# A uniform chain of mass m and length l is lying on the table such that its one-fifth part is hanging from the edge of the table. what maximum work is done in lifting up the chain?

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by lobsang777

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by lobsang777

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length = l

mass per unit length, M = (m/l)

One fifth is hanging from the edge.

So length of hanging part = l/5

Centre of mass of the hanging part will be at mid point of hanging part

So CM will be at = (l/5) / 2 = l/10.

mass of hanging part = M×(l/5) = (m/l) × (l/5) = m/5

So the problem is equivalent to lifting an object of mass m/5 to a height of l/10.

Work done = mgh = (m/5)×g×(l/10) =