Manipal MET2013PhysicsLaws of Motion
A car is moving along a straight horizontal road with a speed v₀ . If the coefficient of friction between the tyres and the road is , the shortest distance in which the car be stopped is
Options
- Av₀^2
- B( v₀ _g )^2
- Cv₀^2 _g
- Dv₀^2 2 _g
Correct answer
D. v₀^2 2 _g
Step-by-step 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= 1 2 m v^2 , this energy is utilized in doing work against the frictional force between the tyres of the car and road. kinetic energy = work done against friction force 1 2 mv^2= mgs where s is the distance in which the car is stopped and is coefficient of kinetic friction. Given V=V₀s= v₀^2 2 g