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JAMB Physics 2024 Paper

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Question : 102

Total: 125

A force of 10N extends a spring of natural length 1m by 0.02m, calculate the length of the spring when the applied force is 40N.

0.08m
0.09m
1.08m
1.80m

Solution:

F = ke

k = $\;\frac{F_1}{e_1}$

F $_1$ = 10N, e $_1$ = 0.02m

F $_2$ = 40N, e $_2$ = ?

$\;\frac{F_1}{e_1}$ = $\;\frac{F_2}{e_2}$

$\;\frac{10}{0.02}$ = $\;\frac{40}{e_2}$

e $_2$ = $\;\frac{40 \times 0.02}{10}$ = 0.08m

The new length = 1 + 0.08 = 1.08m

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