An 810-kg race car can drive aroundan unbanked turn at a maximum speed of 61 m/s

Question

An 810-kg race car can drive aroundan unbanked turn at a maximum speed of 61 m/s without slipping. The turn has a radius ofcurvature of 175 m. Air flowingover the car's wing exerts a downward-pointing force (calledthe downforce) of 11500 N on the car. (a) What is the coefficient of static friction between the trackand the car's tires?


(b) What would be the maximum speed if no downforce acted on thecar?
m/s (a) What is the coefficient of static friction between the trackand the car's tires?


(b) What would be the maximum speed if no downforce acted on thecar?
m/s

Explanation / Answer

centripetal force is only from static friction, so .         u n = mv2 / r .      u (mg + downforce) = mv2 / r .       u (810 * 9.80 + 11500 ) = 810 * 612 / 175 .    u =   0.886    is the coeff ofstatic friction . (b) if there was no downforce, we use the same equationand solve for v... .       u (mg + 0 ) = mv2 /r             u m g = m v2 / r .      v = u g r = 0.886 * 9.80 * 175 =     39.0m/s   is the new max speed

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