Coefficient of Kinetic Friction, Normal Force.

[Laura EK]

Homework Statement

"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."

Homework Equations

F_k=\mukFN
V²-V₀² / 2x = a

The Attempt at a Solution

The first thing I did was write down all of the known variables:
x = 6.00m
V_0x = 3.00m/s (initial)
V_x = 0m/s (final)

The next thing I did was draw a free body diagram, showing the weight (mg) pointing down, the f_k 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 \muk by dividing the found acceleration (-0.75m/s²) by g (-9.8m/2²) to come up with 0.0765 for \muk.

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

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 :)

 

[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.