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

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Question

(A) The work done in pulling the hanging part of the chain onto the table is equal to the increase in the gravitational potential energy of that part.

Vedclass · Accepted Answer

$\frac{m g l}{72}$

Vedclass · Answer

(A) The work done in pulling the hanging part of the chain onto the table is equal to the increase in the gravitational potential energy of that part.Let the mass of the hanging part be $m' = \frac{m}{6}$ and its length be $l' = \frac{l}{6}$.The center of mass of the hanging part is at a distance $h = \frac{l'}{2} = \frac{l/6}{2} = \frac{l}{12}$ below the edge of the table.The work done $W$ is equal to the potential energy required to raise this center of mass to the level of the table:$W = m' g h$$W = \left(\frac{m}{6}\right) g \left(\frac{l}{12}\right)$$W = \frac{m g l}{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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