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Work-Kinetic Energy Theorem Applicability

  1. Jun 24, 2013 #1
    Hello there.

    I've yet to take a theoretical mechanics course (next Spring) but I've been reviewing my intro level class and I've run into some confusion regarding the work-kinetic energy theorem.

    My textbooks state that
    [itex]W_{net}=ΔKE[/itex]

    And this makes sense to me: the net work being zero implies zero net force which implies zero acceleration. However, the books then go on to state that

    [itex]W_{ext}=ΔE_{mech}=ΔKE+ΔPE_{g}[/itex]

    and I interpret this to mean that the net work done by external forces on a body is equal to the total change in mechanical energy of that body, including both kinetic energy and potential energy (in this case, gravity).

    What I can't wrap my head around is why the kinetic-energy theorem gives the net work as the change in kinetic energy but the second equation gives the net work as the change in the entire mechanical energy. I believe that the first theorem applies in all cases, but when is the second applicable?

    Thanks for any clarification!
     
  2. jcsd
  3. Jun 24, 2013 #2

    WannabeNewton

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    Science Advisor

    That second equation actually doesn't make sense. If I have a system in free fall within the uniform gravitational field of the Earth going from position ##z_0## to ##z_1## then that second equation is saying ##W_{z_0 z_1} = E(z_1) - E(z_0) = 0## because ##E(z_1) = E(z_0)## for a system in free fall (the mechanical energy is constant along the trajectory of a particle in free fall). But ##W_{z_0 z_1} = -\int_{z_0}^{z_1}mgdz = -mg\Delta z## is obviously not zero.
     
  4. Jun 24, 2013 #3
    Okay, that makes sense to me. So you're saying that the total energy of the mass doesn't change (because gravitational force is conservative, implying no loss of energy) but the net external work done on the mass is non-zero, so the equation is incorrect in such a situation.

    Am I correct in guessing that the second equation is true for non-conservative external forces? For example, if I push a ball up a frictionless incline so that the ball's speed increases, my external force is contributing to the increase of both kinetic and potential energy.
     
  5. Jun 24, 2013 #4

    jtbell

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    Staff: Mentor

    The two equations don't use the same "net work."

    Start with the basic work-KE theorem (your first equation):

    $$W_{net} = \Delta KE$$

    Separate the work into two pieces: the work done by conservative forces (e.g. gravity), Wc, and the work done by non-conservative forces (everything else), Wnc. Rearrange a bit:

    $$W_c + W_{nc} = \Delta KE \\
    W_{nc} = \Delta KE - W_c$$

    By definition, ##\Delta PE = - W_c##, therefore:

    $$W_{nc} = \Delta KE + \Delta PE \\
    W_{nc} = \Delta E_{mech}$$

    This is what your second equation should look like. The work here includes only the work done by non-conservative forces, whereas the first equation includes the work done by all forces.
     
  6. Jun 24, 2013 #5
    Ah, I see. So the general change of energy formula for non-conservative forces follows from the work-kinetic energy theorem. And that ##ΔKE=-ΔPE## for free fall follows easily from this. Great explanation!

    Thank you both a lot! It makes so much more sense now.
     
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