Posted by Anonymous on .
When you take your 1300-kg car out for a spin, you go around a corner of radius 57.7 m with a speed of 16.5 m/s. The coefficient of static friction between the car and the road is 0.93. Assuming your car doesn't skid, what is the force exerted on it by static friction?
F(fr) = k•N =k•m•g =0.93•1300•9.8 =11848 N,
friction force = centripetal force to keep the car on road
mv^2/R =1300•(16.5)^2/57.7 =6134 N
Net force = 11848 – 6134 =5714 N