Question

In the given figure, a ball of mass $m$ is connected to an elastic spring of force constant $K=mg/L$ through an inextensible string. Now the ball is released from rest. The maximum potential energy stored in the spring during the fall is $mgL[p+\sqrt{q}]$. Find $(p+q)$.

Answer 1

Answer: 5

Using energy conservation between initial moment and equilibrium moment when $T=mg$.
$\frac{1}{2}\frac{mg}{L}\times L^2+\frac{1}{2}m{v}^2=mg\times 2L$
$v=\sqrt{3gL}$
If ball falls a maximum distance of $L+x$, then
$mg(L+x)=\frac{1}{2}\frac{mg}{L}{x}^2$
$x=L(1+\sqrt{3})$
$\therefore P{E}_{\max }=mgL(2+\sqrt{3})$