A 1500kg car is moving on a flat road travels on a road having a radius of curvature of 60 meters and the car is traveling at 15m/s. The coefficient of friction between the tires and the road is 0.50

The car hits a wet spot in the road and the coefficient of friction drops to 0.2. Will the car remain on the road in the circular path? Show all your work.

For the car to remain on the road while taking a turn along a circular path, the frictional force should be equal to or greater than the centripetal force acting on the car.

The Centripetal force acting on the car = m v ² / R.

Given, m = 1500 kg, R = 60 m, v = 15 m/s

F = 1500 x 15² / 60

F = 5625 N.

Friction in the wet part of the road, f = μ m g

= 0.2 x 1500 x 9.8

f = 2940

N. The car would skid, because friction is less than the centripetal force acting on the car, F = 5625N