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Help with friction

  1. Feb 5, 2009 #1
    1. A curling stone with a mass 20.0kg leaves the curler's hand at a speed of 0.885m/s. It slides 31.5m down the rink before coming to rest. Find the average force of friction acting on the stone.



    2. So far I have (I think) net force, acceleration, average velocity and time. I need either the force of friction, or the force applied to work out the force of friction.



    3. So far my combined formula looks like this, have not put any numbers in yet. Fnet = [m(V2-V1)(V1+V2)]/2d.
     
  2. jcsd
  3. Feb 5, 2009 #2

    Nabeshin

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    Thinking about the problem in terms of energy might help to simplify things for you ;)
     
  4. Feb 5, 2009 #3
    Care to explain a little more please? I can find the initial kinetic energy.... can I get from there to the force of friction knowing the distance it took to stop?
     
  5. Feb 5, 2009 #4

    Nabeshin

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    Well you know that it comes to a complete stop so the final kinetic energy should be zero. Do you know how to find the work friction (or a force in general) does over a given distance?
     
  6. Feb 5, 2009 #5
    Hmm, isn't work force multiplied by distance? But I don't know the force, or the work done.
     
  7. Feb 5, 2009 #6

    Nabeshin

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    Right, work is equal to the force multiplied by the distance. In this case the force is the force of friction, and I assume you remember the formula for the force of friction?

    So then you should be able to write down an equation so that [tex]\Delta E = 0 [/tex], including the work done by friction. Does this make sense to you? Give it a try and see if you can come up with a number.
     
  8. Feb 5, 2009 #7
    Ok, so Force of friction = coefficient of friction * mass * 9.81
    I know mass, but not the coefficient of friction.
    How can Delta E = 0 if it starts with some kinetic energy and ends with no energy?
    Kinetic energy at the moment of release is 0.5 * 20 * 0.885^2 = 7.83 N*m which is a watt? I think? So if work is force over distance the work is 7.83 N*m (whatever that unit is) So 7.83 = Force of friction * 31.5m
    Force of friction is.... 0.25N?
     
  9. Feb 5, 2009 #8

    Nabeshin

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    Correct, the average force of friction is .25N.

    Two things to clarify though, I didn't realize that you were looking for force of friction when I made mention of the formula for it, sorry about that!

    Also when I said [tex]\Delta E = 0 [/tex] What I meant was something like [tex] \frac{1}{2}mv^{2}+F_{f}x=0 [/tex]
    Which I should have written instead as:
    [tex]E_{0}+W=E_{f}[/tex]. Sorry, my notation is a little sloppy ^^ But you got it anyways.
     
  10. Feb 5, 2009 #9
    Ahh I see. Thanks for all your help :)
     
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