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O Level Physics Energy Power Quiz

Free O Level Physics Energy Power quiz, Gemma31B Exam version, with questions, answers, and O Level-style practice for Singapore students.

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O Level Physics From Real Exams Generated by Gemma 4 31B Updated 2026-08-17

Questions

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Answers

Answer Key - O-Level Physics Quiz: Energy Power

  1. Energy cannot be created or destroyed; it can only be transformed from one form to another. (1)
  2. Gravitational potential energy \rightarrow Kinetic energy. (2)
  3. Power is the rate of energy transfer (or rate of doing work). (1)
  4. Useful energy=0.40×500=200 J\text{Useful energy} = 0.40 \times 500 = 200\text{ J}. (2)
  5. Some energy is dissipated/wasted as heat due to friction in bearings or electrical resistance in windings. (2)
  6. Renewable: Resources that are replenished at the same rate they are used (e.g., solar). Non-renewable: Resources that are finite and will run out (e.g., coal). (2)
  7. Work is defined as force ×\times distance in the direction of the force. Since there is no distance moved, no work is done. (2)
  8. Elastic potential energy. (1)
  9. Ep=mgh=200×10×15=30,000 JE_p = mgh = 200 \times 10 \times 15 = 30,000\text{ J} (or 3.0×104 J3.0 \times 10^4\text{ J}). (2)
  10. Ek=12mv2=0.5×1200×202=240,000 JE_k = \frac{1}{2}mv^2 = 0.5 \times 1200 \times 20^2 = 240,000\text{ J} (or 2.4×105 J2.4 \times 10^5\text{ J}). (2)
  11. Total Energy=P×t=2500×4=10,000 J\text{Total Energy} = P \times t = 2500 \times 4 = 10,000\text{ J}. h=Emg=1000050×10=20 mh = \frac{E}{mg} = \frac{10000}{50 \times 10} = 20\text{ m}. (3)
  12. E=P×t=60×(2×3600)=432,000 JE = P \times t = 60 \times (2 \times 3600) = 432,000\text{ J}. (2)
  13. KE at bottom=GPE at top12mv2=mgh\text{KE at bottom} = \text{GPE at top} \rightarrow \frac{1}{2}mv^2 = mgh. h=v22g=1022×10=5.0 mh = \frac{v^2}{2g} = \frac{10^2}{2 \times 10} = 5.0\text{ m}. (3)
  14. m=100 kgm = 100\text{ kg}, h=10 mh = 10\text{ m}, t=60 st = 60\text{ s}. Work=100×10×10=10,000 J\text{Work} = 100 \times 10 \times 10 = 10,000\text{ J}. P=1000060=166.67 W167 WP = \frac{10000}{60} = 166.67\text{ W} \approx 167\text{ W}. (3)
  15. W=Fd=5×2.0=10 JW = Fd = 5 \times 2.0 = 10\text{ J}. (2)
  16. E=P×t=1200×(30×60)=2,160,000 JE = P \times t = 1200 \times (30 \times 60) = 2,160,000\text{ J} (or 2.16×106 J2.16 \times 10^6\text{ J}). (2)
  17. (a) ΔEp=mg(h1h2)=500×10×(205)=75,000 J\Delta E_p = mg(h_1 - h_2) = 500 \times 10 \times (20 - 5) = 75,000\text{ J}. (2) (b) 12mv2=75,000v2=150000500=300v=30017.3 m/s\frac{1}{2}mv^2 = 75,000 \rightarrow v^2 = \frac{150000}{500} = 300 \rightarrow v = \sqrt{300} \approx 17.3\text{ m/s}. (3)
  18. (a) Gravitational PE \rightarrow Kinetic Energy \rightarrow Electrical Energy. (2) (b) Eff=7.5×1051.0×106×100%=75%\text{Eff} = \frac{7.5 \times 10^5}{1.0 \times 10^6} \times 100\% = 75\%. (2)
  19. (a) 15 N15\text{ N} (since velocity is constant, net force is zero). (1) (b) W=Fd=15×4.0=60 JW = Fd = 15 \times 4.0 = 60\text{ J}. (2)
  20. (a) W=mgh=2×10×1.0=20 JW = mgh = 2 \times 10 \times 1.0 = 20\text{ J}. (2) (b) Energy input=P×t=100×2=200 J\text{Energy input} = P \times t = 100 \times 2 = 200\text{ J}. Eff=20200×100%=10%\text{Eff} = \frac{20}{200} \times 100\% = 10\%. (2)