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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
2m2g2tanθk
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B
m2g2tanθk
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C
m2g2tanθ2k
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D
m2g2tan2θ2k
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Solution

The correct option is D m2g2tan2θ2k
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,

Nsinθ=kx ........(1)

Ncosθ=mg ........(2)

By (1) ÷ (2) we get,

tanθ=kxmg

x=mgtanθk

PE stored in the spring is,

U=12kx2=m2g2tan2θ2k

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

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