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, then the work required to pull the hanging part on to the table is ?
Please explain the concept.
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in the given problem, the mass of the hanging part will be 1/3M.
Now, the centre of gravity will be located at the midpoint of the mass, that is in L/6.
Therefore the work done will be given by the formula W= mass x acceleration due to gravity x displacement of the centre of gravity.
Which comes to be, 1/3M x g x L/6 = MgL/18
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