71W = \frac{mgL}{2\eta^2} , where \eta is the fraction of the chain hanging from the table.
q) A chain is held on a frictionless table with (1/n)th of its length hanging over the edge. If the chain has a lenth L and mass M, how much work is required to slowly pull the hanging part back on the table?
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3 Answers
71
21Let fraction of chain hanging be 1/n.
Now it is slowly lifted up,i.e. without any acceleration.
Now work done=inc. in PE [as inc. in KE=0]
The centre of mass of that part is lifted by a distance of L/2n.
Mass of this portion=M/n
Now assuming the mass to be concentrated in CM
Inc. in PE=mgh=M/n*L/2n*g=mgL/2n2