A 1 500-kg car rounds an unbanked curve with a radius of 52 m at a speed of 12 m/s.What minimum coefficient of friction must exist between the road and tires to prevent the car from slipping? (g = 9.8 m/s²) A) 0.18 B) 0.30 C) 0.28 D) 0.37

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Quizplus · Accepted Answer

The centripetal force required for the car to make the turn is provided by the force of friction acting on the tires. Using the formula F = mv^2/r, we get F = (1500 kg)(12 m/s)^2/52 m = 5181 N. The force of friction is equal to the coefficient of friction times the normal force, where the normal force is equal to the weight of the car, which is mg = (1500 kg)(9.8 m/s^2) = 14700 N. Therefore, the coefficient of friction required is F/mg = 5181/14700 = 0.352. The closest answer choice is C, 0.28.

A 1 500-Kg Car Rounds an Unbanked Curve with a Radius