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Homework Help: Simple friction problem

  1. Apr 6, 2009 #1
    1. The problem statement, all variables and given/known data

    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: Apr 6, 2009
  2. jcsd
  3. Apr 6, 2009 #2


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    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: Apr 6, 2009
  4. Apr 6, 2009 #3
    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 * 50m

    Solving 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?
  5. Apr 7, 2009 #4


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    Science Advisor
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    Gold Member

    you have written that
    distance = average velocity*time^2 . Correct this and you'll be OK.
  6. Apr 7, 2009 #5
    okay, that makes everything better. That was a silly mistake. Thanks for all the help.
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