Question

A small mass slides down an inclined plane of inclination $\theta$ with the horizontal. The co-efficient of friction is $\mu =\mu x$ where x is the distance through which the mass slides down and $\mu$ a constant. Then the speed is maximum after the masscovers a distance of:

• A. $\frac{cos\theta }{{\mu }_0}$
• B. $\frac{sin\theta }{{\mu }_0}$
• C. $\frac{tan\theta }{{\mu }_0}$
• D. $\frac{2tan\theta }{{\mu }_0}$

Answer 1

The correct option is C $\frac{tan\theta }{{\mu }_0}$

Let the particles mass be $m$. Let the distance it slides down to attain maximum velocity be $l$. Let its maximum velocity be $v$.
We'll use the equation ${KE}_1+{PE}_1+W_{ext}={KE}_2+{PE}_2$
${KE}_1=0$
${PE}_1=mgh=mgl\sin \theta$
${PE}_2=0$
${KE}_2=\frac{1}{2}m{v}^2$
The only external force acting on the particle is friction, $f$.
$f=\mu N={\mu }_{0}xmg\cos \theta$
$W_{ext}={\int }_0^{l}f\cdot dx={\int }_0^l{\mu }_{0}mg\cos \theta x\cdot dx={\mu }_{0}mg\cos \theta \frac{l^2}{2}$
Substituting the values, we get
$\frac{1}{2}m{v}^2={\mu }_{0}mg\cos \theta \frac{l^2}{2}+mgl\sin \theta$
$v^2=2[{\mu }_{0}g\cos \theta \frac{l^2}{2}+gl\sin \theta ]$
When we differentiate the R.H.S of the above equation wrt $l$, we will get the values if $l$ for which $v^2$ and hence $v$ will attain extrema values. By substituting those values, we can find out for which value of $l$ the maximum value of $v$ will be attained.
The value of $l$ for which $v$ will be maximum when $l=\frac{\tan \theta }{{\mu }_0}$