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PhysicsW.E.P.Calculation of workMedium2 minPYQ_2020

PhysicsMediumsingle choice

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 $μ$ . The work done by friction during the period, the chain slips of the table is

Options:

Answer:

Solution:

The linear mass density is $\frac{M}{L}$ . The small work done for the slippage on the small distance $d l$ is given by

$d W = - μ [\frac{M}{L}] g l d l$

Now, total work done

$W = \int_{0}^{\frac{2 L}{3}} - \frac{μ M g}{L} l d l$

$= - \frac{μ M g}{L} {[\frac{l^{2}}{2}]}_{0}^{\frac{2 L}{3}} = - \frac{μ M g}{2 L} [\frac{4 L^{2}}{9} - 0]$

$\Rightarrow W = - \frac{2}{9} μ M g L$

Stream:JEESubject:PhysicsTopic:W.E.P.Subtopic:Calculation of work

⏱ 2mℹ️ Source: PYQ_2020

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