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# A small mass slides down an inclined plane of inclination $\theta$ with the horizontal. The coefficient friction is $\mu ={\mu }_{o}x$ where $x$ is the distance through which the mass slides down and ${\mu }_{o}$​ is a positive constant. Then the distance covered by the mass before it stops is

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Solution

## Srep1: Given dataThe coefficient of friction is $\mu ={\mu }_{o}x$.The angle of the inclined plane is $\theta$.DiagramStep2: Frictional force and accelerationThe frictional force is the force that resists the motion of a body over any surface contact with the body.The frictional force is defined by the form, $f=\mu N$, where, N is the normal force exerted on a body and $\mu$ is the coefficient of friction.Acceleration is the time rate of change in velocity. Acceleration is defined as $a=\frac{dv}{dt}$, where, v is the velocity of the body.Step4: Finding the accelerationFrom the figure, it is clear that the normal force exerted on the body is $N=mg\mathrm{cos}\theta$.As we know, the frictional force is $f=\mu N$.So, the equation of the force of the body from the figure is$F=\left(mg\mathrm{sin}\theta -frictionalforce\right)\phantom{\rule{0ex}{0ex}}orF=mg\mathrm{sin}\theta -\mu N\phantom{\rule{0ex}{0ex}}orF=mg\mathrm{sin}\theta -{\mu }_{o}x.mg\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}orma=mg\mathrm{sin}\theta -{\mu }_{o}x.mg\mathrm{cos}\theta \left(\mathrm{sin}ce,F=mass\left(m\right)×acceleration\left(a\right)\right)\phantom{\rule{0ex}{0ex}}ora=g\left(\mathrm{sin}\theta -{\mu }_{o}x.\mathrm{cos}\theta \right)........\left(1\right)$Step4: Finding the distanceWe know the acceleration, $a=\frac{dv}{dt}$or $a=\frac{dv}{dt}=\frac{dv}{dx}.\frac{dx}{dt}\phantom{\rule{0ex}{0ex}}ora=\frac{dv}{dx}.v\left(\mathrm{sin}ce,v=\frac{dx}{dt}\right)\phantom{\rule{0ex}{0ex}}ora=\frac{dv}{dx}.v................\left(2\right)$Comparing equations 1 and 2 we get, $g\left(\mathrm{sin}\theta -{\mu }_{o}x.\mathrm{cos}\theta \right)=\frac{dv}{dx}.v\phantom{\rule{0ex}{0ex}}org\left(\mathrm{sin}\theta -{\mu }_{o}x.\mathrm{cos}\theta \right)dx=vdv.............\left(3\right)$Integrating equation 3${\int }_{0}^{{x}_{m}}g\left(\mathrm{sin}\theta -{\mu }_{o}x.\mathrm{cos}\theta \right)dx={\int }_{0}^{0}vdv\phantom{\rule{0ex}{0ex}}org{\left[\left(\mathrm{sin}\theta .x\right)-{\mu }_{o}.\mathrm{cos}\theta .\frac{{x}^{2}}{2}\right]}_{0}^{{x}_{m}}=0\phantom{\rule{0ex}{0ex}}org\left[\left(\mathrm{sin}\theta .{x}_{m}\right)-{\mu }_{o}.\mathrm{cos}\theta .\frac{{{x}_{m}}^{2}}{2}\right]=0\phantom{\rule{0ex}{0ex}}or\frac{g{\mu }_{o}.\mathrm{cos}\theta }{2}.{x}_{m}=g\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}or{x}_{m}=\frac{2g\mathrm{sin}\theta }{{\mu }_{o}.\mathrm{cos}\theta .}=\frac{2}{{\mu }_{o}}.\mathrm{tan}\theta .$So, acceleration of the body is $\frac{2}{{\mu }_{o}}\mathrm{tan}\theta$

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