posted by Anonymous on .
6)A 700kg car is travelling at a constant velocity of 35m/s when it slams on the brakes and comes to a stop. How far did the car travel during braking if the coefficient of kinetic friction is 0.62 ? (Hint: if the accelerator is no longer being pressed, is there an applied force by the engine?)
kinetic friction= μk(normal force) 35m/s = .62(700X9.8N/kg)
35m/s=.62(6860) .. is this the correct way to do it?
No, that is not the correct way. You don't multiply μk by velocity * normal force to get friction force.
(Friction force)*X = μk*M*g*X
= Intitial K.E. = (1/2)M(Vi)^2
M cancels out. Vi is the initial velocity.
Solve for X.
X = (1/2)(Vi)^2/(μk*g)