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 fig. System is in equilibrium. Assuming that all surfaces are smooth, the potential energy stored in the spring is:

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$\frac{m g^{2} \tan θ}{2 k}$

$\frac{m^{2} g \tan^{2} θ}{2 k}$

$\frac{m^{2} g^{2} \tan^{2} θ}{2 k}$

$\frac{m g \tan θ}{2 k}$

answer is D.

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Detailed Solution

1. FBD OF M
Question Image
At Equilibrium
$N \sin θ = K x \to (1)$
FBD of m

$N \cos θ = m g \to (2)$
From (1) & (2)
$\frac{(1)}{(2)} \Rightarrow \tan θ = \frac{k x}{m g}$
$x = \frac{m g \tan θ}{k}$
Change in PE $Δ U = \frac{1}{2} k ({X_{f}}^{2} - {X_{i}}^{2})$ $= \frac{1}{2} k \frac{m^{2} g^{2} + \tan^{2} θ}{k^{2}} - {(0)}^{2}$
$∴ Δ U = \frac{m^{2} g^{2} \tan^{2} θ}{2 k}$

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