Question

When a body of mass m is suspended from a spiral spring of natural length L, the spring gets stretched through a distance h. The potential energy of the stretched spring is-

• A. $\frac{mg{h}^2}{2}$
• B. mgh
• C. $\frac{1}{2}mgh$
• D. $\frac{1}{2}mg(L+h)$

Answer 1

Answer: (C)

$\frac{1}{2}mgh$

$P.E=0.5k{h}^2$

But when will the spring is in equilibrium, F=kh=mg;k=mg/h

$P.E=0.5mg{h}^2/h=mgh/2$