Question

A body is moving down a long incline of slope ${37}^{\circ }$. The coefficient of friction between the body and plane varies as $\mu =0.3x$, where $x$ (in $\text{m}$) is distance travelled down the plane. The body will have maximum speed at $x=\frac{k}{2}\text{m}$. Then the value of $k$ is

$(\sin {37}^o=3/5,g=10{\text{m/s}}^2)$

Answer 1

Given,
Coefficient of the friction, $\mu =0.3x$,
Inclination angle, $\theta ={37}^{\circ }$

Free body diagram:


From free body diagram,
Net force acting downwards parallel to the inclined plane is,
$F_{net}=mg\sin \theta -f_k$

$\Rightarrow F_{net}=mg\sin \theta -\mu N$

$\Rightarrow F_{net}=mg\sin \theta -\mu mg\cos \theta$

Thus, $F_{net}=mg\sin \theta -0.3xmg\cos \theta ....\left(1\right)$

From Newton's second law of motion, $F_{net}=ma$

$\Rightarrow mg\sin \theta -0.3xmg\cos \theta =ma$

$\Rightarrow g\sin \theta -0.3xg\cos \theta =a$

At maximum speed $F_{net}=0$ that means acceleration $a$, also becomes zero and after this net force will become negative and body deaccelerate, thus its speed will also decrease.

From equation $\left(1\right)$, $F_{net}=0$, we get,

Thus, $mg\sin \theta -0.3xmg\cos \theta =0$

$\Rightarrow x=\frac{\tan \theta }{0.3}.....\left(2\right)$

From equation $\left(2\right)$,we get
$x=\frac{3/4}{0.3}=2.5\text{m}[\because \tan {37}^{\circ }=\frac{3}{4}]$

According to problem, $x=\frac{k}{2}$
$\Rightarrow k=2x$
$\Rightarrow k=2\times 2.5=5$

Accepted answer : $5$