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(C) Let the top of the wedge be the reference level for potential energy $(U = 0)$.In the initial state (Figure-$A$),the chain is divided into two hal

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$\frac{mgL \sin 	heta}{4}$

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(C) Let the top of the wedge be the reference level for potential energy $(U = 0)$.In the initial state (Figure-$A$),the chain is divided into two halves,each of length $L/2$ and mass $m/2$,hanging on either side. The center of mass of each half is at a distance $L/4$ from the top along the incline.The initial potential energy is $U_i = 2 	imes [-(m/2)g(L/4) \sin 	heta] = -\frac{mgL}{4} \sin 	heta$.In the final state (Figure-$B$),the entire chain of mass $m$ is on the left side. The center of mass of the chain is at a distance $L/2$ from the top along the incline.The final potential energy is $U_f = -mg(L/2) \sin 	heta = -\frac{mgL}{2} \sin 	heta$.By the law of conservation of mechanical energy,the change in potential energy equals the change in kinetic energy: $K_f - K_i = U_i - U_f$.Since the chain starts from rest,$K_i = 0$.$K_f = U_i - U_f = -\frac{mgL}{4} \sin 	heta - (-\frac{mgL}{2} \sin 	heta) = \frac{mgL}{4} \sin 	heta$.

$A$ uniform chain of mass $m$ and length $L$ is originally placed mid-way on the top of a fixed smooth double-sided wedge. The length of each side of the wedge is $L$. It is then given a slight push. The kinetic energy of the chain when the whole chain has just slid to the left side of the wedge is:

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