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

Answer 1

Answer: (D)

Work dorce against frictional force equals the kinetic energy uf the body.
When a body of mass $m$, mioves 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
$\frac{1}{2}m{v}^2=\mu mgs$
where $s$ is the distance in which the car is stopped and $\mu$ is coefficient of kinetic friction.
Given $v=v_0$
$\therefore s=\frac{v_0^2}{2\mu g}$