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?
-
A.
4.05m -
B.
4.08m -
C.
5.05m -
D.
6.08m
Correct Answer: Option B
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