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JAMB Physics 2023 Paper

Question : 18

Total: 65

A 200 kg load is raised using a 110 m long lever as shown in the diagram above. The load is 10m from the pivot P. If the efficiency of the the lever is 80%, find the effort E required to lift the load.

[Take g = 10ms-2]

Validate

Solution:

Concept:

A lever compares useful work done on the load with work done by the effort, using efficiency.

Explanation:

Load weight:

W = mg = 200 \times 10 = 2000 \, \text{N}

Load arm from pivot:

d_L = 10 \, \text{m}

Effort arm from pivot:

d_E = 110 - 10 = 100 \, \text{m}

Work done on the load:

2000 \times 10 = 20{,}000 \, \text{J}

Work done by effort:

E \times 100 = 100E

Efficiency formula:

\text{Efficiency} = \frac{\text{Work on load}}{\text{Work by effort}} \times 100\%

Substitute values:

80 = \frac{20{,}000}{100E} \times 100

Simplifying gives

80 = \frac{20{,}000}{E}

Therefore,

E = \frac{20{,}000}{80} = 250 \, \text{N}

Answer:

The effort required is

250 \, \text{N}

. Correct option: A. 250.

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