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A uniform rope of length $l$ and mass 'm' hangs over a horizontal table with two third part on the table. The coefficient of friction between the table and the chain is $μ$ . The work done by the friction during the period the chain slips completely off the table is

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$\frac{1}{3} μmgl$

$\frac{1}{9} μmgl$

$\frac{2}{3} μmgl$

$\frac{2}{9} μmgl$

answer is A.

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Detailed Solution

$\frac{m}{L}$ is the linear mass density. The little work done for the little distance slippage is provided by

$dW = - μ [\frac{M}{L}] gldl$

thus the total work done,

$W = \int_{0}^{\frac{2 L}{3}} - \frac{μmg}{L} ldl$

$W = - \frac{μ m g}{L} {[\frac{L^{2}}{2}]}_{0}^{\frac{2 L}{3}}$

$W = - \frac{μMg}{2 L} [\frac{4 L^{2}}{9} - 0]$

$W = - \frac{2}{9} μMgL$

Hence the correct answer is $- \frac{2}{9} μ M g L .$

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