Question

The initial velocity of a body moving on a rough horizontal surface having coefficient of kinetic friction $0.5$ is $15\text{m/s}$. Distance after which it will come to rest is $s_1$. If coefficient of kinetic friction is reduced to $0.3$, distance travelled by the body before coming to rest is $s_2$. Find $s_1+s_2$. (Take $g=10{\text{m/s}}^2$)

• A. $52.5m$
• B. $57m$
• C. $60m$
• D. None of the above

Answer 1

The correct option is C $60m$

Due to friction, the body undergoes retardation of $a=\mu g$
where $\mu$ = coefficient of kinetic friction
Also $u=15\text{m/s}$ and $v=0$ (as it comes to rest)

From kinematics,
$v^2=u^2-2as$
$\Rightarrow 0=u^2-2as$
$\Rightarrow s=\frac{u^2}{2\mu g}$
Therefore,
$s_1=\frac{{15}^2}{2\times 0.5\times 10}=22.5\text{m}$
$s_2=\frac{{15}^2}{2\times 0.3\times 10}=37.5\text{m}$

$\therefore$ $s_1+s_2=60\text{m}$