Question

A small body was launched up an inclined plane set at an angle $\alpha ={15}^{\circ }$ against the horizontal. Find 100k if the coefficient of friction is k and if the time of the ascent of the body is $\eta =2.0$ times less than the time of its descent.

Answer 1

When the body is launched up:

Let $k$ be the coefficient of friction, $u$ the velocity of projection and $t$ the distance traversed along the incline. Retarding force on the block $=mg\sin \alpha +kmg\cos \alpha$ and hence the retardation $=g\sin \alpha +kg\cos \alpha$.

Using the equation of particle kinematic along the incline,

$0=u^2-2(g\sin \alpha +kg\cos \alpha )l$

or, $l=\frac{u^2}{2(g\sin \alpha +kg\cos \alpha )}$

and $0=u-(g\sin \alpha +kg\cos \alpha t)$

or, $u=(\sin \alpha +kg\cos \alpha )t$

Using $\left(2\right)$ in $\left(1\right)l=\frac{1}{2}(g\sin \alpha +kg\cos \alpha )t^2$

Case $\left(2\right)$. When the block comes downward, the net force on the body

$=mg\sin \alpha -kmg=cos\alpha$ and hence its acceleration $=g\sin \alpha -kg\cos \alpha$

Let, $t$ be the time required then,

$i=\frac{1}{2}(g\sin \alpha -kg\cos \alpha )t^{\prime 2}$

From Eqs $\left(3\right)$ and $\left(4\right)$

$\frac{t^2}{t^{\prime 2}}=\frac{\sin \alpha -k\cos \alpha }{\sin \alpha +k\cos \alpha }$

But $\frac{t}{t^{\prime }}=\frac{1}{\eta }$

Hence on solving we get

$k=\frac{({\eta }^2-1)}{({\eta }^2+1)}\tan \alpha =0.16$

Therefore 100k =16