A device raises 12 kg through 5 m and later lowers it to its starting height. Use g = 10 N/kg. Its lifting efficiency is 60%; on lowering, 30% of the gravitational energy is returned to its battery. Control electronics take another 80 J from the battery per full cycle. The load starts and ends at rest; ignore all other energy stores. Calculate lifting input, recovered energy and net battery-energy decrease. Explain where the net decrease goes despite the load returning to its original position.

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Lifting input 1000 J; recovery 180 J; net battery decrease 900 J, ultimately transferred to the surroundings. The gravitational energy gained on lifting is mgh = 12 × 10 × 5 = 600 J. This is useful lifting output, not the larger electrical input when efficiency is below one. Convert lifting efficiency to 0.6, then divide: battery energy used for lifting = 600/0.6 = 1000 J. The lifting loss is 400 J. On lowering, the specified recoverable fraction applies to gravitational energy: 0.3 × 600 = 180 J returns to the battery. The unrecovered lowering energy is 420 J. Include the separate electronics consumption and subtract the recovered amount: net battery decrease = 1000 + 80 − 180 = 900 J. Recovery reduces consumption but does not make the cycle self-powering. At the end, load height and speed are unchanged from their initial values. The battery's lost energy has therefore been dissipated through motor resistance, friction and electronics, consistent with energy conservation.

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