Calculating Coefficient of Kinetic Friction in a Hockey Game

  • Thread starter Thread starter xenograftsoul
  • Start date Start date
  • Tags Tags
    Friction
Click For Summary

Homework Help Overview

The problem involves calculating the coefficient of kinetic friction for a puck sliding on ice after being given an initial speed. The puck slides a distance of 50 m before coming to a stop, and the context includes a quiz question that the original poster feels unprepared for due to a lack of prior lessons on friction.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster questions whether the problem can be solved without additional information such as mass or normal force. Some participants suggest using energy concepts, noting that the initial kinetic energy is converted into work done against friction. Others explore the relationship between acceleration, force, and the coefficient of friction, while questioning their calculations and understanding of the concepts involved.

Discussion Status

Participants are actively engaging with the problem, sharing their reasoning and calculations. Some guidance has been offered regarding the relationships between kinetic energy, work, and frictional force. There is recognition of mistakes in calculations, and participants are working through their understanding of the concepts without reaching a consensus on the final answer.

Contextual Notes

There is a mention of the original poster's concern about missing lessons on friction, which may impact their ability to solve the problem effectively. Additionally, there is a correction regarding the use of equations related to distance and time, indicating potential misunderstandings in the setup of the problem.

xenograftsoul
Messages
3
Reaction score
0

Homework Statement



During a hockey game, a puck is given an initial speed of 10 m/s. It slides 50 m on the ice before it stops. What is the coefficient of kinetic friction between the puck and the ice?

A. 0.090
B. 0.10
C. 0.11
D. 0.12 I'm taking an online class and unless I've missed something friction hasn't come up yet in the lessons but is in the quiz. If I remember correctly though (from high school physics) you need some mass or normal force for it.

Is this answerable as is?

Thanks for any help.
 
Last edited:
Physics news on Phys.org
xenograftsoul said:

Homework Statement



During a hockey game, a puck is given an initial speed of 10 m/s. It slides 50 m on the ice before it stops. What is the coefficient of kinetic friction between the puck and the ice?

A. 0.090
B. 0.10
C. 0.11
D. 0.12 I'm taking an online class and unless I've missed something friction hasn't come up yet in the lessons but is in the quiz. If I remember correctly though (from high school physics) you need some mass or normal force for it.

Thanks for any help.

Well, here are some hints: The initial kinetic energy of the puck goes into the work to overcome friction. Work is equal to force times distance. The frictional force is equal to the coefficient of friction times the normal force acting between the puck and the ice. That force is given by the mass of the puck times the acceleration of gravity. Now assemble all these pieces and you will find that you can solve for the coefficient of friction. (There will be a fortuitous cancellation along the way)
 
Last edited:
okay, so far I believe
50m=10m/s / 2 *Time^2 solve for Time gives the time it takes to stop (sqrt10)
10m/s divided by Time= average acceleration (again sqrt10 or 3.16)
3.16m/s^2 * mass = the frictional force = the coefficient of friction * mass * g

I believe that force is also equal to the force in work (equal but opposite).
work = force * distance
work = 3.16m/s^2 * mass * 50mSolving for the coefficient here though still seems fruitless:
3.16m/s^2 * mass =

yup. there was that "and now I feel like an idiot" moment.

3.16m/s^2 * mass = coefficient of friction * mass * 9.8m/s^2

there, now the masses cancel and 3.16m/s^2 divided by 9.8m/s^2 gets rid of the m/s^2
(I am correct in my understanding that the coefficient should be without units)

but that leaves me with sqrt10/9.8 = .32

Have I fumbled somewhere or was this all a red herring?
 
xenograftsoul said:
okay, so far I believe
50m=10m/s / 2 *Time^2 solve for Time gives the time it takes to stop
you have written that
distance = average velocity*time^2 [/color]. Correct this and you'll be OK.
 
okay, that makes everything better. That was a silly mistake. Thanks for all the help.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
1
Views
4K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 16 ·
Replies
16
Views
6K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K