Question

A body of mass $m$, having momentum $p$ is moving on a rough horizontal surface. If it is stopped in a distance $x$, the coefficient of friction between the body and the surface is given by $\mu =\frac{p^2}{zg{m}^{2}x}$. Then find the value of $z$.

Answer 1

We know,
$u=\frac{p}{m}$

As the frictional force causes the deceleration of the body, we can write:
$f=\mu N$
$m|a|=\mu mg$
$|a|=\mu g$

Using 3rd equation of kinematics, we can write:
$v^2-u^2=2as$
Here, $a$ is negative and $v=0$, as the body stops after covering $x$ distance. Hence $s=x$.
$0-\frac{p^2}{m^2}=-2\mu gx$

$\mu =\frac{p^2}{2g{m}^{2}x}$