Forces!!!! please helpp!!! 1) 2500kg car stops from 14 m/s to full stop in a distance of 25.0 m. what is the coefficient of kinetic friction?q 2ad= v squared - v(starting) squared 2a25 = - (14) squared 50a = -196 a = - 3.92 m/s i got this far and then to find Ffriction = u(coefficient)x normal force....and here i got lost. i know that to stop the car a force of - 9800 N is needed. (F = ma = 2500(-3.92) = -9800N but how do i get the friction of force? can somebody help me...please? do i use Fg at all? Fg=2500(9.8)=24500N??? and then 2) box of 5 kg is moving at velocity of 3 m/s due to Force of 10N. What is the normal force? I dont understand normal force. i know its a weight of an object and any other force added to it. so in this case would it be 10N + 5(9.8)N which equals to 59N?