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

`1.172 xx 10^(4) N ; 0.8` .

Explanation

Here, `m = 1500kg ,upsilon = 12.5 m//s, r = 20m`
Frictional force = required centripetal force
` f = (m upsilon^(2))/(r) =(1500 xx 12.5 xx 12.5)/(20) = 1.172 xx 10^(4) N`
`mu = (f)/(R) = (f)/(mg) = (1.172 xx10^(4))/(1500 xx9.8) = 0.8` .