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WAEC Physics 2021 Paper

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Question : 15

Total: 50

The handle of a screw jack is 35 cm long and pitch of the screw is 0.5 cm. What force must be applied at the end of the handle to lift a load of 2000N, if the efficiency of the jack is 30% [π = 22/7]?

152.0N
132.0N
15.2N
440.0N

Solution:

Concept:

For a screw jack, efficiency is given by

\eta = \frac{\text{load} \times \text{pitch}}{\text{effort} \times 2\pi r}

where

r

is the length of the handle.

Explanation:

Convert the given values to metres:

Length of handle,

r = 35\ \text{cm} = 0.35\ \text{m}

Pitch,

p = 0.5\ \text{cm} = 0.005\ \text{m}

In one complete turn, distance moved by effort is

2\pi r = 2 \times \frac{22}{7} \times 0.35 = 2.2\ \text{m}

Substitute into the efficiency formula:

0.30 = \frac{2000 \times 0.005}{F \times 2.2}

Simplify:

0.30 = \frac{10}{2.2F}

So,

F = \frac{10}{0.30 \times 2.2} = 15.15\ \text{N}

Therefore, the applied force is approximately

15.2\ \text{N}

.

Answer:

C.

15.2\ \text{N}

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