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Work, Energy & Power confusion

  1. Jan 9, 2015 #1
    I have quite a hard time understand how these things relate and where i go wrong. I think this fits in better here since what i'm having trouble with is the theory and not a specific exercise problem.

    So the book I'm using derived this connection between work and potential energy when a partikle is moved from point 1 to point 2
    [tex]W_{1-2} = V_1 - V_2 [/tex]
    Now imagine i lift an object a distance h i would get
    \begin{cases}V_1 = 0 \\ V_2 = mgh\end{cases}
    and I end up with a negative work
    [tex]W_{1-2} = -mgh [/tex]
    Is the work supported to be negative when i lift something? Also if i did this work over a certain time ##\Delta t## i get
    [tex]P = \frac{-mgh}{\Delta t}[/tex] a negative power. Which gives me a sign error on every exercise i try while in the book's examples it calculate the work to be negative.

    Then to make things even more confusing for me theres this second law of kinetic energy
    [tex]W_{1-2} = T_2 - T_1[/tex]
    So before i start lifting something it got a velocity of 0 just sitting there at a certain height while after I'm done with the lift and just hold it a bit higher up the velocity is again 0. so i get
    [tex]T_2 = T_1 = 0 \Longrightarrow W_{1-2} = 0[/tex]
    how is this consistant, where am i going wrong?
     
  2. jcsd
  3. Jan 9, 2015 #2
    To answer your question briefly, no, the work done to lift something should not be negative. The confusion here is due to the fact that work is done by someone or something. You must ask who (or what) is doing the work?

    For your problem here, the formula you quote ## W_{1-2} = V_1 - V_2 ##, applies to the work done by gravity. From the work-kinetic energy theorem, the total work done on object is equal to its change in kinetic energy. So, if you drop a particle from a height ##y=h##, it will have kinetic energy of ##mgh## when it hits the ground at ##h=0##. Conversely, if you fire the particle upward from ##y=0## with a kinetic energy of ##mgh##, it will reach a maximum height of ##h##.

    Things get more complicated when you enter the picture and start doing work on the particle. You are doing work on the object, and so is gravity. To lift the object, you must supply a constant force against gravity. You can talk about the total work done by you on the particle, but that will not equal the change in its kinetic energy, since you are not the only thing doing work on the object.

    I remember having similar problems when considering an ideal gas undergoing isobaric/isothermal/adiabatic expansion(compression). When the gas expands(compresses), you can think of the piston as doing positive(negative) work on its environment, or you could call this negative(positive) work done on the gas.
     
  4. Jan 9, 2015 #3
    Thanks for replying! I think i may be starting to understand the first part.
    So since gravity is doing ##W_{1-2}## it must really me doing ##-W_{1-2}## work on it since there are no other forces involved.

    The problem with kinetic energy ##W_{1-2} = T_2-T_1##. I'm not sure i understood this right but is it that what i'm doing is actually calculating the total work of the system. So while the total work is zero i have work generated by both me ##-W_{1-2}## and by gravity ##W_{1-2}## so i get ##0 = W_{1-2} -W_{1-2} = T_2-T_1 = 0## which basicly says ##0=0## or if i didn't already know it proves the work i'm generating is the minus the work generated by gravity.
     
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