AS Daily A Level Physics question
A 12.0 V battery with internal resistance 1.0 Ω powers a small lab hoist that lifts a 2.0 kg mass vertically by 0.80 m in 1.6 s at steady speed. The current is 2.0 A throughout the lift. Neglect any change in kinetic energy. Which estimate is closest to the efficiency of the motor–hoist system during the lift?
Answer
The correct answer is A.
Correct: A — About 49%. Terminal p.d. across the motor is 12 − (2.0 × 1.0) = 10 V, so electrical input to the motor in 1.6 s is 10 × 2.0 × 1.6 = 32 J; the load gains mgh ≈ 2.0 × 9.81 × 0.80 ≈ 15.7 J, giving efficiency ≈ 15.7/32 ≈ 0.49 ≈ 49%. A uses the energy gained (mgh) over the electrical energy delivered to the motor (Vterminal × I × t), which is the correct comparison. B uses the full cell e.m.f. (12 V) so overestimates the electrical input to the motor and underestimates efficiency (≈ 41%). C effectively halves the mechanical power by using an average speed of h/(2t) despite the motion being at steady speed, giving ≈ 25%. D assumes no losses (ignores internal resistance and motor heating), implying ≈ 100%, which is unrealistic here.