A lift raises a total mass of 600 kg through 12 m in 20 s. Take g = 10 N/kg. The motor draws 90 kW while rising. During descent, 50% of the gravitational energy is returned to the battery. (a) Calculate the gravitational energy gained and useful lifting power. [4] (b) Calculate the electrical energy supplied while rising and the lifting efficiency. [3] (c) Calculate the energy returned during descent and the net battery-energy decrease for the up–down cycle. [3] (d) Explain why energy is conserved although the battery does not recover all its energy. [2]

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The lift gains 72 kJ at 3.6 kW. It receives 1.8 MJ electrically, so lifting efficiency is 4.0%. Descent returns 36 kJ and the net battery decrease is 1.764 MJ. (a) Gravitational energy gained is ΔE = mgh = 600 × 10 × 12 = 72,000 J = 72 kJ. (a) Useful lifting power is 72,000/20 = 3600 W = 3.6 kW. Power must use the same 20 s interval as the energy transfer. (b) Electrical energy supplied is Pt = 90,000 × 20 = 1,800,000 J = 1.8 MJ. Efficiency = 72,000/1,800,000 × 100 = 4.0%. (c) Returned energy is 0.50 × 72,000 = 36,000 J. Net battery decrease = 1,800,000 − 36,000 = 1,764,000 J = 1.764 MJ. (d) The unrecovered energy has not disappeared: it is transferred to thermal energy in the motor, electronics, cables, brakes, air and mechanism, plus sound. The selected battery–lift store is not an isolated system.

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