Solve the following :

A uniform chain of length $L$ and mass $M$ overhangs a horizontal table with its two third part on the table. The friction coefficient between the table and the chain is 11 . Find the work done by the friction during the period the chain slips off the table.


Work done for small element is:

$d \mathrm{~W}=\mu \mathrm{Rx}$

$d w=\mu(M / L d x) g(x)$

Total work done is :

$W={ }_{2 L / 3} \int^{0} \mu \mathrm{M} / \mathrm{L} \mathrm{g}(x) d x$

$\mathrm{W}=(2 \mu \mathrm{MgL}) / 9$


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