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 the friction during the period the chain slips off the table?

Answer 1

See diagram.

Force of friction is against the movement (sliding) of the chain on the table.

$dF_f\ on\ element\ dx\ of\ mass\ dm\ due\ to\ friction=\mu*dm*g\\\\=\mu\ g\ (\frac{m}{L})\ dx\\\\Element\ dx\ gets\ displaced\ by\ x,\ when\ the\ chain\ slides\ off.\\\\So\ Work\ done\ on\ dx\ by\ friction=dF_f*x=\mu\ g\ (\frac{m}{L})\ dx * \ x\\\\Total\ work= \int\limits^{2L/3}_{0} {dF_f*} \, x$

$= \frac{\mu\ g\ m}{L} \int\limits^{2L/3}_0 {x} \, dx =\frac{\mu\ g\ m}{L}[x^2/2]_0^{2L/3}\\\\=\frac{2}{9}\ \mu\ mgL$

Answer 2

Answer:

-2newMg/L

Explanation:

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