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 figure- nbsp-System nbsp-is in equilibrium and at rest- Assuming that all surfaces are smooth- the potential energy stored in the spring is-

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Standard XII

Physics

Work Energy Theorem Application

A wedge of ma...

Question

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 figure. System is in equilibrium and at rest. Assuming that all surfaces are smooth, the potential energy stored in the spring is:

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Solution

Force due to spring, $F_{s} = k x$
Force due to gravitation, $F_{g} = m g sin θ$
In equilibrium
$F_{s} = F_{g}$
$k x = m g sin θ$
$x = \frac{m g sin θ}{k}$
Potential energy stored in the spring,
$U = \frac{1}{2} k x^{2}$
$U = \frac{k (m g sin θ)^{2}}{2 k}$
$U = \frac{m^{2} g^{2} {sin}^{2} θ}{2 k}$

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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 figure. System is in equilibrium and at rest. Assuming that all surfaces are smooth, the potential energy stored in the spring is:

Force due to spring, Fs=kxForce due to gravitation, Fg=mgsinθIn equilibriumFs=Fgkx=mgsinθx=mgsinθkPotential energy stored in the spring,U=12kx2U=k(mgsinθ)22kU=m2g2sin2θ2k