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AS Daily A Level Physics question

2026-07-06 OCR A Cross-topic problem solving (mechanics ↔ circuits energy) 3.2 Work, energy and power (P = Fv; efficiency) 4.2 Energy, power and resistance (electrical power) 4.3 Electrical circuits (emf, internal resistance, terminal pd)

An electric hoist is powered by a 9.0 V battery with internal resistance 3.0 Ω. While lifting a 20 N load vertically at steady speed, the current is 1.5 A. Only 60% of the electrical power delivered to the motor becomes useful lifting of the load; the rest is lost. What is the steady lifting speed?

  1. A 0.34 m/s
  2. B 0.41 m/s
  3. C 0.68 m/s
  4. D 0.20 m/s (correct)

Answer

The correct answer is D.

Correct: D — 0.20 m/s. Terminal pd = 9.0 − (1.5 × 3.0) = 4.5 V, electrical power to the motor = 4.5 × 1.5 = 6.75 W; useful mechanical power = 0.60 × 6.75 = 4.05 W, so speed v = P/F = 4.05/20 = 0.20 m/s (2 s.f.). A ignores efficiency, taking all terminal electrical power as useful: 6.75/20 ≈ 0.34 m/s. B ignores internal resistance, using 9.0 V × 1.5 A before applying efficiency: (0.60 × 13.5)/20 ≈ 0.41 m/s. C ignores both internal resistance and efficiency, using 13.5/20 ≈ 0.68 m/s. D matches the correct chain: account for internal resistance, then efficiency, then use P = Fv.

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