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(B) The total mass of the chain is $M$ and the total length is $L$. The mass per unit length is $\lambda = M/L$.The length of the hanging part is $l =

Vedclass · Accepted Answer

$\frac{MgL}{32}$

Vedclass · Answer

(B) The total mass of the chain is $M$ and the total length is $L$. The mass per unit length is $\lambda = M/L$.The length of the hanging part is $l = L/4$.The mass of the hanging part is $m = \lambda 	imes l = (M/L) 	imes (L/4) = M/4$.The center of mass of the hanging part is located at a distance $h = l/2 = (L/4)/2 = L/8$ below the edge of the table.To pull the hanging part back onto the table,the external force must overcome the gravitational potential energy of the hanging part.The work done $W$ is equal to the change in potential energy of the hanging part,which is $W = mgh$.Substituting the values: $W = (M/4) 	imes g 	imes (L/8) = \frac{MgL}{32}$.Therefore,the work done by the external force is $\frac{MgL}{32}$.

$A$ chain of mass $M$ and length $L$ is placed on a table such that $L/4$ of its length hangs over the edge. What is the work done by an external force to pull the hanging part back onto the table?

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