Question

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

• A. $\frac{m{g}^{2}ta{n}^2\theta }{2K}$
• B. $\frac{m^{2}gta{n}^2\theta }{2K}$
• C. $\frac{m^2{g}^{2}ta{n}^2\theta }{2K}$
• D. $\frac{m^2{g}^{2}ta{n}^2\theta }{K}$

Answer 1

The correct option is C $\frac{m^2{g}^{2}ta{n}^2\theta }{2K}$

Make free body diagram of forces on both the blocks one by one. For wedge of mass $M$ the forces acting along the floor are $Nsin\theta$ (normal reaction from block of mass $m$) and $kx$. Next, make free body diagram of block of mass $m$. Forces acting on it perpendicular to the floor are $Ncos\theta$ and $mg$. Solving these equations

$Nsin\theta =kx,$

$Ncos\theta =mg,$

We get the value of $x$,

$x=\frac{mgtan\theta }{k}$

value of potential energy can be evaluated

$E=\frac{1}{2}k{x}^2=\frac{1}{2}k{\left(\frac{mgtan\theta }{k}\right)}^2=\frac{m^2{g}^2{tan}^2\theta }{2k}$