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Conservative forces

  1. Feb 13, 2009 #1
    1. The problem statement, all variables and given/known data
    A 4.5 kg block slides along a track from one level to a higher level after passing through an intermediate valley. The track is frictionless until the block reaches the higher level. There a frictional force stops the block in a distance d. The block's initial speed is v0 = 5.3 m/s, the height difference is h = 1.0 m, and μk = 0.613. Find d.


    2. Relevant equations
    E-mec 1 (object moving) = K1 + U1
    E-mec 2 (object stopped) = K2 + U2 + energy lost to friction


    3. The attempt at a solution
    K1 + U1 - energy lost to friction = K2 + U2
    I said that K1=(1/2)mv^2 and U1=mgy and the energy lost to friction is (fk)d; however, I don't know how to get the right side of the equation. I tried using the same formulas and I came up with 1.6313, but apparently that is incorrect.
     
  2. jcsd
  3. Feb 13, 2009 #2

    Hootenanny

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    This looks good. So you explicitly we have:

    [tex]\frac{1}{2}mv_1^2 + mgh_1 = \frac{1}{2}mv_2^2 + mgh_2 + R\mu_k d[/tex]

    And you want to find the distance, i.e. you want to solve this equation for d. What's the first thing you usually do when solving an equation?
     
  4. Feb 13, 2009 #3
    The first thing I would do is plug in all the values I know.

    m=4.5kg
    v1=5.3m/s
    g=9.8m/s^2
    h1=1m (not entirely sure about this one...)
    v2=(I would think it to still be 5.3m/s, but I could be wrong)
    h2=0 (I think...)
    R=(no clue.)
    Uk=0.613
     
  5. Feb 13, 2009 #4

    Hootenanny

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    Good.
    Don't worry about these, we'll get to these later.
    Hint: You are looking for the distance d when the object has stopped.
    What is the expression for the maximum frictional force on an object?
    Good.

    Personally, I wouldn't plug in any of the values in just yet. Since we're solving for d I would try to make d the subject of the expression.
     
  6. Feb 13, 2009 #5
    So would v2 be 0 and R=(Us)(Fn)?
     
  7. Feb 13, 2009 #6
    And d=((1/2)m(v1)^2+mg(h1)-(1/2)m(v2)^2-mg(h2))/R(Uk).
     
  8. Feb 13, 2009 #7

    Hootenanny

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    Sounds good to me :approve:
    Yup. However, you can write it in a somewhat nicer fashion:

    [tex]\begin{align*}
    d & = \frac{1}{R\mu_k}\left\{\frac{1}{2}mv_1^2 + mgh_1 - \frac{1}{2}mv_2^2 -mgh_2\right\} \\
    & = \frac{1}{R\mu_k}\left\{\frac{1}{2}m\left(v_1^2 - v_2^2\right) + mg\left(h_1-h_2\right)\right\} \\
    & = \frac{1}{R\mu_k}\left\{\frac{1}{2}m\left(v_1^2 - v_2^2\right) + mg\Delta h\right\}\end{align*}[/tex]

    Does that help?
     
  9. Feb 13, 2009 #8
    One last thing, what values do we plug in for (Us) and (Fn)? I got (Fn) to be 44.1 (although I could be wrong), but I don't know how to find (Us).
     
  10. Feb 13, 2009 #9

    Hootenanny

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    Sorry, I misread your previous post. R is simply Fn= 44.1 N.

    As a side note, take care with the sign of [itex]\Delta h[/itex], recall it's definition [itex]\Delta h = h_1-h_2[/itex].
     
  11. Feb 13, 2009 #10
    Oh, and you never got back to clarifying what h1 and h2 are.
     
  12. Feb 13, 2009 #11

    Hootenanny

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    What do h1 and h2 represent?
     
  13. Feb 13, 2009 #12
    The change in y, right? So h1 is 1 and h2 is 0?
     
  14. Feb 13, 2009 #13

    Hootenanny

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    Almost. h1 and h2 represent the initial and final heights respectively. Hence, [itex]\Delta h = h_1-h_2[/itex] represent the change in height. However, you should note that the question states that the final height is greater than the initial height.
     
  15. Feb 13, 2009 #14
    Oh, so it's the other way around? Meaning (delta)h=-1?
     
  16. Feb 13, 2009 #15

    Hootenanny

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    Yes. :approve:
     
  17. Feb 13, 2009 #16
    Thank you so much for your help!
     
  18. Feb 13, 2009 #17

    Hootenanny

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    A pleasure :smile:
     
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