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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 μ . Find the work done by friction during the period the chain slips off the table.
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Solution
Let x length of the chain be on the table at a particular instant.
Consider a small element of length 'dx' and mass 'dm' on the table.
dm = \[\frac{M}{L}\text{ dx}\] Work done by the friction on this element is \[\text{ dW }= \mu \text{ Rx } = \mu \left( \frac{M}{L} \times \text{ gx } \right) \text{ dx }\]
Total work done by friction on two third part of the chain,
\[W = \int_{2L/3}^0 \mu\frac{M}{L} \text{ gx dx }\]
\[ \therefore W = \mu\frac{M}{L}\text{ g } \left[ \frac{\text{x}^2}{2} \right]_0^{2\text{L}/3} \]
\[ = - \mu\frac{M}{L}\text{g} \left[ \frac{4 L^2}{18} \right]\]
\[ = - \frac{2\mu \text{ MgL} }{9}\]
The total work done by friction during the period the chain slips off the table is \[- \frac{2\mu \text{ MgL }}{9}\]
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