JAMB - Physics (2007 - No. 39)
A string of length 4m is extended by 0.02m when a load of 0.4kg is suspended at its end. What will be the length of the string when the applied force is 15N?
4.05m
4.08m
5.05m
6.08m
Explanation
from hooke's law F=Ke
\(
\frac{F_1}{e_1} - \frac{F_2}{e_2} \text{therefore}
\frac{4}{0.02} = \frac{15}{e_2} \)
e1 = \( \frac{0.02 x 15}{4} = 0.075M \)
The new length of the string is 4 + 0.075
= 4.075
= 4.08
\(
\frac{F_1}{e_1} - \frac{F_2}{e_2} \text{therefore}
\frac{4}{0.02} = \frac{15}{e_2} \)
e1 = \( \frac{0.02 x 15}{4} = 0.075M \)
The new length of the string is 4 + 0.075
= 4.075
= 4.08
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