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Energy Lost by Friction

  1. Oct 14, 2009 #1
    1. The problem statement, all variables and given/known data
    A 22.9-kg girl slides down a playground slide with a vertical drop of 3.29 m. When she reaches the bottom of the slide, her speed is 1.34 m/s. How much energy was dissipated by friction?


    2. Relevant equations
    I used [tex]\Delta[/tex]KE + [tex]\Delta[/tex]PE = 0



    3. The attempt at a solution
    I got KEinitial = 0
    KEfinal = 1/2 * 22.9kg * (1.34)2 = 20.56

    I got PEinititial = m * g * h = 22.9 * 9.8 * 3.29 = 738.34
    PEfinal = 0

    20.56 (Friction) = 738.34

    Is that on the right track? Where would I go from there?
     
  2. jcsd
  3. Oct 14, 2009 #2

    gabbagabbahey

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    Hi Sw0rDz, welcome to PF!:smile:

    If you are using PE to represent gravitational potential energy, then there is another term that should be included in this equation; the work done by friction Wf:

    [tex]\Delta\text{KE}+\Delta\text{PE}+W_f=0[/tex]

    make sense?

    P.S. In the future, problem like this should go in the Introductory physics folder:wink:
     
  4. Oct 14, 2009 #3

    Doc Al

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    This would be true if mechanical energy was conserved--there was no friction--but that's not the case here. How would you modify this to include energy dissipated by friction?

    This looks good.

    I don't know where you got this or what it means.

    Hint: Compare the initial mechanical energy with the final.
     
  5. Oct 14, 2009 #4
    Thanks! That is what I needed.

    Second part of the problem is:
    If the slide is inclined at 20.5° with the horizontal, what is the coefficient of kinetic friction between the girl and the slide?

    KE doesn't depend on the angle, so I left it the same.

    I took PE to be m * g * cos(20.5) *h

    Is that right?
     
  6. Oct 14, 2009 #5

    gabbagabbahey

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    No, neither energy term will depend on the angle....what will depend on the angle are the components of the gravitational and frictional forces parallel and normal to the slide.:wink:
     
  7. Oct 15, 2009 #6
    Energy lost due to work will be the same. Just when trying to calculate the coefficient of friction will depend on the angle?
     
  8. Oct 15, 2009 #7

    gabbagabbahey

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    What direction does the force of friction act in? What is the formula for finding its magnitude?
     
  9. Oct 15, 2009 #8
    In my diagram, have it acting left (negative)

    Previously I got.
    Wfriction = 717.79J
    (Was right according to LON-CAPA and previous equation)

    Now I set. 717.79 = [tex]\upsilon[/tex]KF * M * G * Cos([tex]\Theta[/tex])
     
  10. Oct 15, 2009 #9

    gabbagabbahey

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    Where did you get that formula from?:confused:

    What is the general formula for frictional force? What is the general formula (involves an integral) for work done by a force?
     
  11. Oct 15, 2009 #10
    Ffriction = (Coefficient of KF) * (Force)
    W = [tex]\int[/tex] F * dl

    Where dl = infinitive displacement.
     
  12. Oct 15, 2009 #11

    gabbagabbahey

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    Right, and what is the normal force in this case?
     
  13. Oct 15, 2009 #12
    [tex]F_n_o_r_m_a_l = M *G * Cos (\theta) [/tex]
     
  14. Oct 15, 2009 #13

    gabbagabbahey

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    You should probably use a lowercase 'g' so as not to confuse it with the universal gravitational constant, but yes.

    So, [itex]F_f=\mu_f mg\cos\theta[/itex]...what about its direction?
     
  15. Oct 15, 2009 #14
    Negative because it acts against the positive kinetic energy.

    Maybe..
    [tex]
    -F_f=\mu_f mg\cos\theta
    [/tex]
     
  16. Oct 15, 2009 #15

    gabbagabbahey

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    "Negative" isn't a direction, it's just a sign... does the frictional force act to the left? to the right? up? down? tangent to the slide? normal to the slide?
     
  17. Oct 15, 2009 #16
    In my diagram. Positive direction is to the right. Negative is to the left. Friction goes to the left.
    Tangent to the slide, I think..
     
    Last edited: Oct 15, 2009
  18. Oct 15, 2009 #17

    gabbagabbahey

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    Directly to the left? Shouldn't it be going tangent to the slide, opposite to the girl's velocity?
     
  19. Oct 15, 2009 #18
    Yes because friction is based of the Normal Force. And Normal force is perpendicular to the tangent line of the slide.
     
  20. Oct 15, 2009 #19

    gabbagabbahey

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    Okay, so wha soes that tell you about [itex]\textbf{F}_f\cdot d\textbf{l}[/itex]? And what do you get when you integrate it over the length of the slide? How long is the slide?
     
  21. Oct 15, 2009 #20
    [tex]
    \int_0^{3.2} \left( ce_{KF} \cdot m\cdot g\cdot cos(\theta)\cdot h\right) dh
    [/tex]


    [tex]
    \left[ ce_{KF} \cdot m\cdotg\cdot g\cdot cos(\theta) \cdot \frac{1}{2} h^{2} \right]_{0}^{3.2}

    [/tex]

    ??
     
    Last edited: Oct 15, 2009
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