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Coefficient of kinetic friction on hockey puck

  1. Sep 25, 2007 #1
    The way I approached the problem I found acceleration using V^2=vo^2 + 2ax. Then I set uk*mg = ma. The masses cancel... so its uk*g = a. I keep getting the wrong answer. What am I doing wrong?




    1. The problem statement, all variables and given/known data

    A hockey puck on a frozen pond with an initial speed of 22.3 m/s stops after sliding a distance of 249.3 m. Calculate the average value of the coefficient of kinetic friction between the puck and the ice.

    2. Relevant equations

    uk*m*g = ma

    3. The attempt at a solution

    0.0093
     
    Last edited: Sep 25, 2007
  2. jcsd
  3. Sep 25, 2007 #2
    I know the masses cancel... What am I missing?
     
  4. Sep 25, 2007 #3

    Kurdt

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    How did you work out the acceleration?
     
  5. Sep 25, 2007 #4
    0 = 22.3 ^2 + 2 * a * 249.3

    and i solved for a
     
  6. Sep 25, 2007 #5

    Kurdt

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    Ok. Now you know the masses cancel, which leaves you with:

    [tex] \mu_k = \frac{a}{g} [/tex]

    What do you get for your answer, knowing that g is the acceleration due to gravity?
     
  7. Sep 26, 2007 #6
    Try this:

    1 ∑F_{x}=-f_{k}=ma_{x}
    2 ∑F_{y}=n-mg=0
    note: n=mg becomes(n/g)=m
    3 if puck is moving right then:
    4 -u_{k}n=-u_{k}mg=ma_{x}
    5 a_{x}=-u_{k}g
    6 V_{xf}²=V_{xi}²+2a_{x(X_{f}}-x_{i)}
    7 so now 0=V_{xi}²+2a_{x}x_{f} becomes
    8 V_{xi}²-2u_{k}gx_{f}
    9 u_{k}=((V_{xi}²)/(2gx_{f}))
    10 u_{k}=(((22.3(m/s))²)/(2(9.80(m/(s²)))(249.3m))) =0.1017727
    11 how's that?
     
  8. Sep 26, 2007 #7

    Kurdt

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    Please don't give away full solutions. It defeats the object of helping students to learn for themselves.
     
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