The surface of the exit road is horizontal,not banked.(See figure.)If the radius of curvature of a highway exit is 82.5 m and the maximum speed with which the car can exit the highway safely without sliding is 25.2 m/s,what is the coefficient of static friction between the tires and the surface of the road?
A)0.69
B)0.85
C)0.78
D)0.56
E)0.72

Question

The surface of the exit road is horizontal,not banked.(See figure.)If the radius of curvature of a highway exit is 82.5 m and the maximum speed with which the car can exit the highway safely without sliding is 25.2 m/s,what is the coefficient of static friction between the tires and the surface of the road? 
A)0.69 
B)0.85 
C)0.78 
D)0.56 
E)0.72

Quizplus · Accepted Answer

The maximum speed that the car can take the exit safely without sliding can be found using the formula:

vmax = sqrt(μrg)

where μ is the coefficient of static friction, r is the radius of curvature, and g is the acceleration due to gravity.

Plugging in the given values, we get:

25.2 = sqrt(μ x 9.8 x 82.5)

Squaring both sides and rearranging, we get:

μ = (25.2/9.8)^2/82.5 = 0.78

Therefore, the coefficient of static friction between the tires and the surface of the road is 0.78. The closest answer choice is C.

The Surface of the Exit Road Is Horizontal,not Banked