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Why does work done by a conservative force = 0 in a closed path?

  1. May 19, 2010 #1
    Why does work done by a conservative force = 0 in a closed path?
    I know this sounds foolish :uhh: but how can some forces have such a property?
    Can anybody give a satisfactory physical explanation?:confused:
     
  2. jcsd
  3. May 19, 2010 #2
    Think of a frictionless circular roller-coaster loop and a mass right in the top of it. You give this mass a very very kick to the left, and it then performs the whole loop, arriving to the top with just exactly that little velocity you transmitted to it during the kick.

    This means that with conservative forces, an object can make a loop (or any strange looking path). When the object comes back to its initial position, it will have exactly the same velocity it had when it began its path. And you know that the work done in a object is the change of its kinetic energy: if velocity in beginning and in the end are the same, the kinetic energy hasn't changed when the object performed its loop, so the work gravity has done in it is zero.

    This won't happen (in the general case) with non conservative forces. If there were any friction in that roller-coaster, than it can't come back to the initial position with the same velocity.

    The physical interpretation of it is that "conservative forces" are associated with potential energy. This means the object may gain kinetic energy spending its potential energy, and also it may store kinetic energy as potential energy (it's like converting some amount of money between two currencies: you may transform some dollars into euros and some euros into dollars, but there's no change on the amount you have overall). The amount of dollars you may have at a particular position is fixed: during your path you may exchange it for euros, but when you come back to your original position, you MUST have the very same qunatity of dollars.

    Is it clear?
     
    Last edited: May 19, 2010
  4. May 19, 2010 #3
    What did you mean by a "non-conservative field" ?
    Can you give a mathematical correlation between a general conservative force and potential energy? (for better understanding)
    Are "fields" related to conservative forces only?
     
  5. May 19, 2010 #4
    I'm sorry - I've just realised I had commited a mistake in that sentence. What I meant was "conservative field" but it is clearer to explain things in terms of forces. Delay the study of force fields when you get deeper in this subject.

    Please read my post again. I've only changed that part, you won't have to read much.

    Hmmm, it is somewhat mathematically complicated to explain the correlation between the forces and potential energy (I don't know how your math skills are). Force is the negative gradient of the potential energy. If you are considering 1D systems, then it's minus the derivative of potential energy with respect to x (where x is the position of the object) - but I think this sentence won't make things clear for you, since it appears you are only beggining to study Physics.

    A simpler way, although not strictly mathematical as you wanted, is to say that the force always points to the direction of decreasing potential energy (i.e. the potential increases in the opposite direction to the force). Also, if the potential is rapidly changing over a direction, force will have a larger intensity.

    It will perhaps be easier to give you a reason for that: when you just put a object in some place (with no initial speed) and there's a conservative force (gravity, electrostatic, spring force, for example) acting upon it, than the object will accelerate in the direction of the force. If the object is accelerating, than it's gaining kinetic energy. This means that the object is loosing potential energy in order to change it for kinetic energy - so the potential increases in the direction opposite to the force.

    Also, if potential changes a lot along a particular direction, than the object will win (or loose) quickly a lot of kinetic energy when it travels along that path. If there's a large variation of kinetic energy, there was a large acceleration - so forces have bigger intensities if potential changes significantly over a small distance.

    Any doubts?
     
  6. May 19, 2010 #5
    conservative forces have many forms, you can write one of them as
    (qualitatively for understanding the orm is excellent, but i dont want to get into a fist fight with nabla operator it makes things worse to understand so dont assume it as general.),

    [tex]\vec{F}=f(|r|)\hat{r}[/tex]

    when you take the integral over the closed path:

    [tex]W_{closed path}=\int_{l}\vec{F}.d\vec{l} =\int_{l}f(|r|)\hat{r}.d\vec{l}[/tex]
    the aove integral canbe reduced through various techniques,
    l being the linear region over which your force moves on or rotates sweeps et cetera..
    if you consider the point of action of the force, in one complete rotationdoes the point of action move at all..????
    the answer is a smooth,cheesy no which expains things that there is no net displacement of the action point so how is there any work done
    although this is not always true for all forces but consevative forces are considered special because of the above form and this resulting property.
     
  7. May 19, 2010 #6
    here it is:
    [tex]\vec{F}_{consevative}= \vec{\nabla} (U) [/tex]
    or for a one dimensional case,
    [tex]\vec{F}_{consevative}=-\frac{\partial}{\partial x} U [/tex]

    so are you preparing for IIT-JEE or are you through with it..(i gave JEE2010 waiting for results)?
     
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