Question

A small mass starts sliding down a fixed inclined plane of inclination ${45}^{\circ }$ with the horizontal. The coefficient of friction is $\mu ={\mu }_{0}x$ where $x$ is the distance through which the mass slides down and ${\mu }_0$ is a constant. Find the heat produced during half the distance travelled by the particle before it stops.

• A. $\frac{mg}{\sqrt{2}{\mu }_0}$
• B. $\frac{mg}{2\sqrt{2}{\mu }_0}$
• C. $\frac{mg}{{\mu }_0}$
• D. $\frac{mg}{2{\mu }_0}$

Answer 1

The correct option is B $\frac{mg}{2\sqrt{2}{\mu }_0}$



From the FBD:
Applying equilibrium condition along $y-$ axis,
$mg\cos \theta =N...\left(i\right)$
Along $x-$ axis,
$mg\sin \theta -f=ma...\left(ii\right)$
Given $\mu ={\mu }_{0}x$
Then, $f=\mu N={\mu }_{0}xmg\cos \theta ...\left(iii\right)$

From Eq $\left(ii\right)\&\left(iii\right)$:
$\Rightarrow mg\sin \theta -{\mu }_{0}xmg\cos \theta =ma$
$\therefore a=g\sin \theta -{\mu }_{0}xg\cos \theta ...\left(iv\right)$
We know that $a=v\frac{dv}{dx}$
Substituting in Eq $\left(iv\right)$,
$v\frac{dv}{dx}=g\sin \theta -{\mu }_{0}xg\cos \theta$
Grouping terms and integrating on both sides
${\int }_{v_i=0}^{v_f=0}vdv={\int }_0^x(g\sin \theta -{\mu }_{0}xg\cos \theta )dx$
(limits of velocity are both zero, since block starts from rest and finally comes to rest)
$\Rightarrow {\left[\frac{v^2}{2}\right]}_0^0=g\sin \theta {\left[x\right]}_0^x-{\mu }_{0}g\cos \theta {\left[\frac{x^2}{2}\right]}_0^x$
$\Rightarrow 0=g\sin \theta x-\frac{{\mu }_{0}g\cos \theta x^2}{2}$
$\Rightarrow x=\frac{2\sin \theta }{{\mu }_0\cos \theta }=\frac{2}{{\mu }_0}\tan \theta$
Putting $\theta ={45}^{\circ }$,
$x=\frac{2}{{\mu }_0}\tan {45}^{\circ }=\frac{2}{{\mu }_0}$
is the distance travelled along the inclined plane before the mass stops.

Magnitude of work done by friction in small displacement $\left(dx\right)$ is
$dW=-fdx$
$=-\mu mg\cos \theta dx=-\left({\mu }_{0}x\right)mg\cos \theta dx$

$\text{Heat produced}=\text{Work done by friction}$
$\therefore \Delta H={\int }_0^{\frac{1}{{\mu }_0}}\left({\mu }_{0}xmg\cos \theta \right)dx$
[For half distance, limits are $x=(0\to \frac{1}{{\mu }_0})$ ]
$\Delta H={\mu }_{0}mg\cos \theta {\int }_0^{\frac{1}{{\mu }_0}}xdx={\mu }_{0}mg\cos {45}^{\circ }{\left[\frac{x^2}{2}\right]}_0^{\frac{1}{{\mu }_0}}$
$\Delta H={\mu }_{0}mg\times \frac{1}{\sqrt{2}}\times \frac{1}{2{\mu }_0^2}$
$\therefore \Delta H=\frac{mg}{2\sqrt{2}{\mu }_0}$