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Work done by air on a baseball

  1. Nov 12, 2009 #1
    1. The problem statement, all variables and given/known data

    The following is data from a computer simulation for a batted baseball with mass 0.145 kg, including air resistance:
    t x y v_x v_y
    0 0 0 30.0 m/s 40.0 m/s
    3.05 s 70.2 m 53.6 m 18.6 m/s 0
    6.59 s 124.4 m 0 11.9 m/s -28.7 m/s

    How much work was done by the air on the baseball as it moved from its initial position to its maximum height?

    How much work was done by the air on the baseball as it moved from its maximum height back to the starting elevation?


    Man, I am clueless about this problem. The baseball has to move in a parabolic motion though, right? It's a batted ball.

    I made a bar graph first of all


    (Initial) KE = PE+KE(Final)

    Initially, the Vx and Vy are 30.0 m/s and 40.0 m/s respectively. I did the Pythagorean theorem and obtained V = 8.37m/s

    Now

    1/2mv^2 = mgh + 1/2mv^2

    .5(.145kg)(8.37m/s)^2 = (.145kg)(9.80m/s^2)(53.6m) +1/2(.145kg)(18.6m/s)2 + [tex]Delta[/tex]Uinternal

    -96.17J = Uinternal

    ?
     
  2. jcsd
  3. Nov 12, 2009 #2

    PhanthomJay

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    right.
    you forgot to include the work done by the air in this equation.
    try it again, you have an error here, it's v^2 = vx^2 + vy^2
    correct your math error for v, and you should be OK. How about the downward journey? Watch your plus and minus signs.
     
  4. Nov 13, 2009 #3
    Where does the work done by air go to? Initial? How would it be expressed in an equation?
     
  5. Nov 13, 2009 #4

    PhanthomJay

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    you will end up with the correct answer in the end when you correct the math, but your equations left off the work term or U_therm term initially, then you added it in later.
    There are several ways to write the general Conservation of Energy equation. One of them is

    [tex]KE_i + PE_i + W_{nc} = KE_f + PE_f [/tex] where [tex]W_{nc}[/tex] is the work done by all nonconservative forces like air resistance, friction, applied forces, etc. Another way is to write it

    [tex] KE_i + PE_i + W_{applied forces} = KE_f + PE_f + U_{therm}[/tex] where [tex]U_{therm}[/tex] is the internal heat energy caused by the the work done by friction, air resistance, or other heat producing non-conservative forces. I like the first one better. There are other ways, like

    [tex] W_{nc} = \Delta KE + \Delta PE [/tex], which is the same as the first equation in another form. As always, be sure to mind your plus and minus signs.
     
    Last edited: Nov 13, 2009
  6. Nov 13, 2009 #5
    Doing the math correctly now


    .5(.145kg)(50m/s/s)^2 = (.145kg)(9.80m/s^2)(53.6m) +1/2(.145kg)(18.6m/s)2 + LaTeX Code: Delta Uinternal

    181.25J=101.2477J + [tex]\Delta[/tex]Uinternal

    [tex]\Delta[/tex]Uinternal = 80.0023J

    But my hw accepts it at -80.0023J. I'm guessing it is because friction always opposes so it would be negative work?


    For the second part of the question: How much work was done by the air on the baseball as it moved from its maximum height back to the starting elevation?

    Initial (Top) KE + PE = Final (Bottom) KE+ Uinternal

    101.2477J = .5(.145kg)(31.069m/s)^2 + Uinternal

    I got the 31.069m/s by

    (11.96)^2+(28.7)^2=V^2

    101.2477J = 69.98300017J + Uinternal

    31.26J = Uinternal ?
     
    Last edited: Nov 13, 2009
  7. Nov 13, 2009 #6

    PhanthomJay

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    yes, that's one reason why i like the other form of the equation that uses work instead of thermal or internal energy; the minus sign takes care of itself that way.
    Oh there you go again, right number, wrong sign. What happened to the minus sign? :cry: The work done is negative, but i guess you consider the internal energy as positive, but the problem asks for the work, not energy.
     
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