Question

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

• A. $\frac{2m^2{g}^2\tan \theta }{k}$
• B. $\frac{m^2{g}^2\tan \theta }{k}$
• C. $\frac{m^2{g}^2\tan \theta }{2k}$
• D. $\frac{m^2{g}^2{\tan }^2\theta }{2k}$

Answer 1

The correct option is D $\frac{m^2{g}^2{\tan }^2\theta }{2k}$

Considering extension in the spring as $x$, the FBD of the wedge is shown in figure.


Given, the system is at equilibrium and is at rest. Hence, balancing the forces,

$N\sin \theta =kx........\left(1\right)$

$N\cos \theta =mg........\left(2\right)$

By $\left(1\right)÷\left(2\right)$ we get,

$\tan \theta =\frac{kx}{mg}$

$\therefore x=\frac{mg\tan \theta }{k}$

$PE$ stored in the spring is,

$U=\frac{1}{2}k{x}^2=\frac{m^2{g}^2{\tan }^2\theta }{2k}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option $\left(D\right)$ is the correct answer.