Question

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

Answer 1

Srep1: Given data

1. The coefficient of friction is $\mu ={\mu }_{o}x$.
2. The angle of the inclined plane is $\theta$.

Diagram

Step2: Frictional force and acceleration

1. The frictional force is the force that resists the motion of a body over any surface contact with the body.
2. 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.
3. 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 acceleration

From the figure, it is clear that the normal force exerted on the body is $N=mg\cos \theta$.

As we know, the frictional force is $f=\mu N$.

So, the equation of the force of the body from the figure is

$$\begin{gathered} F=\left(mg\sin \theta -frictionalforce\right) \\ orF=mg\sin \theta -\mu N \\ orF=mg\sin \theta -{\mu }_{o}x.mg\cos \theta \\ orma=mg\sin \theta -{\mu }_{o}x.mg\cos \theta (\sin ce,F=mass\left(m\right)\times acceleration\left(a\right)) \\ ora=g\left(\sin \theta -{\mu }_{o}x.\cos \theta \right)........\left(1\right) \end{gathered}$$

Step4: Finding the distance

We know the acceleration, $a=\frac{dv}{dt}$

or $$\begin{gathered} a=\frac{dv}{dt}=\frac{dv}{dx}.\frac{dx}{dt} \\ ora=\frac{dv}{dx}.v\left(\sin ce,v=\frac{dx}{dt}\right) \\ ora=\frac{dv}{dx}.v................\left(2\right) \end{gathered}$$

Comparing equations 1 and 2 we get,

$$\begin{gathered} g\left(\sin \theta -{\mu }_{o}x.\cos \theta \right)=\frac{dv}{dx}.v \\ org\left(\sin \theta -{\mu }_{o}x.\cos \theta \right)dx=vdv.............\left(3\right) \end{gathered}$$

Integrating equation 3

$$\begin{gathered} {\int }_0^{x_m}g\left(\sin \theta -{\mu }_{o}x.\cos \theta \right)dx={\int }_0^{0}vdv \\ org{\left[\left(\sin \theta .x\right)-{\mu }_o.\cos \theta .\frac{x^2}{2}\right]}_0^{x_m}=0 \\ org\left[\left(\sin \theta .x_m\right)-{\mu }_o.\cos \theta .\frac{{x_m}^2}{2}\right]=0 \\ or\frac{g{\mu }_o.\cos \theta }{2}.x_m=g\sin \theta \\ or{x}_m=\frac{2g\sin \theta }{{\mu }_o.\cos \theta .}=\frac{2}{{\mu }_o}.\tan \theta . \end{gathered}$$

So, acceleration of the body is $\frac{2}{{\mu }_o}\tan \theta$