# Newton's 2nd Law Problem

1. Oct 10, 2007

### AnnieD

1. The problem statement, all variables and given/known data
The driver of a car whose mass is 1200kg is traveling 45km/h[W] on a slippery road when he applies the breaks. The car skids to a stop in 35m. Determine:
a) the car's acceleration
b) the net force acting on the car
c) the coefficient of friction between the tires and the road.

2. Relevant equations
a = Fnet/m
Fg= m(g)
Ff= u(Fn)

3. The attempt at a solution

It's been a year since I've had physics class, so we're on review and I'm a little rough. This is what I have so far, can anyone tell me if I'm right or wrong?

Given:
m= 1200kg
v1= 45km/h = 12.5m/s
v2= 0km/h = 0m/s
d = 35m

a) v2 ^2 = v1^2 + 2ad
0 = 12.5^2 + 2a(35)
-156.25 = 70a
a = - 2.23m/s^2

b) a = Fnet/m
Fnet = m x a
= 1200kg x -2.23m/s^2
= -2 676N

c) Fg = m x g
= 1200kg x 9.8m/s^2
= 11 760N
Fg= Fn

This is where I'm stuck.. I know the equation is Ff = u x Fn .. but how do I figure out Ff (Force of friction)? Use Fnet = Ff + Fapplied.. ?

Thanks! :)

Last edited: Oct 10, 2007
2. Oct 10, 2007

### ptr

When breaks are applied, the only forces on the car are gravitational forces and frictive forces, and of these only the frictive force acts in the horizontal direction in which the car is travelling. Hence the sum of forces in the x direction is $$F_{f}=\mu F_{n}$$, where mu is the co-efficient of friction, and you know the sum of forces in the x direction because that's the only force causing any net acceleration.

3. Oct 10, 2007

### AnnieD

Okay, so then that means that the Fnet is really the same value as the force of friction since the velocity at the time is 0.

So Ff = u x Fn
-2676N = u(-2676)
u = 1

?

4. Oct 10, 2007

### bob1182006

Fn =\= Ff

Fn=mg, equal magnitude, opposite direction to the weight.