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Work done by a spring.

  1. Jun 5, 2010 #1
    I know that the work of a system is the rate that a force is translated across a certain distance, mathematically written as
    9949eba0cfc92f8fa6cc4c72581c7b05.png
    when the force and/or position vector are not constant, and my understanding is that potential energy is the energy able to be translated to motion due to a conservative force. So my question is why is the potential energy of a spring
    sprp2.gif (I understand the derivation) but when the force and position are constant then the formula for work is just [PLAIN]http://upload.wikimedia.org/math/6/a/a/6aae8efcc88d5ea092cf032c93511b1b.png, [Broken] so by that definition the energy available due to an outside for is just W=kx2, why is the second definition wrong?
     
    Last edited by a moderator: May 4, 2017
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  3. Jun 5, 2010 #2

    rock.freak667

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    The variation of force and extension is such that F∝x.

    If you use W=Fxcosθ; then you are saying that the variation of Force is constant with extension, which is not true. As only half of this energy is stored in the spring.
     
  4. Jun 6, 2010 #3

    Cleonis

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    You are trying to ask a question, but what is the question? I cannot reconstruct what question you are trying to ask.

    There is multiple self-contradiction in what you wrote down:
    - You suddenly mention 'when the position is constant'. But when postion is constant then no work is being done.
    - Your question starts with mentioning a spring. More specifically, you are mentioning the spring constant 'k'. Evidently what you have in mind is a spring that is described by Hooke's Law: The force necessary to displace away from unstressed state is proportional to the displacement. More displacement => more force. The factor 'k' is the proportionality between displacement and force.

    The case of _constant force_ is another case entirely. When the force is constant then there is no factor 'k'.
     
  5. Jun 6, 2010 #4
    That doesn't sound right to say "there is no factor 'k'". If you have a constant applied force F, the spring will start deforming, and that displacement will stop at some position x. The spring will respond to its deformation by exerting a constant restoring force. These two forces kx and -kx acting on a piece of the spring are in equilibrium, therefore a piece of the spring is not accelerating. If you put a scale in line with the spring it will display a magnitude of a force, and the ratio of that reading to length x will be k.
     
  6. Jun 6, 2010 #5

    Cleonis

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    You have an interesting point here.

    Imagine a very weak spring; it takes next to no force at all to compress it. It is actually impossible to exert a large force upon a weak spring.

    Assume that the motion of compressing the spring is slow. The reason to compress slowly is to avoid effects that arise from inertia. Now, in compressing a spring slowly the amount of force you can exert upon the spring depends on the counterforce from the spring

    Let's say you are using a hydraulic jack to exert force upon the spring. Let's say you removed one of the wheels of a car, you have placed a jack under the coil spring of that wheel's suspension. You are using a jack with a force gage. As you jack up, what will the force gage read?

    Initially, when the jack just touches the spring, the force gage will read a very low number. Next you jack up a fraction of a milimeter: the amount of force that the force gage reads will still be small, since a small force suffices to compress the spring just a fraction of a milimeter. As the spring is compressed more and more, more and more force is needed.

    At all times the amount of force that the jack is exerting upon the spring is the same as the counterforce from the spring. (More precisely: in order to compress the force upon the spring needs to be a fraction larger than the spring's counterforce. Just a fraction.)

    Therefore to evaluate the work done upon the spring the force at each point along the way is the force as described by Hooke's law. (If you use Hooke's law for the approximation.)


    Possibly what you have in mind is a scenerio where a heavy object that is in motion makes contact with a spring, for example a moving train making contact with a spring buffer at the end of a railway track.
    Does the train start exerting full force right at the very instant that it makes contact with the buffer spring? No, it doesn't. In order to exert a force the train must be decelerating (it takes a force to decelelate). In the instant after contact is made the compression of the spring is still minute, so the force exerted by the spring upon the train is still minute, hence just a minute deceleration, hence just a minute force from the train upon the spring. Of course, as the compression of the spring proceeds the force becomes larger and larger, and the train does come to a stop.
     
    Last edited: Jun 6, 2010
  7. Jun 6, 2010 #6
    Why is only half of the energy due to the force stored in the spring though? That's what I don't understand is why the two formulas conflict. It seems that I have a flawed understanding of the delta x term in each and what it represents.
     
  8. Jun 7, 2010 #7
    E=∫F dx that means the surface between F and x axis.

    For spring F=kx which means that the force is in linear relationship with the displacement x. If you draw the force-displacement diagram you will see that only the surface of the lower triangle represents the energy stored in the spring.

    Look what happens if the force is constant over the displacement x, F=const=k ( which is not the characteristic of a spring ).Then the whole square represents the energy.
     
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