Question

A 1400kg automobile moving at a maximum speed of 23m/s on a level circular track of readius of 95m. What is the coefficient of friction?

Answer 1

The coefficient of friction is 0.56

Step-by-step explanation:

It is given that,

Mass of the automobile, m = 1400 kg

Speed of the automobile, v = 23 m/s

Radius of the track, r = 95 m

The automobile is moving in a circular track. The centripetal force is given by :

`F_c=(mv^2)/(r)`............(1)

Frictional force is given by :

`F_f=\mu mg`...................(2)

`\mu` = coefficient of friction

g = acceleration due to gravity

From equation (1) and (2), we get :

`(mv^2)/(r)=\mu mg`

`\mu=(v^2)/(rg)`

`\mu=((23\ m/s)^2)/(95\ m* 9.8\ m/s^2)`

`\mu=0.56`

So, the coefficient of friction is 0.56. Hence, this is the required solution.