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
Giải thích
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