# Coeffiecient of Kinetic Friction problem? (Shuffleboard and Distance traveled)

1. Oct 14, 2008

### glindawantsme

Coeffiecient of Kinetic Friction problem??? (Shuffleboard and Distance traveled)

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

A shuffleboard disk is accelerated to a speed of 5.8m/s and released. If the coefficient of kinetic friction between the disk and the concrete court is 0.31, how far does the disk go before it comes to a stop? The courts are 15.8m long.

Known info:
vi = 5.8m/s
vf = 0m/s
$$\mu$$k = 0.31

2. Relevant equations

I have no idea!

3. The attempt at a solution

So, I originally tried vf2 = vi2+ 2a$$\Delta$$d, but that would make no sense because I used 15.8 as the distance but then the problem would be solved....

2. Oct 15, 2008

### tiny-tim

Welcome to PF!

Hi glindawantsme! Welcome to PF!

(have a delta: ∆ )

Yes, that is the right equation …

So what is vf? what is vi? and what is a?

Put them all in, and find ∆d.