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Suppose there exists a Force $F$, acting on a body, which is initially having a velocity $v$, in a direction opposite to that of it's motion. Let the point of application of force produce a displacement of $x$ in the body. Let there be a loss of kinetic energy of $\Delta KE$(body is not necessarily brought to rest, just that its velocity reduces) and heat produced by $H$. Further let $\Delta KE + H = 0$.

By work energy theorem $Work_{F} = \Delta KE + H = 0$.

But on doing $Work_{F} = \int_{0}^{x} F \cdot dx = -\int_{0}^{x} Fdx \neq 0$..

Why are these two coming approaches giving different result?

Where is there a flaw in my argument? I feel that the WE theorem gives the correct answer. Is there some constraints/assumptions to be made before applying $W=\int F \cdot dx$?

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  • $\begingroup$ Where did you find this form of work energy theorem? $\endgroup$ – nasu Oct 15 '19 at 18:17
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Work done by a force is always given by the line integral in your question.

The work energy theorem, as you have written it, is basically a statement of a conservation of energy. Work is a mechanism of moving energy from one place or form to another. In a system where the only places for the energy to go is into kinetic energy or heat, then the total work done must equal the amount of energy that shows up in those two compartments. Not every system satisfies this criterion: for example, some of the work could go into compressing a spring (sticking some of the energy into the "elastic potential energy" compartment).

Basically, you have overdetermined the action that you describe, which is why it appears to create a contradiction. The requirements you put on the various quantities cannot all be true. Assuming $F$ is the only force acting on the body, and that the only available energy compartments are heat and kinetic energy you cannot independently specify both the work done by $F$ and the total energy change of those compartments. I mean, I suppose you can specify them both, but that doesn't make it a physically realizable situation.

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  • $\begingroup$ Actually, I was working on a problem and this situation had occurred. The problem was basically regarding a resistor connected to a movable conducting rod in the horizontal plane. The system is present in a uniform magnetic field perpendicular to the circuit plane. The rod is given an initial velocity horizontally. The question was to determine the work done by magnetic field until the velocity reduced to a certain value. Using the line integral the answer was coming out to be non zero. But using the work energy equation as mentioned in the question, I got the the answer as 0(which is correct $\endgroup$ – user600016 Oct 15 '19 at 15:40
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    $\begingroup$ Work done by magnetic field is always zero, or so I was told by Griffiths. $\endgroup$ – Ben51 Oct 15 '19 at 15:42
  • $\begingroup$ is that true for the field acting a current carrying wire as well?(I know it is true for a point charge though) $\endgroup$ – user600016 Oct 15 '19 at 15:45
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    $\begingroup$ I'd say if you're being pedantic, then yes: the magnetic field deflects the charge carriers, which then transfer momentum to the bulk wire via electrical interaction. So in the end the work is actually done by electric field. $\endgroup$ – Ben51 Oct 15 '19 at 15:47
  • $\begingroup$ Oh I see thank you. $\endgroup$ – user600016 Oct 15 '19 at 15:50
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I believe the issue is that you are including heat in your work calculation when you shouldn't.

The formula $W_F = \Delta KE + H$ is not correct.

The change in internal energy ($\Delta U$) according to the first law of thermodynamics is given as $$ \Delta U = Q - W$$ where $Q$ is the same as your $H$.

Work should just be $W = \Delta KE$, the heating does not count as work, but is factored into the conservation of energy through the first law of thermodynamics, and it's effect on the thermal energy of the system.

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  • $\begingroup$ Isn't the general statement of work energy theorem $W_{ext}+W_{int \quad conservative}+W_{int \quad non-conservative} = \Delta KE$. and we take $ W_{int \quad conservative} = \Delta PE, W_{int \quad non-conservative} = - Heat$? $\endgroup$ – user600016 Oct 15 '19 at 14:59
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    $\begingroup$ @user600016 That doesn't seem like any version of the work-energy theorem I'm familiar with. As far as I'm aware, Heat doesn't fit into the definition of work. $\endgroup$ – JMac Oct 15 '19 at 16:21
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The work done is the change in KE of the body. You have caused the contradiction by assuming that the change in KE is exactly counterbalanced by the generation of heat- that is a spurious term.

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