Efficiency

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A motor takes 5 kJ electrical energy to lift a 100 kg mass by 4 m (g=10). Efficiency?

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50%

60%

75%

80%

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Subject: physics
Chapter: work, energy and power
Topic: efficiency
Difficulty: Medium
Year: 2025

Solution

Correct Answer:

80%

Definition of Efficiency ((\eta)): Ratio of useful work output to total energy input.
Energy Input ( $E_{in}$ ): $5$ kJ $= 5000$ J.
Useful Work Output ( $W_{o u t}$ ): Lifting a weight requires work against gravity.
- $W_{o u t} = m g h = 100 \times 10 \times 4 = 4000$ J.
Calculation:
- $η = \frac{4000}{5000} = 0.8$ .
Result: Efficiency is $80%$ .

This medium difficulty physics question is from the chapter work, energy and power, covering the topic of efficiency. It appeared in the 2025 exam.

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Efficiency

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