkan versi bahasa Inggris.

Rumus

  • Kinetic energy

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

    Berlaku bila：A scalar, never negative, independent of the direction of motion

  • Work–energy theorem

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

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

Cara ujian sebenarnya menguji konsep ini

Soal dari ujian CSCA sebenarnya yang menguji konsep ini. Kerjakan sendiri dulu sebelum membuka jawabannya.

  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
    Lihat jawaban & pembahasan

    Jawaban: 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}.
    Lihat di CSCA Physics December 2025 Past Paper lengkap
  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
    Lihat jawaban & pembahasan

    Jawaban: 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}.
    Kerjakan CSCA Physics April 2026 Past Paper lengkap — berwaktu & dinilai
  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
    Lihat jawaban & pembahasan

    Jawaban: 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.
    Kerjakan CSCA Physics March 2026 Past Paper lengkap — berwaktu & dinilai

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Konsep terkait

Work and PowerConservation of Mechanical Energy