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Hingga ujian berikutnya 14–15 November 2026:
Semua konsep

Conservation of Mechanical Energy

机械能守恒定律

Gravitational potential energy and its relation to the work done by gravity, testing whether mechanical energy is conserved, and applying the conservation law.

7 soal ujian asli · 4 kali ujian12 soal tryout16 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

  • Gravitational potential energy

    Definisi
    Ep=mghE_p=mgh

    Berlaku bila：h is measured from the chosen zero level; the sign of Eₚ depends on that choice

  • Conservation of mechanical energy

    Hukum
    Ek+Ep=constE_k+E_p=\text{const}

    Berlaku bila：Only gravity (or elastic force) does work; all other forces do zero net work

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
    Question: A basketball with mass m=0.5 kg\mathrm { m } = 0.5 \mathrm {~kg} bounces from the ground to a height h=1 m\mathrm { h } = 1 \mathrm {~m} before falling back to the ground. The change in gravitational potential energy during this entire process is ( ). (g=10 m/s2)\left( \mathrm { g } = 10 \mathrm {~m} / \mathrm { s } ^ { 2 } \right)
    • A．10 J
    • B．0.5 J
    • C．5 J
    • D．0 J
    Lihat jawaban & pembahasan

    Jawaban: D

    1 Initial and final positions are both on the ground, so gravitational potential energy change over the whole bounce cycle is ΔU=0\Delta \mathrm { U } = 0. 2 Therefore the net change in gravitational potential energy is 0 J .
    Lihat di CSCA Physics December 2025 Past Paper lengkap
  2. CSCA Physics March 2026 Past Paper
    Mass mm is thrown upward at v from ground level, no air resistance. Mechanical energy at the highest point is (ground == zero PE)
    • A．12mv2\frac { 1 } { 2 } m v ^ { 2 }
    • B．mgh
    • C．1/2mv2+mgh1 / 2 m v ^ { 2 } + m g h
    • D．0
    Lihat jawaban & pembahasan

    Jawaban: A

    No air resistance → mechanical energy conserved. At launch: E=1/2mv2+0=1/2mv2\mathrm { E } = 1 / 2 \mathrm { mv } ^ { 2 } + 0 = 1 / 2 \mathrm { mv } ^ { 2 }. At top: KE=0\mathrm { KE } = 0, PE=mgh\mathrm { PE } = \mathrm { mgh }, but total E still =1/2mv2= 1 / 2 \mathrm { mv } ^ { 2 }. B and C are not correct expressions; D is clearly wrong. Answer: A．
    Kerjakan CSCA Physics March 2026 Past Paper lengkap — berwaktu & dinilai
  3. CSCA Physics April 2026 Past Paper
    A 0.2 kg ball falls freely from a height of 2 m to the ground. The change in its gravitational potential energy is (g=10 m/s2)\left( g = 10 \mathrm {~m} / \mathrm { s } ^ { 2 } \right)
    • A．0 J
    • B．- 4 J
    • C．4 J
    • D．2 J
    Lihat jawaban & pembahasan

    Jawaban: B

    As the ball descends, its gravitational PE decreases. ΔEp=−mgh=−0.2×10×2=−4 J\Delta E _ { p } = - m g h = - 0.2 \times 10 \times 2 = - 4 \mathrm {~J}.
    Kerjakan CSCA Physics April 2026 Past Paper lengkap — berwaktu & dinilai

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Tersedia 16 soal latihan terkaitLatihan gratis: Usaha dan Energi

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