# Newton's Laws and the coefficient of kinetic friction

1. Oct 29, 2007

### petern

Here is the problem:
A 12.5 Kg wooden crate with an initial velocity of 2.5m/s slides across a rough cement floor for 1.7 m before coming to rest. Find the coefficient of kinetic friction. (Use GUESS method.) The answer is 0.19.

The equation you use is: coefficient = kinetic friction/normal force

I worked backwards so I got 0.19 = kinetic friction/(9.8 gravity x 12.5 kg)

That means the kinetic friction has to be 23.275 N.

Can someone show me how to work this one correctly. I don't know where the 23.275 came from.

2. Oct 29, 2007

### Staff: Mentor

Start by finding the crate's acceleration. Then apply Newton's 2nd law.

3. Oct 29, 2007

### petern

OK, so I figured out that acceleration is -.075 m/s^2. So friction is equal to m x a. This would mean the plugged in values would be: coefficient = (12.5 kg x -.75 m/s^2)/(12.5 kg x 9.8 m/s^2). I got the answer -.077 but this is wrong because the correct coefficient is .19. What am I doing wrong?

EDIT: I just realized that I forgot to square the velocity as I was solving for a. So the acceleration should have actually been -1.838 m/s^2. Thank you so much for the help. I finally figured it out.

Last edited: Oct 29, 2007
4. Oct 29, 2007

### petern

I do have one last question, is the working equation (v^2/x)(m) / (g)(m)?

5. Oct 29, 2007

### eyehategod

is the mew exactly .19?

Last edited: Oct 29, 2007
6. Oct 29, 2007

### eyehategod

i havent done these types of problems in awhile but this is what i have.

to find acceleration use:
v$$^{2}_{f}$$=v$$^{2}_{i}$$+2a$$\Delta$$x//solve for a
a=(v$$^{2}_{f}$$-v$$^{2}_{i}$$)/2$$\Delta$$x//substitute known values
a=-1.838m/s$$^{}2$$

now use:
$$\Sigma$$F=ma//the sum of the forces=ma;the only force acting here is friction
-$$\mu$$mg=ma//m's cancel out and you get
$$\mu$$=-a/g
$$\mu$$=.187

Last edited: Oct 29, 2007
7. Oct 29, 2007

### petern

Yeah, I got that exact answer too. Thanks for the help.