Answer:
The coefficient of friction is 0.56
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 :
[tex]F_c=\dfrac{mv^2}{r}[/tex]............(1)
Frictional force is given by :
[tex]F_f=\mu mg[/tex]...................(2)
[tex]\mu[/tex] = coefficient of friction
g = acceleration due to gravity
From equation (1) and (2), we get :
[tex]\dfrac{mv^2}{r}=\mu mg[/tex]
[tex]\mu=\dfrac{v^2}{rg}[/tex]
[tex]\mu=\dfrac{(23\ m/s)^2}{95\ m\times 9.8\ m/s^2}[/tex]
[tex]\mu=0.56[/tex]
So, the coefficient of friction is 0.56. Hence, this is the required solution.