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Quick work question

  1. Dec 15, 2004 #1
    does work equal change in potential energy or change in kinetic energy?
     
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  3. Dec 15, 2004 #2

    arildno

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    If you don't bother to make the distinction between conservative and non-conservative forces, then the work equals the change of kinetic energy.
     
    Last edited: Dec 15, 2004
  4. Dec 15, 2004 #3

    Doc Al

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    Just to add to arildno's note: The work done by the net force on object will equal the change in its KE.
     
  5. Dec 15, 2004 #4
    I read that something =(-dPE)/(dt)
    not I for got what. Any ideas? Power=(-dPE)/(dt)?
     
  6. Dec 15, 2004 #5

    Tjl

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    Power is [tex]\frac{work}{time}[/tex]

    What you are reading is a calculus based derivative, not relavent to what you are asking.
     
  7. Dec 15, 2004 #6
    I'm asking what (-dPE)/(dt) is equal to.

    if you integrate it, it will become PE/t, which is work/time which is power.

    What variable do you integrate to get power?

    also, is this all correct?
    ∫F(x)=w(x)
    ∫P(t)=w(t)=Fv=F(at)=∆KE/t
     
  8. Dec 15, 2004 #7

    arildno

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    I think a DETAILED explanation is in order here:
    1) Let us consider Newton's 2. law for a system (particle if you like) with constant mass m under influence of a SINGLE force [tex]\vec{F}[/tex]:
    [tex]\vec{F}=m\vec{a} (1)[/tex]
    2) Let us take the dot product of (1) with the velocity [tex]\vec{v}[/tex]:
    [tex]\vec{F}\cdot\vec{v}=m\vec{a}\cdot\vec{v} (2)[/tex]
    Now, note that the right-hand side can be re-written as:
    [tex]m\vec{a}\cdot\vec{v}=\frac{d}{dt}(\frac{1}{2}m\vec{v}^{2})=\frac{dK.E.}{dt}[/tex]
    Or, introducing the POWER [tex]\mathcal{P}=\vec{F}\cdot\vec{v}[/tex]
    [tex]\mathcal{P}=\frac{dK.E.}{dt} (3)[/tex]

    (If we have a system under influence of several forces, (3) reads NET power equals rate of change of kinetic energy)

    3) Let us integrate (3) between two arbitrary time values:
    [tex]\int_{0}^{T}\vec{F}\cdot\vec{v}dt=K.E\mid_{t=T}-K.E\mid_{t=0}(4)[/tex]
    But, since [tex]\vec{v}=\frac{d\vec{x}}{dt}[/tex]
    we may rewrite the time integral with the line integral along the object's trajectory I between positions [tex]\vec{x}(0),\vec{x}(T)[/tex]
    That is:
    [tex]\int_{0}^{T}\vec{F}\cdot\vec{v}dt=\oint_{I}\vec{F}\cdot{d\vec{x}}=W (5)[/tex]
    That is, we have:
    [tex]W=\bigtriangleup{K.E}(6)[/tex]
    (where W is the work done)
    (A notation which ought to be clear)
    And also:
    [tex]\frac{dW}{dt}=\mathcal{P}(7)[/tex]
    Clear so far?
     
  9. Dec 15, 2004 #8
    okay, so

    ∫F(x)=w(x)
    ∫P(t)=w(t)
    P=Fv=F(at)=∆KE/t
    W= ∆KE

    does ∫W=-dPE/dt?
     
  10. Dec 15, 2004 #9

    arildno

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    Your notation is rather obscure, but since you "okayed" my answer, I'll continue:
    Do you know the difference between conservative and non-conservative forces?
     
  11. Dec 15, 2004 #10
    Does this make sence:
    [tex]\int\mathcal{F}(x)=w(x)[/tex]
    where x is in terms of meters
    [tex]\int\mathcal{P}(t)=w(t)[/tex]
    where t is in terms of time

    energy is lost in motion?
     
  12. Dec 15, 2004 #11

    Tjl

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    Energy is not lost in motion. That is a dissipative force. Energy is never lost, it is transformed into another form. Such as heat, wind resistance, etc.
     
  13. Dec 15, 2004 #12
    P=Fv=F(at)=∆KE/t
    this is:

    Power=Force*velocity
    Power=force*(acceleration*time)
    Power=work/time
    Power=change in kinetic energy/time
     
  14. Dec 15, 2004 #13

    arildno

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    "energy lost in motion?"
    No, and your notation is still not good.

