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.

\frac{2 m^{2} g^{2} tan θ}{k}

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\frac{m^{2} g^{2} tan θ}{k}

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\frac{m^{2} g^{2} tan θ}{2 k}

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\frac{m^{2} g^{2} {tan}^{2} θ}{2 k}

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Solution

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

$N sin θ = k x . . . . . . . . (1)$

$N cos θ = m g . . . . . . . . (2)$

By $(1) \div (2)$ we get,

$tan θ = \frac{k x}{m g}$

$∴ x = \frac{m g tan θ}{k}$

$P E$ stored in the spring is,

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

 Hence, option $(D)$ is the correct answer.

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Limiting Friction

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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 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.