Question:
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.
Solution:
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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