JAMB - Physics (2017 - No. 26)
A string of length 5cm is extended by 0.04m when a load of 0.8kg is suspended at the end. How far will it extend if a force of 16N is applied? [g = 10ms\(^{-2}\)]
0.04m
0.12m
0.01m
0.08m
Explanation
From Hoke's Law, F = ke, K = Constant of Force
For he first case, Force = mg,
= 0.8 × 10 = 8N
∴ 8 = k × 0.04
k = \(\frac{8}{0.04}\)
= \(\frac{200N}{m}\)
Since K is constant
For the second case, F = ke
F = 200 × e
∴ e = \(\frac{F}{200}\)
= \(\frac{16}{200}\)
= 0.08m
For he first case, Force = mg,
= 0.8 × 10 = 8N
∴ 8 = k × 0.04
k = \(\frac{8}{0.04}\)
= \(\frac{200N}{m}\)
Since K is constant
For the second case, F = ke
F = 200 × e
∴ e = \(\frac{F}{200}\)
= \(\frac{16}{200}\)
= 0.08m
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