    A conservative force has the following property:
    We say that a force is CONSERVATIVE if the work done by that force ONLY DEPEND UPON THE END POINTS OF THE TRAJECTORY, and NOT on the particular trajectory connecting those two points.
    That is, suppose your "initial point" is [tex]\vec{x}_{0}[/tex] and your final point is [tex]\vec{x}_{T}[/tex] (that is, just two points in space)

    Now, let I and J be two different trajectories between [tex]\vec{x}_{0},\vec{x}_{T}[/tex]
    For a conservative force it is then true that:
    [tex]\oint_{I}\vec{F}\cdot{d\vec{x}}=\oint_{J}\vec{F}\cdot{d\vec{x}}[/tex]
    for ANY choice of connecting trajectories I, J.
    That is, the value of the work done can ONLY depend on [tex]\vec{x}_{0}[/tex] and [tex]\vec{x}_{T}[/tex]
    Okay with that?
     
  15. Dec 15, 2004 #14
    so what are some examples of non conservative forces vs conservative forces?
     
  16. Dec 15, 2004 #15

    arildno

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    Good question!
    Take for example the force of gravity (a CONSERVATIVE force):
    [tex]\vec{F}=-mg\vec{k}[/tex]
    Now,
    [tex]\vec{F}\cdot\vec{v}=-mg\frac{dz}{dt}[/tex]
    (where z(t) is the vertical position as a function of time)
    Hence, the work done by gravity can be written as:
    [tex]W=\int_{0}^{T}-mg\frac{dz}{dt}dt=-mgz(T)+mgz(0)=-mgz_{T}+mgz_{0}[/tex]
    That is, IRRESPECTIVE of the path the object has undertaken, the work done by gravity is simply found as the difference between final vertical position and initial vertical position multiplied with (-mg)
    AGREED?
     
  17. Dec 15, 2004 #16
    yes, so [tex]\oint_{I}\vec{F}\cdot{d\vec{x}}=\oint_{J}\vec{F}\cdot{d\vec{x}}[/tex] is essentially saying that the work done in one position is conserved and is the same as the work done in the final position?
     
  18. Dec 15, 2004 #17

    arildno

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    It means that the value of the WORK doesn't depend upon the particular PATH you may have gone along, but ONLY on the initial and final POSITIONS.
    In particular, it implies that if you end up where you started NO NET WORK has been produced by that force!

    To see this, I just state that we mathematically can prove, that a conservative force
    can ALWAYS be regarded as the gradient of a scalar potential -U, or, in the 1-D case:
    [tex]F=-\frac{dU}{dx}[/tex] (in general, [tex]\vec{F}=-\nabla{U}[/tex]
    Let's take the case of gravity:
    We have now found, by equating with the change in kinetic energy:
    [tex]-mgz_{T}+mgz_{0}=\frac{1}{2}m\vec{v}^{2}\mid_{(\vec{x}=\vec{x}(T)=\vec{x}_{T})}- \frac{1}{2}m\vec{v}^{2}\mid_{(\vec{x}=\vec{x}(0)=\vec{x}_{0})[/tex]
    Rearranging, we have:
    [tex]\frac{1}{2}m\vec{v}^{2}\mid_{\vec{x}_{0}}+mgz_{0}=\frac{1}{2}m\vec{v}^{2}\mid_{\vec{x}_{T}}+mgz_{T}[/tex]
    Since 0, T are ARBITRARY, doesn't this merely state CONSERVATION OF MECHANICAL ENERGY?
    (Since only gravity were assumed to be present, the velocities in the horizontal directions haven't changed, so they may be removed from your equation)
     
    Last edited: Dec 15, 2004
  19. Dec 15, 2004 #18
    what is a nonconservative force then?

    When does it depend on the path?
     
    Last edited: Dec 15, 2004
  20. Dec 15, 2004 #19

    arildno

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    A non-conservative force is a force whose value of WORK may well depend upon the particular path you've gone along, not just the initial point and the final point.

    Air resistance is one example (that is also a DISSIPATIVE force, in that it removes mechanical energy from the system (or if, you like, the object loses momentum to the air molecules which get set in motion))
     
  21. Dec 15, 2004 #20
    so if a ball was dropped in water, the buoyant force is a non-conservative force?
    If a ball was allowed to slide down an inclice, then the force of friction is a non-conservative force?

    is the net work = 0?
     
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