Coefficients of kinetic friction, help PLEASE

AI Thread Summary
The discussion revolves around a physics problem involving a hockey puck's motion on ice. The average acceleration of the puck was calculated to be -1.56 m/s², and the distance traveled in 5 seconds was found to be 45.5 m. The main challenge was determining the average coefficient of kinetic friction, which required understanding forces and Newton's second law. Initially, the user struggled with the absence of mass but eventually realized that dividing the acceleration by gravity provided the necessary information. The conversation highlights the importance of applying fundamental physics concepts to solve problems.
myelevatorbeat
Messages
55
Reaction score
0
Here is the problem:

A hockey puck is hit on a frozen lake and starts moving with a speed of 13.0 m/s. Five seconds later, its speed is 5.20 m/s.

(a) What is its average acceleration?

I used the kinematic equation V=Vo+at to find the accerlation which came out to be -1.56 m/s^2 (this answer is right).

(b) What is the average value of the coefficient of kinetic friction between puck and ice?

(c) How far does the puck travel during the 5.00 s interval? m

I used X-Xo=1/2[V+Vo]t to solve this and got 45.5 m, which is right.


So basically, I'm stuck on part B of this problem. I don't remember my professor going over how to solve this particular type of problem in class so I was hoping someone here could get me started. I'm not asking for the answer, just an explanation of how to solve Part B.
 
Physics news on Phys.org
Think about forces and Newton's second law. You have already figured out the acceleration. What's the definition of kinetic friction?
 
I assume I would make F=ma equal to Fk=uN.

However, I don't see how I can do this since the mass wasn't given.
 
OK, but what is the normal force equal to?
 
The normal force is equal to the weight, which is mass x gravity, but again I get lost because I don't have the mass.
 
Wait, nevermind. I got it. I divided acceleration/gravity and came out with the answer.

Thanks for your help though!
 
Good!

You're welcome :smile:
 
Back
Top