Question

An ideal spring is hung vertically from the ceiling and a blovk of mass m is attached to the lower end. Maximum extension is

Answer 1

By using work energy theoram-

=> work done by gravity+ work done by spring=change in K.E.

$mgx - \frac{1}{2} k {x}^{2} = 0$
So, by using this

$mgx = \frac{1}{2} k {x}^{2} \\ x = \frac{2mg}{k}$


so, the maximum extension will be 2mg/k