Question

A wedge of mass M fitted with a spring of stiffness 'k' is kept on a smooth horizontal surface. A rod of mass m is kept on the wedge as shown in the figure. System is in equilibrium and at rest. Assuming that all surfaces are smooth, the potential energy stored in the spring is:

Answer 1

Force due to spring, $F_s=kx$

Force due to gravitation, $F_g=mg\sin \theta$

In equilibrium

$F_s=F_g$

$kx=mg\sin \theta$

$x=\frac{mg\sin \theta }{k}$

Potential energy stored in the spring,

$U=\frac{1}{2}k{x}^2$

$U=\frac{k(mg\sin \theta {)}^2}{2k}$

$U=\frac{m^2{g}^2{\sin }^2\theta }{2k}$