A car of mass 1500 kg is moving with a speed of `12.5 ms^(-1)` on a circular path of radius 20 m on a level raod what should be the frictional force between the car and the raod so that the car does not slip what should be the value of the coefficent of friction to attain this force

Explanation

Given `m= 15000 kg`
`v=12.5 ms^(-1)`
`r=25 m`
Friction force = Required centripetal force
or `f= (mv^(2))/(r ) = (15000 xx 12.5 xx 12.5)/(20)`
`=1.172 xx 10^(4) N`
But `f= muR =mu mg`
coefficient of friction
`mu = (f )/(mg ) =- (1.172 xx 10^(4))/( 1500 xx 9.8 ) = 0.8`