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 :
............(1)
Frictional force is given by :
...................(2)
= coefficient of friction
g = acceleration due to gravity
From equation (1) and (2), we get :
![\dfrac{mv^2}{r}=\mu mg](https://tex.z-dn.net/?f=%5Cdfrac%7Bmv%5E2%7D%7Br%7D%3D%5Cmu%20mg)
![\mu=\dfrac{v^2}{rg}](https://tex.z-dn.net/?f=%5Cmu%3D%5Cdfrac%7Bv%5E2%7D%7Brg%7D)
![\mu=\dfrac{(23\ m/s)^2}{95\ m\times 9.8\ m/s^2}](https://tex.z-dn.net/?f=%5Cmu%3D%5Cdfrac%7B%2823%5C%20m%2Fs%29%5E2%7D%7B95%5C%20m%5Ctimes%209.8%5C%20m%2Fs%5E2%7D)
![\mu=0.56](https://tex.z-dn.net/?f=%5Cmu%3D0.56)
So, the coefficient of friction is 0.56. Hence, this is the required solution.