At what speed will a car round a 52.0-m-radius curve,banked at a 40.0° angle,if no friction is required between the road and tires to prevent the car from slipping? (g = 9.80 m/s²) A)18.1 m/s B)22.6 m/s C)20.7 m/s D)43.6 m/s

Question

Quizplus · Accepted Answer

The centripetal force required to keep the car moving in a circular path is provided by the horizontal component of the normal force, which is given by Ncosθ. The force due to gravity acting on the car is given by mg, where m is the mass of the car and g is the acceleration due to gravity. Since there is no friction between the road and the tires, there is no force acting in the horizontal direction. Therefore, the centripetal force is equal to the force due to gravity.

Ncosθ = mg

The normal force can be resolved into its components:

Nsinθ = mv²/r

where v is the speed of the car while rounding the curve.

Solving the above two equations for v, we obtain:

v = sqrt(gr tanθ)

Substituting the given values, we get:

v = sqrt(9.80 m/s² x 52.0 m x tan 40°) ≈ 20.7 m/s

Therefore, the answer is (C) 20.7 m/s.

At What Speed Will a Car Round a 52