Question

To test the performance of its tires, a car travels along a perfectly flat (no banking) circular track of radius 101 m. The car increases its speed at uniform rate of 3.56 m/s^2 until the tires start to skid. If the tires start to skid when the car reaches a speed of 16.9 m/s what is the coefficient of static friction between the tires and the road?

Answer 1

We know , Centripetal acceleration = v^2/r

= 16.9^2/101 = 2.83 m/s^2

Net acceleration a = sqrt(2.83^2 +3.56^2) = 4.55 m/s^2

Coefficient of friction = friction/normal = ma/mg

= a/g = 4.55/9.8

= 0.464 answer