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Is the work-energy theorem valid for only particles or rigid bodies as well?

Most places where I have read this seem to claim the latter. But an example I thought up has been troubling me.

Consider a block resting on a rough surface. We apply a force on the block. The force is balanced by friction and the block does not move. However, the force is such that the point of application of the force moves on the object. Thus work is done by this force on the object. However, the kinetic energy does not change seemingle violating the theorem.

Is there some flaw in my reasoning or is the work kinetic energy theorem valid really only for particles?

EDIT: To be clear, I am describing the simplest form of the work kinetic energy theorem, the one for a single nonrigid body. I don't believe things like internal energy and potential energy need to be invoked in such a case. Or do they?

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    $\begingroup$ For extended bodies you need to decided how you are going to treat the energy of rotation about the center of mass and that of deformation of the body itself. Life gets more complicated. $\endgroup$ – dmckee --- ex-moderator kitten Jan 24 '15 at 16:32
  • $\begingroup$ @dmckee: yes, minor complications do not matter. What I mean is if the theorem is still valid in spirit? As in, if it is still functional with minor adjustments? Moreover, since we are talking of rigid bodies, I don't think deformation of the body would play a role. $\endgroup$ – Gerard Jan 24 '15 at 17:00
  • $\begingroup$ Well, the short-short answer is yes, but explaining it entails sorting out all those complications. The example you use (as I'm interpreting it) involves dealing with deformation of the extended object, which means involving elastic energies at the very least. $\endgroup$ – dmckee --- ex-moderator kitten Jan 24 '15 at 17:14
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Is there some flaw in my reasoning or is the work kinetic energy theorem valid really only for particles?

It does not matter whether the system is modeled as number of particles or rigid body. What matters is whether you want the work-energy theorem to refer to macroscopic kinetic energy only or allow other kinds of energy.

In your example, when you rub some solid object, you do work on it and this work does not manifest in increase of macroscopic kinetic energy. If you only allow this kind of energy, then the work-energy theorem is not valid.

However, thanks to Joule, Mayer, Helmholtz and others we know doing work on a body by friction does not necessarily mean energy is lost, but may be thought of as transformation of macroscopically visible energy into other kind of energy - in thermodynamics it is called internal energy. Most easily recognizable indication of this is increase in temperature of the body.

If we allow this internal energy into the work-energy theorem, we believe it gets valid again (based on Joule's and others' experiments). It is then named "the first law of thermodynamics": increase in total energy of a body is sum of work and equivalent heat transferred to it.

In mechanics of micro-scale motions of such body the work-energy theorem can be made valid too. The total mechanical energy is not only that due to macroscopically visible motion, but includes contributions due to chaotic motions and potential energies of the microscopic particles forming the body.

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However, the force is such that the point of application of the force moves on the object. Thus work is done by this force on the object.

I don't agree here. Just by moving a force, gradually or not, to different spots on an object doesn't change the work done if no movement (rotational or translational) is happening. It is like pushing, then letting go, then pushing at a new spot, then letting go, then pushing at a new spot, then letting go etc. None of these many forces applied on different points do work on the object.

But if you drag a particle through the object, work is done on that particle.

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  • $\begingroup$ Yes, but you are not just pushing then letting go. The force is being continuously applied as the point of application moves. A real life example would be if you were rubbing the object in one direction with your hand. $\endgroup$ – Gerard Jan 24 '15 at 16:58
  • $\begingroup$ @Gerard, is the object being rubbed over (moving over) the surface, or is your hand being rubbed over (moving over) the object? The latter case is just as described. The kinetic friction force between the object's surface and your hand might do (negative) work in your hand though, and this will appear as heat in your hand (as when rapidly scrubbing your hand over a blanket, the hand gets hot). $\endgroup$ – Steeven Jan 24 '15 at 22:29
  • $\begingroup$ Yes, but if we talk of the work energy theorem for a single particle, namely for the block, work is being performed but the block's KE does not change. $\endgroup$ – Gerard Jan 25 '15 at 2:26
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    $\begingroup$ @Gerard work is being done, but just not on the block. Instead it's being done on the atoms that make up your hand and the block, and it is indeed increasing their kinetic energy, which you feel as heat. $\endgroup$ – Nathaniel Jan 25 '15 at 2:36
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Work energy theorem suggests:
work done on an object = change in the kinetic energy of the object.
So, to determine work done on an object (particle or rigid body) we just need to know the change in its kinetic energy. Thus, this theorem is perfectly valid in either cases.
P.S. Large bodies are treated as point objects in-order to carry out calculations conveniently. And the good thing is, it works.

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First of all, if a body is rigid it will never deform under any force. So the title is not describing your problem properly. Now, about the problem I must say that work-energy theorem is only applicable to particles. For general purposes there is an extended version of work-energy theorem which states:

$W_{external}=\Delta K + \Delta V_{internal} + \Delta K_{internal}$

$W_{external}$ is the work done by external forces, $\Delta K$ is the overall kinetic energy, $\Delta V_{internal}$ is the internal potential energy and $\Delta K_{internal}$ is the internal kinetic energy. Point masses don't have the extra terms in their work-energy theorem because they don't have any internal structure.

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  • $\begingroup$ Why must a body be nonrigid for the point of application of the force to move on the object? If I rub a block with my hand, clearly the point of application of the force is moving whereas the object is rigid . $\endgroup$ – Gerard Jan 25 '15 at 2:21
  • $\begingroup$ Because that's the very definition of a rigid body. Rigid body is an ideal body which has infinite stiffness. Read a mechanics of materials book to get a clear picture of what stiffness is. However, if what you mean by rigid is "relatively rigid" that's another story. $\endgroup$ – QuantallicA Jan 25 '15 at 6:39

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