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Work Done

  1. Nov 30, 2004 #1
    Work Done Sorry, Still have some Qs

    Work done refers to a force x distance.
    Or a force (continuous) x distance?
    I think it is the second one when I think of the potential energy, right?
    For PE=mgh,
    How to prove it? And when an object falls, it loses potential energy, why?
    Where is the negative sign? :surprised
     
    Last edited: Nov 30, 2004
  2. jcsd
  3. Nov 30, 2004 #2
    How to determine F=Weight? Isn't that if you apply mg upward, the distance will be 0?
    So we can only think of it is falling with a force mg and the distance travelled but not think of it being elevated for the proof of the equation?
     
  4. Nov 30, 2004 #3
    you know that the work done by a conservative force does not depend on path

    this is how to prove it:
     

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  5. Nov 30, 2004 #4
    Potential energy has no sign since it is not a vector.
    Work= constant force (continuous)x distance.
    If it is added by a continuous force, the object should not stopped no matter there is friction(smaller than applied force). So not continuous force?
     
    Last edited: Nov 30, 2004
  6. Nov 30, 2004 #5
    In [MATH] KE=1/2mv^2 [/MATH]
    Why the proof is using u=0 but not v=0? Isn't after the work is done, v=0?
     
  7. Nov 30, 2004 #6

    James R

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    Work done by a constant force [itex]\vec{F}[/itex], acting over a displacement [itex]\vec{s}[/itex] is:

    [tex]W = \vec{F}.\vec{s}[/tex]

    Take a constant gravitational force acting on a mass: [itex]\vec{F} = m\vec{g}[/itex]. The force acts to make the mass fall through a displacement of [itex]\vec{s} = \vec{h}[/itex], where [itex]\vec{h}[/itex] is directed downwards. The work done by gravity on the mass as it falls is:

    [tex]W = m\vec{g}.\vec{h} = mgh[/tex],

    which is positive since the force and the displacement vectors are in the same direction.

    Gravity is a conservative force, and the change in potential energy associated with such a force is the negative of the work done by the force. i.e.

    [tex]\Delta U = -W[/tex]

    For gravity

    [tex]\Delta U = -mgh[/tex],

    so the potential energy decreases as the mass falls.
     
  8. Nov 30, 2004 #7

    HallsofIvy

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    What are you talking about?

    It would help if, when you ask "Why the proof is using u=0 but not v=0? Isn't after the work is done, v=0?", you tell us what u and v are. I assume that v is the speed of some object. In that case, yes, the kinetic energy of an object of mass m is [itex]\frac{1}{2}mv^2[/itex]. I have no idea what "u" is. IF work is done to bring the object to a halt, then, yes, v= 0 when it is no longer moving. I don't know why whatever problem you are talking about doesn't take v= 0 because I don't know what problem you are talking about.

    "If it is added by a continuous force, the object should not stopped no matter there is friction(smaller than applied force). So not continuous force?"

    I assume you are talking about a problem in which there is a friction force, fe, and an "applied force", fa, opposite to the friction. The net force will be fa-fe. Assuming that fa> fe, the not come to a halt. In fact it will keep accelerating. How about posting a specific problem so we can see what you are talking about?
     
  9. Dec 1, 2004 #8
    I think the force in the equation of work done is continuously applied,like weight, right?
    And the object would not stop finally, right?

    For the PE, a weight of 100N is lifted up by a man up to a height of 10m.
    Why the force applied by the man continuously is 100N?
    Isn't 100 N causing no effect on the weight as it is compensated by the mg of the weight, right?
     
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