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

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 :

$F_c=\dfrac{mv^2}{r}$............(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 :

$\dfrac{mv^2}{r}=\mu mg$

$\mu=\dfrac{v^2}{rg}$

$\mu=\dfrac{(23\ m/s)^2}{95\ m\times 9.8\ m/s^2}$

$\mu=0.56$

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