Question

A 2.00 × 103 kg car rounds a circular turn of radius 20.0 m. If the road is flat and the coefficient of static friction between the tires and the road is 0.70, how fast can the car go without skidding?

Answer 1

11.7 m/s

Step-by-step explanation:

In order for the car to remain in circular motion along the road and not to skid, the frictional force between the tires and the road must be equal to the centripetal force.

Therefore, we can write:

`\mu mg = m(v^2)/(r)`

where:

m = 2000 kg is the mass of the car

`\mu=0.70` is the coefficient of friction

`g=9.8 m/s^2` is the acceleration due to gravity

v is the speed of the car

r = 20.0 m is the radius of the turn

Substituting and solving for v, we find the speed of the car:

`v=√(\mu gr)=√((0.70)(9.8)(20.0))=11.7 m/s`