A uniform chain has a mass m and length l. It | vedclass

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(A) The mass of the overhanging part of the chain is $m' = \frac{m}{6}$.The length of the overhanging part is $l' = \frac{l}{6}$.The center of gravity

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$\frac{mgl}{72}$

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(A) The mass of the overhanging part of the chain is $m' = \frac{m}{6}$.The length of the overhanging part is $l' = \frac{l}{6}$.The center of gravity of this overhanging part is at a distance $h = \frac{l'}{2} = \frac{l}{12}$ below the edge of the table.The work done to pull the chain back onto the table is equal to the change in potential energy of the hanging part,which is $W = m'gh$.Substituting the values,$W = (\frac{m}{6}) \cdot g \cdot (\frac{l}{12}) = \frac{mgl}{72}$.

$A$ uniform chain has a mass $m$ and length $l$. It is held on a frictionless table with one-sixth of its length hanging over the edge. The work done in just pulling the hanging part back on the table is

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