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.
A
0.08m
B
0.09m
C
1.08m
D
1.80m
Verified Correct Answer: C.
Step-by-Step Mathematical 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
Source: WAEC / JAMB / NECO Classroom Archive (Record #70671)