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(D) Let the mass per unit length of the chain be $\lambda = M/L$.The length of the hanging part is $l = L/n$.The mass of the hanging part is $m = \lam

Vedclass · Accepted Answer

$MgL/2n^2$

Vedclass · Answer

(D) Let the mass per unit length of the chain be $\lambda = M/L$.The length of the hanging part is $l = L/n$.The mass of the hanging part is $m = \lambda l = (M/L) 	imes (L/n) = M/n$.The center of mass of the hanging part is at a distance $y = l/2 = L/(2n)$ below the edge of the table.To pull the chain back onto the table,we must lift the center of mass of the hanging part to the level of the table.The work done $W$ is equal to the change in potential energy of the hanging part:$W = mgh = (M/n) 	imes g 	imes (L/2n) = MgL/(2n^2)$.

$A$ chain of mass $M$ and length $L$ is placed on a smooth table such that $1/n$ of its length is hanging over the edge of the table. The hanging part of the chain is pulled back onto the surface of the table. Find the work done to pull it up.

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