# Kinetic energy and friction force?

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1. Oct 8, 2014

### Jessica Sweet

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

A car with a weight of 15,000 N moves horizontally at 30 m/s.
a) What is the car's KE?
b) What will be the magnitude friction force to stop the car over a 60m distance?

2. Relevant equations
F=ma
KE=1/2*m*v2

3. The attempt at a solution
a) mass of car = 1531 kg
KE = 1/2(1531)(30)2 = 690,000 J

b) This is where I hit trouble.
I found acceleration by (0-30)/2 = -15 m/s2
Then I used F=ma, so Ff = (1531)(-15) = -22,965 N

Last edited: Oct 8, 2014
2. Oct 8, 2014

### Staff: Mentor

What equation will you use that relates velocity, acceleration, and distance, to allow you to correctly calculate the acceleration (i.e., deceleration)?

3. Oct 8, 2014

### Jessica Sweet

Oh okay! I totally forgot about that equation.
v2 = vi2 + 2a(delta x)
That'll give me acceleration, and then I use F=ma, right?

4. Oct 8, 2014

### Staff: Mentor

Sounds right.

5. Oct 8, 2014

### Jessica Sweet

I'm just curious, is there a way to find that without needing that equation??

6. Oct 8, 2014

### Staff: Mentor

You could use energy conservation. The car starts out with a certain KE (which you calculated), and that energy will be lost to friction acting over the given stopping distance.

7. Oct 8, 2014

### Jessica Sweet

Could someone demonstrate this? Because I'm a little lost.

8. Oct 8, 2014

### Staff: Mentor

What is the work done by a force F acting over a distance d?

9. Oct 9, 2014

### Jessica Sweet

I'm thinking of W = Fdcosθ

10. Oct 9, 2014

### Staff: Mentor

Sure. Here θ is 180° since the force is acting against the direction of travel. So you can write

W = -Fd

So the friction force will be "stealing" energy from the car as it moves.

Now, the car starts out with a store of KE. This will be lost to friction according to the work done by that friction force. Write an equation that equates the starting KE to the energy lost to friction over distance d.