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Work energy theorem clarification.

  1. Nov 12, 2012 #1
    I just wanted to start a thread about the basic concepts of work and energy to clarify some things I have been wondering about.

    Firstly, work and kinetic energy are concepts independent of force fields; right? Am I right to think that the relationship of force and kinetic energy with respect to distance is analogous to the relationship of force and momentum with respect to time? Because if I apply a constant force for a certain time period I change my momentum from an initial momentum to a final one ([itex] F \Delta t = \Delta P [/itex]). Similarly; if I apply a constant force for a certain distance I change my kinetic energy ([itex] F \Delta x = \Delta E_k [/itex]). Work is like if I gave a name to the quantity [itex] F \Delta t = \Delta P [/itex], which as far I know has no name and is only know as change in momentum or F times distance. Is all of that too far off?

    Anyway.. potential energy only exists in the presence of a force field; correct? So in the absence of a force field, work done on a system will always equal the kinetic energy of that system. Like if there was no force fields in the universe there wouldn't even be a concept called potential energy. So is it that in the presence of a force field [itex] W=E_p + E_k [/itex]?

    I guess the only thing that is confusing me at this point is the fact that the expression for potential energy depends on what force field you're dealing with. Is that correct to assume or is there a general expression for potential energy? Is it the integral of force with respect to position?

    EDIT: Also, what would be the analogue to potential energy in my momentum analogy? Maybe some special case since force fields are not necessarily dependent on time?
    Last edited: Nov 12, 2012
  2. jcsd
  3. Nov 12, 2012 #2


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    Hi V0ODO0CH1LD! :smile:
    i see what you're getting at … ∫ F dt = momentum, and ∫ F.dx = work
    yes, potential energy is just another name for minus the work done by a conservative force

    (so there has to be a force everywhere, ie a field)
    yes (assuming we say that heat and radiation are also forms of kinetic energy) :smile:
    yes, it's -∫ F.dx, the work done :wink:
    not following you :confused:
  4. Nov 13, 2012 #3
    What is confusing me is:

    First the formula for kinetic energy ([itex] \frac{1}{2} mv^2 [/itex]). I realize that in the processes of deriving the formula they always assume that the force doing work on the system is constant. Is that because the kinetic energy is only dependent on the final speed that the object reaches? Kind of like it doesn't care how it got to that state of kinetic energy, so it might as well assume it was via a constant force?

    Could I specify that kinetic energy is the amount of work, done by a constant force, needed to accelerate a body of a given mass from rest to its stated velocity? And I don't mean that work can only be done by a constant force. But if a system has a certain kinetic energy and it got to that state though some variant force acting on it for a certain distance, the theoretical force that acted on it, by definition, is whatever constant force had to act on it for that specified distance to get it to that state of kinetic energy? Is that correct? Is it because by the end of the day, one force is the actual variant force times distance and the other is the average force times distance and the quantity "work" is the same for both?

    The only interaction with potential energy, in the other hand, is with movement in the direction of the force field. So if I apply force that keeps a point mass in a certain potential but moves it perpendicular to the force field, even though I have to apply a certain force to keep my point mass in that potential, the only component of the force who's changing the energy of the particle is the one perpendicular to the force field which is increasing the kinetic energy; right?

    But when I lift a point mass up in the air with some constant velocity I am not changing the kinetic energy of that point mass, because acceleration is zero therefore force is zero so work has to be zero. But I am giving that point mass some potential; what is happening?

    EDIT: Also, I just though of something else: When I say (potential energy) + (kinetic energy) = (constant); is the kinetic energy I am talking about also due to the force field? Like (kinetic energy due to force field) + (potential energy due to force field) = constant? Or am I talking about any exterior forces that may be acting on the system inside the force field as well as the actual forces of the force field?

    Is it as if the conservation of energy theorem exists with respect to a force field only? But, in the case with gravity, if my hand is laying on the floor with a rock in it, there is no potential/kinetic energy in the hand/rock system. But I can lift it, and give it some potential energy. Where did that come from? Did my hand have potential energy on the floor because I could lift the rock if I wanted to?
    Last edited: Nov 13, 2012
  5. Nov 13, 2012 #4


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    Hi V0ODO0CH1LD! :smile:
    No, they do not assume F is constant …

    1/2 mv2 = ∫ mv dv = ∫ mv dv/dx dx = ∫ ma dx = ∫ F dx

    kinetic energy is the integral of F wrt distance, even if F is not constant :wink:

    (F includes all forces, even those with an associated potential energy, such as gravity or spring force)
    sentence too long: brain hurts :redface:
    what is happening is that there are two forces, the force from your hand, and the force of gravity

    they are (obviously) equal and opposite, so F here is 0 :wink:
    kinetic energy is 1/2 mv2 … that's it! …

    it comes from all the forces
    it came from chemical potential energy in your muscles :smile:
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