# Homework Help: Friction problem

1. Jan 20, 2008

### Rusk1

[SOLVED] Friction problem

1. The problem statement, all variables and given/known data
Car is moving down a 10degree slope and applies the breaks. The net force acting on the car as it stops is -2.0 x 10^4 N. coefficient of kinetic friction is .797. What is the mass? what is the magnitude of the normal force?

2. Relevant equations
Fnet=M*Anet=Fapplied - Fk
Fk= muk*m*a = muk*Fn
M= (Fapplied/Anet + mukG)
i think that would be it

3. The attempt at a solution
i tried doing this many ways but always seem to get it wrong. I quess I don't understand what the net force actually is. It is not equal to Fk. This is probably a lot simpler and i'm putting too much thought into it.

2. Jan 20, 2008

### DylanB

The only forces acting on the car are the force of kinetic friction between the wheels and the ground, the normal force and gravitational force.

These forces make up the net force, I'm unsure what you mean by Fapplied? You should break these forces into components acting parallel and perpenticular to the slope. The forces acting perpendicular will sum to zero (no acceleration in that direction), and the sum of the parallel forces will add to the net force.

I suggest drawing a free body diagram, and writing down these two equations to start.

3. Jan 20, 2008

### Rusk1

ok so what i'm trying to ask is how do i find the mass? i would need the mass to find the normal force and the gravitational force. i have drawn a FBD and everything i try i always come up with the wrong answer. I'm just out of ideas.

4. Jan 20, 2008

### rl.bhat

From FBD find the component of the weight of the car along the inclined plane and normal to the inclined plane. Frictional force acts along the inclind plane but in the opposite direction to the motion of the car. Net force applied by the break to stop the car is equal to the difference of the forces acting along the inclined plane. Write the equations. You will get the mass.

5. Jan 20, 2008

### Rusk1

i finally figured it out thanks.