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

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Solution

The work required to be done in lifting the chain up will be equal to change in the potential energy of the hanging part.

The mass of the hanging chain is $\frac{m}{6}$

The change in potential energy will be same as bringing the center of mass of the chain up on the table.
The center of mass of the chain lies at the $\frac{l}{3}$ below the table.

So, the change in the potential energy is
$E = M g h$

$E = \frac{m}{6} \times g \times \frac{l}{3}$

$= \frac{m g l}{18}$

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A uniform chain has a mass m and length l. It is held on a frictionless table with two third of its length hanging over the edge. Find the work done in just pulling the hanging part back on the table.

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