From Real Exams Quiz

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.

These static practice materials are generated from the site's syllabus and paper-generation workflow, with source and model context shown so students and parents can evaluate the material before use.

O Level Physics From Real Exams Generated by Gemma 4 31B Updated 2026-09-16

Questions

Free quiz and exam paper access

Enter your details to view this paper

Your access is remembered on this device.

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 top→12mv2=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 W≈167 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(h1−h2)=500×10×(20−5)=75,000 J\Delta E_p = mg(h_1 - h_2) = 500 \times 10 \times (20 - 5) = 75,000\text{ J}. (2) (b) 12mv2=75,000→v2=150000500=300→v=300≈17.3 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)