Coefficient of kinetic friction for a hockey puck on ice

[WCU]

ok, i have a problem and w/o a given mass i am lost on how to solve it.

"What is the coefficient of kinetic friction between the ice and the puck in the problem above?"

problem above states, "A Carolina Hurricanes player hits a puck with his hockey stick during a practice, giving the puck an initial speed of 5.50m/s. The puck decelerates uniformly and comes to rest after traveling 25.0m. What is the rate of acceleration for the puck?"
-answer is 0.605m/s^2

Please Help Me!

 

[Parth Dave]

Why don't you draw yourself a free body diagram of the puck? Try to figure out what forces are working in the horizontal direction. And remember, F = ma.

 

[WCU]

yes, i am aware of that, but the MASS is UNKNOWN

if anyone knows an alternate formula (one that doesn't require mass) i would be most grateful

thanks for your help...

 

[stunner5000pt]

you don't need the mass

watch

now that you know the acceleration due to the friction (since the friction force stops it)

$$F = ma = \mu F_{n} = \mu mg$$

where Fn is normal force of the puck and mu is the co efficient of friction.

 

[WCU]

thanks for the help, the formula makes sense, sorta, but i was never taught how to get the normal force without the mass, so I'm still lost, but not as lost

i do know the answer is 0.0617 ... i just don't know how to get there ...

 

[stunner5000pt]

thanks for the help, the formula makes sense, sorta, but i was never taught how to get the normal force without the mass, so I'm still lost, but not as lost

i do know the answer is 0.0617 ... i just don't know how to get there ...

normal force without mass?? come again??

in that equality i have simply used Newtons first law that F = ma and frictional force is F = mu Fn = mu mg and equated the two together and eliminate the mass factor

 

[WCU]

I thought m = mass... i need the coefficient of friction (u), and i don't have F, Fn, or m ... what units do i use? i feel like the answer is right in front of me and I'm just too blind to see it

 

[stunner5000pt]

ok realize this FOR ANY FORCE F = ma

secondly Force of friction = u Fn = u mg (because Force of friction is mg when there is no angle involved)

now use the first and second lines and you can equate ma = u mg

a = u g

u = a / g

 

[WCU]

so (0.605m/s^2)/(9.80m/s^2)=u=0.0617, which happens ot be the answer i was looking for.

thankyou so much