A 910-kg race car can drive around an unbanked turn at a maximum speed of 41 m/s

Question

A 910-kg race car can drive around an unbanked turn at a maximum speed of 41 m/s without slipping. The turn has a radius of 130 m. Air flowing over the car's wing exerts a downward-pointing force (called the downforce) of 11000 N on the car. (a) What is the coefficient of static friction between the track and the car's tires? (b) What would be the maximum speed if no downforce acted on the car?

Explanation / Answer

FOLLOW THIS a ) Normal force = 855 * 9.8 + 1400 = 9779 N Max. frictional force = ? * 9779 N mv^2 / r = ? * 9779 => ? = ( mv^2 / 9779 r) = ( 855 * (60)^2 / 9779 *130) = 2.42 [Value of ? has to be less than 1. Value of ? much higher than 1 indicates some error in the problem. The error seems to be with the speed. 60 m/s = 216 km/h seems too high to be achievable on a curved road.]

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