Test Index

JAMB Physics 2011 Paper

Question : 11

Total: 44

Solution:

Concept:

Efficiency compares useful work output to the total work input.

Explanation:

The useful work done in lifting the load is force times height.

\text{Work output} = 20 \, \text{N} \times 10 \, \text{m} = 200 \, \text{J}

Efficiency is given as

40\%

, so:

40\% = \frac{\text{Work output}}{\text{Work input}} \times 100

0.40 = \frac{200}{\text{Work input}}

Therefore:

\text{Work input} = \frac{200}{0.40} = 500 \, \text{J}

This total work input is the work done by the machine.

Answer:

The work done is

500 \, \text{J}

Correct option: B. 500 J

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