Newton's Laws and the coefficient of kinetic friction

[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.

 

[Doc Al]

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

 

[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.

 

[petern]

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

 

[eyehategod]

is the mew exactly .19?

 

[eyehategod]

i haven't done these types of problems in awhile but this is what i have.

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

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

 

[petern]

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