# Solving for Accelleration with only Coeffifcient of Friction, and mass

1. Nov 6, 2012

### Syndrix

1. The problem statement, all variables and given/known data

The coefficient of kinetic friction between rubber tires and wet pavement is 0.50. The brakes are applied to a 750kg car travelling 30.0 m/s and the car skids to a stop. What would be the magnitude and direction of the acceleration on the car?

u = 0.50
m = 750kg
V(i) = 30m/s

2. Relevant equations

F(f) = uF(n)
F(n) = mg
F(net) = ma

3. The attempt at a solutionp

No matter what way I look at this I cannot find a way to calculate acceleration with this set of data. I realize I need F(a) but I can't seem to get it.
I know F(f) = umg, and solved: F(f) = (0.5)(750)(9.81) and got F(f) = 3678.75 N
F(net) = ma => F(f) + F(a) = ma
This is as far as I get because I have two unknowns, F(a) and a

2. Nov 6, 2012

### SHISHKABOB

The only force acting on the car is the force of friction.

3. Nov 6, 2012

### Syndrix

Wouldn't the breaks act as an applied force?

4. Nov 6, 2012

### SHISHKABOB

The brakes are applying a force internal to the system of the car. The brakes force the wheels to stop moving, and then the force of friction between the wheels and the road is what slows down the car.

5. Nov 6, 2012

### Syndrix

So if Force friction is the only force, then;
F(f) = F(net)
F(f) = ma
F(f) = a
M

3678 = a
750

a = 4.9m/s

6. Nov 6, 2012

### SHISHKABOB

don't forget to consider the direction of the acceleration

7. Nov 6, 2012

### Syndrix

Ah, Thank you so much for your help. This will help me a lot with my studies.

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