- Home
- Standard 11
- Physics
5.Work, Energy, Power and Collision
normal
A car is moving on a straight horizontal road with a speed $v.$ If the coefficient of friction between the tyres and the road is $\mu ,$ the shortest distance in which the car can be stopped is
A
$\frac{{{v^2}}}{{2\mu g}}$
B
$\frac{{{v^2}}}{{\mu g}}$
C
${\left( {\frac{v}{{\mu g}}} \right)^2}$
D
$\frac{{{v^2}}}{\mu }$
Solution
Initial kinetic energy of the car $=\frac{1}{2} \mathrm{mv}^{2}$
Work done against friction $=\mu m g s$
From conservation of energy
$\mu m g s=\frac{1}{2} m v^{2} \text { or } s=\left(v^{2} / 2 \mu g\right)$
$\left[\text { Note: } s=\frac{v^{2}}{2 \mu g}=\frac{\frac{1}{2} m v^{2}}{\mu m g}=\frac{K}{F}\right]$
Standard 11
Physics
Similar Questions
normal