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AS Daily A Level Physics question

2026-05-29 OCR A Forces and Motion (M3): Work, energy and power, Electrons, Waves and Photons (M4): Energy, power and resistance; Electrical circuits OCR-A H156 Module 3.3 Work, energy and power (P = Fv, lifting at constant speed) OCR-A H156 Module 4.1.2 Energy, power and resistance (P = VI; proportional reasoning) OCR-A H156 Module 4.1.3 Electrical circuits (current in DC circuits)

A 6.0 V DC motor lifts a small toolbox vertically at steady speed. With a 0.50 kg load the current is 0.80 A. The lifting speed is kept the same when the load is increased to 0.75 kg. Assume the motor’s efficiency stays the same and friction is unchanged. What current would you expect now?

  1. A 0.80 A — unchanged because the speed is the same
  2. B 0.96 A — 20% higher since the mass only increases by 0.25 kg
  3. C 1.00 A — a small fixed increase to overcome extra friction only
  4. D 1.20 A — 50% higher as lifting power is proportional to mass at constant speed (correct)

Answer

The correct answer is D.

Correct: D — 1.20 A — 50% higher as lifting power is proportional to mass at constant speed. A … Wrong because at constant speed the mechanical power needed is P_mech = mgv, so increasing mass increases required power and therefore current. B … Incorrect proportional reasoning: 0.75/0.50 = 1.5, so the current should rise by 50%, not 20%. C … Not appropriate here because we’re told friction and efficiency are unchanged; the dominant change is the increased weight, so current is not a small fixed increment. D … Correct since P_elec ∝ P_mech and V is constant: I_new = I_old × (m_new/m_old) = 0.80 × (0.75/0.50) = 0.80 × 1.5 = 1.20 A.

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