A chain of mass m and length l is hanging fre | vedclass

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Question

(C) In diagram $(I)$,the center of mass of the chain is at a distance of $\frac{l}{2}$ below the edge $A$. The potential energy is $U_I = -mg(\frac{l}

Vedclass · Accepted Answer

$mg\frac{l}{4}$

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(C) In diagram $(I)$,the center of mass of the chain is at a distance of $\frac{l}{2}$ below the edge $A$. The potential energy is $U_I = -mg(\frac{l}{2})$.In diagram $(II)$,the chain is folded such that it hangs to a depth of $\frac{l}{2}$. The center of mass of this folded chain is at a distance of $\frac{l}{4}$ below the edge $A$. The potential energy is $U_{II} = -mg(\frac{l}{4})$.The work done $W$ is equal to the change in potential energy: $W = \Delta U = U_{II} - U_I$.$W = -mg(\frac{l}{4}) - (-mg(\frac{l}{2})) = -mg\frac{l}{4} + mg\frac{l}{2} = mg\frac{l}{4}$.

$A$ chain of mass $m$ and length $l$ is hanging freely from edge $A$ (as shown in diagram $I$). Calculate the work done to fold it as shown in diagram $(II)$.

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