Question

A car is moving along a straight horizontal road with a speed $v_0$. If the coefficient of friction between the tyres and the road is $\mu$, the shortest distance in which the car be stopped is?

• A. $\frac{v_0^2}{\mu g}$
• B. ${\left(\frac{v_0}{\mu g}\right)}^2$
• C. $\frac{v_0^2}{3\mu g}$
• D. $\frac{v_0^2}{2\mu g}$

Answer 1

Answer: (D)

$\frac{v_0^2}{2\mu g}$

Work done against frictional force equals the kinetic energy of the body.
when a body of mass m, moves with velocity v, it has kinetic energy $k=\frac{1}{2}m{v}^2$, this energy is utilized in doing work against the frictional force between the tyres of the car and road.
$\therefore$ kinetic energy $=$ work done against friction force
$\frac{1}{2}{m}^2=\mu$ mgs
where s is the distance in which the car is stopped and $\mu$ is coefficient of kinetic friction.
Given $v=v_0$
$s=\frac{v_0^2}{2\mu g}$.