Question

Spring of force content 100N/m is extended to 240 cm and released. Find out the work done when the extension has reduced to 10 cm.

Answer 1

The work done is stored as elastic potential energy in the spring.

It is given by

$W=\frac{1}{2}\times k(x_2^2-x_1^2)$

$x_1=240cm$

$x_2=10cm$

Force constant, $k=100N/m=\frac{100}{100}N/cm=1N/cm$

$W=\frac{1}{2}\times 1({10}^2-{240}^2)=-28750Ncm$

Work done , $W=-28750Ncm=-287.5Nm=-287.5J$