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Help with Potential Energy vs Kinetic Energy and finding final velocity!

  1. Dec 11, 2012 #1
    Ok before we get to number 3, I need to show you what I did for number 2 because this is kind of a two-part problem:

    Number2EDITED.png

    PEinitial + KEinitial = [STRIKE]PEfinal[/STRIKE] + KEfinal
    [STRIKE]m[/STRIKE]gy + (1/2)[STRIKE]m[/STRIKE]vi2 = (1/2)[STRIKE]m[/STRIKE]vf2
    gy + (1/2)vi2 = (1/2)vf2
    (9.81)(10) + (1/2)(102) = vf2/2
    98.1 + 50 = vf2/2
    148.1 = vf2/2
    2(148.1) = vf2
    296.2 = vf2
    √(296.2) = vf
    17.2 m/s = vf
    (ANSWER "A")

    NUMBER3EDITED.png

    So for this one, I didn't really know what to do :(
    I tried doing this, but didn't get any of the choices above

    PEinitial + KEinitial = [STRIKE]PEfinal[/STRIKE] + KEfinal
    [STRIKE]m[/STRIKE]gy + (1/2)[STRIKE]m[/STRIKE]vi2 = (1/2)[STRIKE]m[/STRIKE]vf2
    gx + (1/2)vi2 = (1/2)vf2
    (9.81)(10) + (1/2)(252) = (1/2)vf2
    98.1 + 312.5 = vf2/2
    410.6 = vf2/2
    2(410.6) = vf2
    821.2 = vf2
    √(821.2) = vf
    28.6 m/s vf

    But that's not right :(
    Help?
     
  2. jcsd
  3. Dec 11, 2012 #2
    Oh wait!!!

    1) If μ = 1.5, then fk = (1.5)(10 * -9.81)
    fk = -147.15

    2) If fk = -147.15, then F (forward) = +147.15

    3) If F (forward) = +147.15, I can find acceleration, knowing that my mass, m, is 10 kg.

    4) 147.15 = (10)a
    147.15/10 = a
    14.7 m/s2 = a


    5) Then I tried plugging into a kinematic equation that had the 3 variables I had the values for (vf = 25 m/s, a = 14.7 m/s2, and Δx = 9.0 m)

    vf2 = vi2 + 2axΔx
    252 = vi2 + 2*14.7*9
    625 = vi2 + 264.6)
    625 - 264.6 = vi2
    √(360.4) = vi
    18.98 = vi
    19.0 m/s = vi


    Am I right? :D
     
    Last edited: Dec 11, 2012
  4. Dec 11, 2012 #3

    SammyS

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    Treat this as an entirely different problem in which the block is moving with a velocity of 25.0 m/s on the horizontal portion of the track, just before encountering the 9 meter long section having non-zero friction.

    In your second post, you did calculate the friction force correctly.

    Since you used energy concepts to solve problem #2, that seems like a good approach for this problem as well.
    I suggest using the Work - Energy Theorem.​

    By the way: The potential energy is constant in problem #3. --- Why ?
     
  5. Dec 11, 2012 #4
    oh well I said it was constant because of this diagram I found online.

    img_full_46603.gif
     
  6. Dec 11, 2012 #5
    I have one more question though...

    This one I have no idea how to solve for it... :(
    It's a continuation of the last 2. It's basically the last part of this question.

    ScreenShot2012-12-11at73118PM.png

    Edit:

    Oh wait!!!

    Given
    k = 50.4 N/m
    vi = 10 m/s
    m = 10 kg

    KEinitial/moving block about to collide with spring in equilibrium = PEspring
    (1/2)mv2 = (1/2)kx2
    (1/2)(10)(102) = (1/2)(54)x2
    500 = 27x2
    500/27 = x2
    18.5 = x2
    √(18.5) = x
    4.3 m = x

    Is that right? :D
     
    Last edited: Dec 11, 2012
  7. Dec 11, 2012 #6

    haruspex

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    Looks good.
     
  8. Dec 11, 2012 #7
  9. Dec 11, 2012 #8

    SammyS

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    That's sort of right .

    Why did you call that vi ?

    That's the speed after passing over the friction region.
     
  10. Dec 11, 2012 #9

    haruspex

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    Which is the starting point of part 3.
     
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