Question

A body is moving down a long inclined plane of slope ${37}^{\circ }$. The coefficient of friction between the body and plane varies as $\mu =0.3x$, where $x$ is the distance travelled down the plane. The body will have maximum speed at x=?$(\sin {37}^{\circ }=\left(\frac{3}{5}\right)\text{and}g=10{\text{m/s}}^2)$

Answer 1

Here, $\mu =0.3x$
From figure,
$mg\sin \theta -\mu mg\cos \theta =ma$
$\Rightarrow g\sin {37}^{\circ }-\mu g\cos {37}^{\circ }=a$
$\Rightarrow a=6-2.4x$
$\Rightarrow v\frac{dv}{dx}=\int (6-2.4x)$
$\Rightarrow \underset{0}{\overset{v}{\int }}vdv=\underset{0}{\overset{x}{\int }}(6-2.4x)dx$
$\Rightarrow v^2=12x-2.4x^2$ ... (1)
For maximum speed, $\frac{dv}{dx}=0$

From eq. (1), after differentiating
$2v\frac{dv}{dx}=12-4.8x=0$
$\Rightarrow x=2.5\text{m}$

Aliter,
v is maximum when direction of $a$ changes
$\Rightarrow g\sin {37}^{\circ }-\mu g\cos {37}^{\circ }=a$
$\Rightarrow a=6-2.4x$
$\Rightarrow x=2.5\text{m}$

Why this question? Visualization: How visualizing when v is maximum saves us a lot of calculation.