# Coefficient of Kinetic Friction, Normal Force.

1. Oct 20, 2009

### Laura EK

1. The problem statement, all variables and given/known data
"A box initially moving at 3.00m/s on a rough, level floor comes to a stop after sliding 6.00m. Find the coefficient of kinetic friction between the box and the floor."

2. Relevant equations
Fk=$$\mu$$kFN
V2-V02 / 2x = a

3. The attempt at a solution

The first thing I did was write down all of the known variables:
x = 6.00m
V0x = 3.00m/s (initial)
Vx = 0m/s (final)

The next thing I did was draw a free body diagram, showing the weight (mg) pointing down, the fk pointing to the left (in the negative x direction), and Normal force (N) pointing up.

I found the acceleration by plugging given values into the second relevant equation:
02 - (3.00m/s)2 / 2 (6.00m) = -0.75m/s2

Then I found $$\mu$$k by dividing the found acceleration (-0.75m/s2) by g (-9.8m/22) to come up with 0.0765 for $$\mu$$k.

According to the solutions page, the answers are correct, but I have no idea why.
Why does a/g = $$\mu$$k?

Also, if you are not given the mass of an object, how can you find the normal force of the object if N=mg?

Any and all feedback is appreciated. Thank you :)

2. Oct 20, 2009

### Delphi51

Welcome to PF, Laura.
Your answer looks good to me. I got .0765 but I used g=9.81.
The equation of motion is Ff = umg = ma so dividing by m gives you
u = a/g. Yes, you can't find the normal force mg but you can still do the problem! Think of it as finding the normal force per kg of mass - you are just doing each kg separately.