A car of mass `1500kg` is moving with a speed of `12.5ms^(-1)` on a circular path of radius `20m` on a level road What should be the frictional force to avoid slipping of car Calculate the cofficient of friction .

Answer

`0.8`

Explanation

Here, m = 1500 kg.
`v= 12.5 "ms"^(-1) r = 20 `
Frictional force = Centripetal force required `F = (mv^2)/r =( 1500(12.5)^2)/20`
` = 1.172 xx 10^4 N`
As `F = muP = mumg`
`mu = F/(mg) = (1.172 xx10^4)/(1500 xx9.8)`
`= 0.797 ~~ 0.8`