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Is Power only defined for nonconservative forces?

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  1. May 7, 2015 #1
    I am searching for power concept in physics, all I see is the sources give frictional or applied forces for power calculation examples. Frictional or applied forces are nonconservative forces.

    Why is that so?

    Is power calculation only valid (or meaningful) for nonconservative systems?
     
  2. jcsd
  3. May 7, 2015 #2

    ShayanJ

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    No, but if you consider a conservative system, which means its total energy is constant, then power will be zero! A quantity which is always zero is not of much interest!
     
  4. May 7, 2015 #3
    So when we are calculating the power, in the formula(P = W / Time), is Work "a work of a system" or what?

    In conservative systems total energy is zero, so we can't talk about for example power of gravity. Right?

    So, here "work" must have a subscript that indicates total energy or something like that and I can't find it in any books. I mean that work is work of what?

    Thank you...
     
  5. May 7, 2015 #4

    ShayanJ

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    The definition of power doesn't have to involve the concept of work. Any change/transfer of energy if divided by the relevant amount of time, gives a power.
    Also I didn't say in a conservative system the total energy is zero, I said its a constant. Which means there is no change/transfer of energy for the system as a whole and a power associated to such a thing will be necessarily zero. I don't say such strong statements as "we can't talk about for example power of gravity", I can only say in the sense you mean, the power associated to any conservative system is zero but some people may mean a different thing than you expect by power. I'm not saying I know such a case, I'm just saying some may say such a thing so I should consider the possibility.

    The best advice I can give you, is that stop arguing on such things so early. If you continue your education and be a good student, you will see enough examples to actually understand it.
     
  6. May 7, 2015 #5

    Dale

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    First, I disagree that applied forces are nonconservative. Usually they are not specified one way or the other. But an external force could be powered by a spring or gravity or some other conservative power source.

    Power calculation is valid and meaningful for all systems. I am not sure what you mean by "conservative systems". Usually I have heard of forces as being conservative rather than systems.
     
  7. May 7, 2015 #6

    ShayanJ

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    True. But consider this case: We pick a box and attach to its inside wall a spring. Then attach a mass to the other side of the spring. Then we put the box in a time-dependent gravitational field. All forces are conservative, true but what if we have an observer inside the box who is studying only the spring-mass system?
    Well, again true. But here we meant a system which has no nonconservative forces acting in it.
     
  8. May 7, 2015 #7

    Dale

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    The power calculation would still be valid even there. Power is power. It doesn't matter if the forces or systems involved are conservative or not.
     
  9. May 7, 2015 #8

    ShayanJ

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    I never said the concept of power is invalid in a given situation. I was saying that a conservative force may appear otherwise to some observers. What I said is that when all forces are conservative and our system is large enough to consider all the effects present, then the power associated to the change of the energy of the system as a whole will be identically zero because the total energy is constant.
     
  10. May 7, 2015 #9

    russ_watters

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    Well I agree with what you said earlier: this isn't a hair that needs to be split. Power/work can be calculated for individual forces in any group of forces and it really doesn't matter if they are conservative or not. I see no reason why this question should matter.

    If I apply a force of 1N to an object and it moves 1m in 1 second, I applied 1W of power. Was it conservative? What opposed me? Gavity? Friction? Acceleration? A spring? It doesn't matter: I don't even have to tell you.
     
  11. May 7, 2015 #10

    ShayanJ

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    Yeah but the OP asked about it and that's why I explained it that way!
     
  12. May 7, 2015 #11

    Dale

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    But that is exactly what the OP was asking. I just wanted to clearly answer the question asked by the OP.

    The power calculation is valid irrespective of the nature of the force.
     
  13. May 7, 2015 #12

    ShayanJ

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    I did answer that question in post #2. But it doesn't matter anyway. One more post and it becomes more clear.
     
  14. May 7, 2015 #13

    Dale

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    I saw that.
     
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