A uniform chain of length L and mass M is lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If g is the acceleration due to gravity, the work required to pull the hanging part on to the table is
Solution:
Consider, the weight of hanging part (L/3) of chain = Mg/3
This weight acts at the centre of gravity of the hanging part, which is at a distance of (L/6) from the table.
Work done = force × distance
Hence,
W = (Mg/3)(L/6)
W = MgL/18
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