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Thought experiment about energy

  1. Aug 18, 2012 #1
    I have a thought experiment for physicists and thermodynamicists and the like. Those who are always looking for 2+2=4 but for some reason avoid discussing certain issues that are very clear.

    Simply put the picture where we have an object of mass m in the ground. The physics math show that we have zero gravitational potential energy there. We need to do work in order to gain potential energy. This work is meant to have a negative sign. Then when the object falls, it does a positive work because kinetic energy, and then we have the plus work minus negative work equal zero..... Ok, enough talk. Now look into reality.... If the object breaks at the time it reach ground, where do we add that energy, the internal energy (energy of deformation?), in the 'perfect' thermodynamic cycle?? And what if instead of the object being a simple rock, it is some mechanism composed of magnet, solenoid, and a spring..... Obviously when such a mechanism hit ground after falling, it presents an electric current for the very basic reason of mechanical vibration and electromagnetic induction.
    Last edited: Aug 19, 2012
  2. jcsd
  3. Aug 19, 2012 #2


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    You can choose any value you like.

    A part of the kinetic energy goes into breaking the object (and another part into heating the environment). You do not have a cycle, you invested work to lift the object and converted it into breaking chemical bonds and heating the rock.
  4. Aug 24, 2012 #3

    You are just doing what physicist like to do when they screwed up with something..... talking and talking qualitatively but avoid equations they used to teach. We know that when an object of mass m falls, it presents a change in kinetic energy, and it is accounted like a work done. We know that work is being done because the object accelerate and it is because a force is applied on it. I saw a nasty trick on an educational site, where they didn't account the change in kinetic energy when object falls like work done, a true representation of physicists effort to confuse the general public.

    Now wikipedians, who can change information in a minute, are trying to apply a 'patch' to thermodynamics, redefining and redefining, like infinitely, the concept of isolated system and the like.

    My conclusion: It appears to be a fake procedure in physics to relate potentials, energy, etc, with position in space. Some dirty trick to avoid discussing the idea of 'perpetual motion' machinery.
    Last edited: Aug 25, 2012
  5. Aug 25, 2012 #4
    Changes in potential energy are all that ever matter physically; the actual value of potential energy is meaningless. Often, though, it’s convenient to establish a reference point at which the potential energy is defined to be zero. When we say “the potential energy U,” we really mean the potential-energy difference between that reference point and whatever other point we’re considering. For a rock climber, for example, might find it convenient to take the zero of potential energy at the base of the cliff. But the choice is purely for convenience; only potential-energy differences really matter.

    Essential University Physics
  6. Aug 25, 2012 #5
    Ok, I understand very well your point, but I was refering the situation of a 'thermodynamic cycle', where we have conservation of energy like iput works, and output works, and the sum of them is equal to zero. It doesn't matter where we define the start point in the trajectory of the object, I understand that, but there supossedly exists negative works (those against the gravity field) and positive works. In the situation I mentioned above, there exist a work not accounted by classical mechanics, at least.
  7. Aug 25, 2012 #6


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    Incorrect. The work done by something, such as a person, to lift an object is positive. However, there is also the force of gravity pulling down. This is represented by "negative work" since it works against the displacement of the object lifted. Once the object starts to fall gravity is now doing positive work on the object, since the direction of force is the same
    as the displacement.

    Similarly, the ground will be doing negative work when it decelerates the object upon impact. (The force acts against the displacement and slows the object down) As would anyone who catches the object before it hits the ground.


    No energy is added. Energy was used to break the bonds in the material and break/deform it. This energy comes from the kinetic energy of the object at the time of impact. I'm not sure why you are asking about a thermodynamic cycle here, there is no cycle as the system never returns to its initial state. http://en.wikipedia.org/wiki/Thermodynamic_cycle

    While I'm sure there may be some stray electrons or ions around, there will be practically no electrical currents generated by anything unless we specifically set up such a situation beforehand.

    As I said above, there is no cycle here. However everything still must obey conservation laws. As for work existing that isn't accounted for by classical mechanics, I can see no such situation. Everything is accounted for.
  8. Aug 25, 2012 #7


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    No, there does not. You are confusing an object doing work against a force and object's potential energy getting converted to kinetic energy in a force field. As a result, you are double-counting. If an object did work on the way down, it arrived at the bottom with a lesser increase in kinetic energy by amount of work done. In case you describe, the kinetic energy gain is precisely zero. So no additional energy is released on impact.
  9. Aug 25, 2012 #8


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    In science, you should always know where a quantitative analysis is required and where a qualitative description is sufficient.
    Anyway, there is an obvious way to transform my post into a set of equations, as soon as you clearly state which process you want to look at.

    Right. So what?

    The fact that you do not understand it does not mean that it is wrong.
  10. Aug 25, 2012 #9
    That in bold is incorrect. If you are going to accelerate an object, for example, in horizontal direction, even in a frictionless surface, you have to apply a force along all the trajectory. If you stop acting on the object with force, then the kinetic energy will be the same and the object will travel with the same velocity after you release it.

