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
The correct answer is: C
Explanation
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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