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(D) The length of the hanging part is $l = \frac{L}{5}$.The mass of the hanging part is $m = \frac{M}{L} 	imes \frac{L}{5} = \frac{M}{5}$.The center

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$\frac{M g L}{50}$

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(D) The length of the hanging part is $l = \frac{L}{5}$.The mass of the hanging part is $m = \frac{M}{L} 	imes \frac{L}{5} = \frac{M}{5}$.The center of mass of the hanging part is at a distance $h = \frac{l}{2} = \frac{L/5}{2} = \frac{L}{10}$ below the edge of the table.The work done to pull the hanging part back onto the table is equal to the change in potential energy of the hanging part.$W = mgh = \left( \frac{M}{5} ight) g \left( \frac{L}{10} ight) = \frac{M g L}{50}$.

$A$ chain of length $L$ and mass $M$ is placed on a frictionless table with one-fifth of its length hanging over the edge. What is the work required to pull the hanging part back onto the table?

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