A 1400-kg car rounds a corner of radius r= 55m . The coefficient of static frict

Question

A 1400-kg car rounds a corner of radius r= 55m . The coefficient of static friction between the tires and the road is 0.83. Part A) At what speed will the force of static friction exerted on the car by the road be equal to half the weight of the car? v= _____ m/s Part B) Suppose that the mass of the car is now doubled, and that it moves with a speed that again makes the force of static friction equal to half the car's weight. Is this new speed greater than, less than, or equal to the speed in part A?

Explanation / Answer

F=mV^2/r

given F=(1/2)Fcar

mV^2/r =mg/2

V=sqrt(rg/2)=sqrt(55*9.8/2)

V=16.4 m//s

b)

Speed is independent of mass ,so speed is still the same

V=16.4 m/s

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