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(D) The mass of the hanging part of the chain is $m = M/3$. The length of the hanging part is $l = L/3$. The center of mass of the hanging part is at

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$MgL/18$

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(D) The mass of the hanging part of the chain is $m = M/3$. The length of the hanging part is $l = L/3$. The center of mass of the hanging part is at a distance $h = l/2 = L/6$ below the edge of the table. The potential energy of the hanging part relative to the table surface is $U = -mgh = -(M/3)g(L/6) = -MgL/18$. To pull the hanging part onto the table, the work done by an external force must be equal to the change in potential energy, which is $W = \Delta U = U_{final} - U_{initial} = 0 - (-MgL/18) = MgL/18$.

$A$ uniform chain of length $L$ and mass $M$ is lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If $g$ is the acceleration due to gravity, the work required to pull the hanging part onto the table is

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