    You are trying to avoid accounting for work done even when the very basic formula of work done relates to an applied force and a displacement.

    I respond the late paragraph for the other people here. Avoid confusing the people with 'special cases' and the like. Your formulas should apply to nature without applying 'patches' to they. The very basic formula of work done tells what I'm refering to. Maybe some physicist apply the conservations laws differently from others.
    Last edited: Aug 25, 2012
  11. Aug 25, 2012 #10


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    O.K., let's try and analyse this step by step.(I'm trying to use only scalars here to make it simpler)

    0. a body is at rest at ground level(r0=0), which we choose to represent gravitational potential enerygy U0=0. Its kinetic energy is Ek0=0.

    1. we add energy to the system by performing work on the body and lifting it up to some height r1. The work performed is W=FΔr, where F is equal but opposite of the force of gravity(F=mg).
    Work equals change in total energy:

    Since we've chosen the force doing work like we did, the body gains no kinetic energy, so the resultant energy at height r is:

    Note that in this step we are adding energy to the system from an external source, so the total energy at the end of step 1 is more by W than it was at step 0. Thus, it is not a closed system, and simple equivalence of kinetic and potential energies doesn't hold.

    1a. Alternatively, instead of "lifting" the body by applying constant force, we may "kick" the body upwards, thus giving it some non-zero kinetic energy(Ek1) at the begining of this step.
    This kinetic energy gets wholy converted into potential energy gain:

    This "kick" is, again, adding energy to the system from outside.

    2. the body falls back to r0=0, gaining kinetic energy and losing potential energy

    Ek2=W (for 1)
    Ek2=Ek1 (for 1a),
    i.e. we can see that the final kinetic energy of the falling body, that may then go into breaking the body, or deforming the surface it hits, or winding up a spring in some mechanism, comes from the energy that was supplied to the system from outside in the form of the work or "kick".
    If there is no such external energy supply, the body remains at rest as in step 0.

    If I follow your train of thought correctly, you seem to be assuming that all the steps represent one closed system, which in fact is true only for 0 and 2.
  12. Aug 25, 2012 #11


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    No, you simply misunderstand what K^2 is saying.
    In fact, it seems like you just have a problem with "scientists" in general, as can be seen by you attempting to blame physicists for confusing the public and such. Such an attitude is not welcome here on PF. It is understandable to be frustrated when confused by something, but to lash out and try to place the blame on others is unacceptable. If you wish to learn then please stay and ask further questions, but if you simply wish to push your "agenda" then please don't come back as it will do no one any good.
  13. Aug 26, 2012 #12


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    That is a mistake. Where do you think there is any work unaccounted for?

    An object is in a gravitational field with PE=0 and KE=0. A non-gravitational force does work raising the object increasing the PE by W and the KE remains 0. The object falls reducing PE to 0 and increasing KE to W. The object hits the ground and deforms plastically losing all the KE. Because the deformation is plastic the area inside the stress strain curve is non-zero, specifically it is equal to W. That work rapidly goes to thermal energy. So in the end, work W was done, the final kinetic and potential energy are the same as the initial, and thermal energy is increased by W. So where is the work not accounted for?
  14. Aug 26, 2012 #13
    No I didn't. And I put below what K^2 wrote, again.

    What????. He screwed up with his ability to relate an increase in kinetic energy to work done. He assumed that an object only do work when loosing velocity, like when an object is subjected to friction in a surface.

    Thanks for your thoughtful, detailed and lovely explanation. I really aprecciated it, but people like K^2 will continue to apply wrongfully their analysis if they don't update their way of thinking. So they will have problems to accept energy breakthroughs if they exists.
  15. Aug 27, 2012 #14


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    From your initial post.
    This right here. A free falling object does no work, because it is not applying force on anything, which is required to do work.

    You insist that the object did work on the way down equal to work initially done on the object. That means that over your Δh the object was doing work against force with average magnitude of mg. That means, rather than letting the object drop, you lowered it gently applying a support force. That means, at the bottom, kinetic energy is zero. Object did not hit ground with any velocity, precisely because the net energy gain of the object through this "cycle" is zero.

    When you say the object did work on the way down and had kinetic energy equal to it's earlier potential energy you are, in fact, double counting. So naturally, you end up with a discrepancy.

    And for future reference, if the topic is basic mechanics, and somebody with a forum title of contributor, helper, or adviser tells you something you disagree with, your first instinct should be to look for errors in your own work. We all make mistakes, but having many years of experience with various fields of physics, we make these mistakes far less often than people new to the subject. If you start acting like everyone around you is wrong, you probably won't last here. If you want to learn something, ask questions. That's how people generally learn. Not by acting like they know it all when they can't comprehend the energy conservation.
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