What is the maximum speed that a 2200 kg car can go around a level circular track with a radius of 30.0 m without slipping if the coefficient of static friction between the tires and the road is 0.900?

The maximum speed of turn on the given circular track is 16.27 m/s.

Explanation:

Given;

mass of the car, m = 2200 kg

radius of the track, r= 30 m

coefficient of static friction between the tires and the road, μ = 0.9

The net vertical force on the car = N = mg

The net horizontal force on the car = Centripetal force

The coefficient of static friction is given as;

where;

v is the maximum speed of turn

Therefore, the maximum speed of turn on the given circular track is 16.27 m/s.

you get 2.3=1.21*\frac{a}2 or 4.6=1.21a or a=3.802 (b)

explanation: