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Frictional Force 2

  1. Feb 17, 2007 #1
    INTRODUCTION:
    This is a problem from my Introduction to Physical Science class using "Conceptual Physics" 10th Ed.by Paul G. Hewitt

    EXACT PROBLEM:
    "A hockey puck slides 20.0m across a frozen pond in a time of 8.0 seconds before coming to a rest."

    PROBLEMS FACED:
    Find the coefficient of kinetic friction between the ice and the puck.

    MY THOUGHTS:
    I don't know what to do here really. I am not sure which formulas to use.
     
    Last edited: Feb 17, 2007
  2. jcsd
  3. Feb 17, 2007 #2
    you use the equation x= x0 + vt +at^2 to determine how quickly the puck deaccelerates and plug that rate into the friction equation.
     
  4. Feb 17, 2007 #3
    That sounds like a plan! So I need to use one of these friction equations?
    https://www.physicsforums.com/showpost.php?p=920984&postcount=4
     
  5. Feb 17, 2007 #4
    so how do I rewrite this to solve for (de)acceleration?

    Does this look correct?
    a = (x-x0-v0t)/(.5t)

    Do the X variables cancel out?
     
    Last edited: Feb 17, 2007
  6. Feb 17, 2007 #5

    hage567

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    Homework Helper

    "you use the equation x= x0 + vt +at^2 to determine how quickly the puck deaccelerates and plug that rate into the friction equation."


    I don't think this the proper equation to use, at least not by itself. V is this equation represents the initial velocity, which we don't know. So you can start by getting an expression for the initial velocity (Vo, by using another kinematic equation), and then put that into this equation for V.

    If you bring arrange the equation so (X-Xo) = Vo + 0.5at^2, you can see that the (X-Xo) is just the displacement of the puck (20 - 0 = 20 m).
     
  7. Feb 18, 2007 #6
    how can I figure out the velocity if I don't know the acceleration?
     
  8. Feb 18, 2007 #7

    hage567

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    Homework Helper

    You can get an equation for Vo in terms of a by using one equation. Then you put that into the other equation you have with Vo and a, and you will only have "a" left over to solve for. It is a two equations, two unknowns kind of thing.
     
  9. Feb 18, 2007 #8
    To me, this is like telling a lost person to just go North, when the lost person doesn't even know which direction north is.

    I need a point of reference and a starting point. I know the forum rules about how you aren't supposed to do my work, but how are you helping me if you only give me a small fraction of advice every time?
     
    Last edited: Feb 18, 2007
  10. Feb 18, 2007 #9
    Well, that's because we want you to try to work it out on your own. Unfortunately people who are lost see these pieces of advice as cryptic hints. But, you should probably know these two equations:

    [tex]x = x_0 + v_0t + \frac{1}{2}gt^2[/tex]

    [tex]v = v_0 + at[/tex]

    Then think about what the problem gives you. The problem says the puck is initially traveling at some speed v0 (which you don't know) and it is decelerating because of friction. It travels a certain distance from the point x0 = 0 to x = 20 in t = 8 seconds. See if you can do it from there!
     
  11. Feb 18, 2007 #10
    I am familiar with these equations, but why did you replace variable a with variable g? Gravity, I take it?
     
  12. Feb 18, 2007 #11
    Oh crap, yes. That's just a typo, I'm so used to using g there it just slipped out :X But yes, the same equation, just use a instead of g.
     
  13. Feb 18, 2007 #12
    so does that mean it is:

    x = x0 + v0t + (1/2)gt^2
    20 = 0 + v0(8) + (1/2)(9.8)(8^2)
    20 = v0(8) + 313.6
    -293.6/8 = v0
    v0 = 36.7 m/s
     
    Last edited: Feb 18, 2007
  14. Feb 18, 2007 #13
    No, it looks like this typo is causing more trouble than it was worth :X Your acceleration is not g, you aren't even considering gravity in this problem. The equations I meant to write are:


    [tex]x = x_0 + v_0t + \frac{1}{2}at^2[/tex]

    [tex]v = v_0 + at[/tex]
     
  15. Feb 18, 2007 #14
    superb. idk then.
     
  16. Feb 18, 2007 #15
    Alright, we have those two equations, and the problem told us:

    x0 = 0 m
    x = 20.0 m
    t = 8.0 s
    v = 0 m/s (if the object comes to rest, it doesn't have velocity!)
    v0 = ?
    a = ?

    We have two equations and two unknowns, so we need to solve a system of equations! Go ahead and write those numbers into those equations and you'll probably see how to solve it. It's just algebra from here!
     
  17. Feb 18, 2007 #16
    Ok then:

    V = v0 + at
    0 = v0 + a8
    v0 = -a8

    x = x0 + v0t +(.5)a(t^2)
    20 = 0 +(-a8) + (.5)a(8^2)
    20 = -a64 + a32
    .625 = -a64 +a
    .0097 = -a +a
    .0097 = 0 ???

    well now, maybe that algebra isn't so simple after all...
     
  18. Feb 18, 2007 #17
    Try to do it without the numbers first, solve for it in terms of variables only. All the numbers is confusing you. When you have the eqn already, then plug in the numbers.
     
  19. Feb 18, 2007 #18
    So far, so good...

    Here is where you made a mistake. If 20 = -a*64 + a*32, isn't this the same as
    20 = -64*a + 32*a
    20 = -32*a
     
  20. Feb 18, 2007 #19
    OHHHHHH! Stupid mistake there!

    Ok then:

    V = v0 + at
    0 = v0 + a8
    v0 = -a8

    x = x0 + v0t +(.5)a(t^2)
    20 = 0 +(-a8) + (.5)a(8^2)
    20 = -a64 + a32
    20 = -32a
    a = -.625

    v0 = -a8
    v0 = .625(8)
    v0 = 5

    so here's all of the variables that we know now:
    x0 = 0 m
    x = 20.0 m
    t = 8.0 s
    v = 0
    v0 = 5
    a = -.625
     
    Last edited: Feb 18, 2007
  21. Feb 18, 2007 #20
    Alright! So we have our acceleration on the puck. Now, remember Newton's second law, F = ma. We can ignore forces in the y-direction because they are mg and N, and they cancel exactly in this case. So, we are only concerned about forces along the x-axis. And that force acting on the puck is...?
     
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