# Braking distance of a car(forces and friction)

1. Feb 15, 2012

### chaotiiic

1. The problem statement, all variables and given/known data
the driver of a 1500 kg car travelling at 12.5 m/s slams on the brakes, locking the wheels, and bringing the car to a stop. If the coefficient of friction is .735, what is the braking distance of the car?

2. Relevant equations
W = K_f - K_i
friction force

3. The attempt at a solution
K_f = (1/2)(1500)(12.5)^2 = 117187.5
K_i = 0
W = 117187.5

F_n = 14700
friction force = 14700 * .735 = 10804.5

distance = 117187.5/10804.5 = 10.84m
correct?

2. Feb 15, 2012

### ehild

Hi, chaotiiic,

The result is correct.

ehild

3. Feb 15, 2012

### chaotiiic

i cant find this in my book but why does normal force need to be multiplied with friction coefficient in order to get friction force? does friction force have a unit?

and why does Work need to be divided by friction force in order to get the distance? is there a name for this equation?

4. Feb 15, 2012

### ehild

The force of friction is proportional with the force that presses two surfaces together. This force is normal to the surface of touch, parallel with it and opposes any relative motion of the objects. It is a force, so the unit is newton (N) There is kinetic friction, when the surfaces slide on each other, and static friction which prevents any relative motion. The magnitude of kinetic friction is FKKN, the static friction is the force needed to prevent relative motion, and FS≤μSN.

Work is defined as the scalar product of the force vector F and displacement vector d. If the force and displacement are parallel, W=Fd. The work is positive when the force and displacement have the same direction, and negative when the force points in opposite direction as the motion.

You certainly find these in your textbook...

ehild