A) \[\frac{v_{0}^{2}}{\mu }\]
B) \[{{\left( \frac{{{v}_{0}}}{{{\mu }_{g}}} \right)}^{2}}\]
C) \[\frac{v_{0}^{2}}{{{\mu }_{g}}}\]
D) \[\frac{v_{0}^{2}}{2{{\mu }_{g}}}\]
Correct Answer: D
Solution :
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 u is coefficient of kinetic friction. Given \[v={{v}_{0}}\] \[s=\frac{v_{0}^{2}}{2\mu \,\,g}\]
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