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Question

A chain is held on friction-less horizontal table with one-fourth of its length hanging over the edge. If the chain gas a length l and a mass M, what is the work required to pull the hanging part back on the table?

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Solution

Length of chain hanging from the table = L/4

Mass of chain (only the hanging part) = m/4

Gravitation force on hanging part of the chain = mg/4

Center of mass of the hanging part of the chain is at L/8 from end.( i.e, half of hanging lenght)

therefore, Work done = F.s

= mg/4 * L/8 cos180(degrees)

=-mgL/32

Hence, external work done required to pull the hanging length of the chain =1/32mgL


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