# 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 .System is in equilibrium and at rest Assuming that all surfaces are smooth, the potential energy stored in the spring is(theta is the angle between the inclined plane of Wedge M and horizontal surface ) Ans- m^2 g^2 tan^2 theta/2K

Force due to spring, F

_{s}= kx

Force due to gravitation, F

_{g}= mg sinϴ

F

_{s}= F

_{g}

kx = mg sinϴ

x = mg sinϴ/k

Potential Energy stored in the spring, U = $\frac{k{x}^{2}}{2}=\frac{k(mg\mathrm{sin}\theta {)}^{2}}{2k}=\frac{{m}^{2}{g}^{2}\mathrm{sin}{\theta}^{2}}{2k}$

Regards

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