A car is moving on a straight horizontal road with a speed v. If coefficient of friction between tyr

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5.Work, Energy, Power and Collision

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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

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