A particle of mass ' $m$ ' at rest on a rough horizontal surface with a coefficient of friction ' $\mu$ ' is given a velocity ' $u$ '. The average power impaired by friction before it stops

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Aicharya · Accepted Answer

Acceleration of block, $\begin{aligned} & a=\frac{-f_r}{m}=\frac{-\mu m g}{m} \\ & \therefore \quad a=-\mu m \end{aligned}$ $\therefore$ Time to stop, $\mathrm{t}=\frac{\mathrm{u}}{-\mathrm{a}}=\frac{\mathrm{m}}{\mu \mathrm{g}}$ By work - energy theorem, $\mathrm{W}_{\mathrm{f}}=\Delta \mathrm{k}=0-\frac{1}{2} \mathrm{mu}^2=-\frac{1}{2} \mathrm{mu}^2$ $\therefore$ Power, $\mathrm{P}_{\mathrm{ar}}=\frac{\left|\mathrm{W}_{\mathrm{f}}\right|}{\mathrm{t}}=\frac{\frac{1}{2} \mathrm{mu}^2}{\frac{\mathrm{u}}{\mu \mathrm{g}}}=\frac{1}{2} \mu \mathrm{mgu}$

A particle of mass ' $m$ ' at rest on a rough horizontal surface with a coefficient of friction ' $\mu$ ' is…

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