# Homework Help: Due Tommorrow:Hockey Puck shot across the ice(trying to find coefficient of friction)

1. Dec 17, 2008

### PvtDonutDJ88

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

A hockey puck is hit on a frozen lake and starts moving with a speed of 12.0m/s. Exactly 5.0sec later, its speed is 6.0m/s. What is the puck's average acceleration? What is the coefficient of kinetic friction between the puck and ice?

2. Relevant equations
VF=VI+at
FN=mg
FF=UFN
FF=ma
EFY=FN+Fgcos(0)
(cos(0)=1)

3. The attempt at a solution
I have the acceleration it's -1.2m/s2 (that was easy been doing that for a while)The problem is I am having trouble on finding the coefficient of Friction. I'm pretty sure the last four equations are correct, do i need to find Fn next? Because i don't have the mass and it's not moving at a constant velocity, so i do need mass right?if and multiply that by gravity to get Fn? All I know i have the info just a little confused on how to use it

Never mind i got it no worry's

Last edited: Dec 17, 2008
2. Dec 17, 2008

### luijsu

Re: Due Tommorrowockey Puck shot across the ice(trying to find coefficient of frict

If you solve the equation for FF=ma, you'll find that the m's cancel out.

3. Dec 17, 2008

### PvtDonutDJ88

Re: Due Tommorrowockey Puck shot across the ice(trying to find coefficient of frict

uh? what do u mean m's cancel out? The mass cancels out? and ya that's one formula forgot (i edited it in)

4. Dec 17, 2008

### luijsu

Re: Due Tommorrowockey Puck shot across the ice(trying to find coefficient of frict

Yes. You'll find that writing the frictional force in terms of m,g, and u and setting it equal to ma will result in the masses canceling out.

5. Dec 17, 2008

### bpaterni

Re: Due Tommorrowockey Puck shot across the ice(trying to find coefficient of frict

edit: yep ^^^^^^^^

I believe he/she means that you can replace m*a for the force of friction, so:

-1.2m = k*-9.81m

By solving this, you see that the mass cancels and you're left with the coefficient of friction.