Posted by Anonymous on Thursday, April 12, 2012 at 3:56pm.
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
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