A chain is held on a frictionless table with 1/ n th of its length hanging over the edge. If the chain has a length L and a mass M, how much work is required to pull the hanging part back on the table?
Text Solution
Verified by ExpertsA
If m is the mass per unit length of the chain, the mass of the chain of length y will be y x m = ym and the force acting on
it due to gravity will be my x g = mgy (assuming that y is the length of the chain hanging over the edge). So, the work
done in pulling the d y length of the chain on the table
[as y is decreasing]
i.e., d W = mgy(-dy) [as F = mgy]
So, the work done in pulling the hanging portion on the table:
or W = MgL/2n 2 [as M = mL]
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