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A uniform chain of length L and mass m is lying on a smooth table. One-third of its length is hanging vertically down over the edge of the table. How much work need to be done to pull the hanging part back to the table? |
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Answer» Solution :Mass of hanging part of chain `= m//3` POSITION of centre of GRAVITY below table `= L//6` `:.` Work done = Change in POTENTIAL energy or Work done `= (m/3) g(L/6) = (mgL)/(18)` or Work required to be done = `(mgL)/(18)` |
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