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I know that by definition, forces for which the work done is independent of the path taken are known as conservative force while the forces for which the work done is path dependent are known as non-conservative forces. My question is why is the work done by some forces path independent while for others it is are path dependent or simply put why do some forces do not dissipate energy while others dissipate energy. What makes them do so?

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  • $\begingroup$ Are you asking about friction?, Sorry, but I am a bit puzzled by your question. $\endgroup$ – user108787 Oct 29 '16 at 19:20
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    $\begingroup$ Your question is a bit weird. Having both conservative and non-conservative forces is the most general scenario possible; why would you expect only one of them to be the case? $\endgroup$ – Javier Oct 29 '16 at 19:22
  • $\begingroup$ I am asking about the reason behing their path independence and dependence. $\endgroup$ – MrAP Oct 29 '16 at 19:26
  • $\begingroup$ As much as I could get , I think you are asking for the proof of the fundamental theorem of line integral. Moreover what you say is also proved by Green's theorem. I wouldn't be able to provide the exact proof but I know for certain that you can get it in both Thomas and Stewart's Calculus. $\endgroup$ – Shashaank Oct 29 '16 at 20:09
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You can see for yourself that if a force depends only on position (not on time and speed), then it is conservative, while if it depends on speed it may not be.

Simple examples should do: take the weight force, $\vec f = m \vec g$: when something falls down from an height $h$, you have $\ W = \vec f \cdot \vec s = mgh.$ Now suppose you pick that thing up and put again at his former place. Now your work is $\ W = \vec f \cdot \vec s = - mgh$. The minus comes from the fact that you are pushing up, while the weight pushes down. Total work is zero, the same of the null path.

Now imagine pushing your cell phone on your desk, from point $A$ to point $B$, say $ |AB| = l$ and examine the work done by friction: now $\vec f = - \mu_d mg \hat v, \ $ where $\hat v$ is the unit vector parallel to $\vec v$: work is $\ W = \vec f \cdot \vec s = - \mu_d mgl$. Now you push your phone back to point $A$: again, friction is opposite to motion, so, again $\vec f = - \mu_d mg \hat v, \ $ so work is $\ W = \vec f \cdot \vec s = - \mu_d mgl$, this time total work along this closed path is not zero, so you see that work now it depends on the path.

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From Conservative & Non Conservative Forces

Sorry for a copy and paste response, if I had more time I would prefer to write my own answer, but see what you think.

In physics, it’s important to know the difference between conservative and nonconservative forces. The work a conservative force does on an object is path-independent; the actual path taken by the object makes no difference. Fifty meters up in the air has the same gravitational potential energy whether you get there by taking the steps or by hopping on a Ferris wheel. That’s different from the force of friction, which dissipates kinetic energy as heat. When friction is involved, the path you take matters — a longer path will dissipate more kinetic energy than a short one. For that reason, friction is a nonconservative force.

Conservative forces are easier to work with in physics because they don’t “leak” energy as you move around a path — if you end up in the same place, you have the same amount of energy. If you have to deal with nonconservative forces such as friction, including air friction, the situation is different. If you’re dragging something over a field carpeted with sandpaper, for example, the force of friction does different amounts of work on you depending on your path. A path that’s twice as long will involve twice as much work to overcome friction.

What’s really not being conserved around a track with friction is the total potential and kinetic energy, which taken together is mechanical energy. When friction is involved, the loss in mechanical energy goes into heat energy. You can say that the total amount of energy doesn’t change if you include that heat energy. However, the heat energy dissipates into the environment quickly, so it isn’t recoverable or convertible. For that and other reasons, physicists often work in terms of mechanical energy.

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  • $\begingroup$ That does not answer my question: My question was why is the work done by some forces path independent while for others is path dependent? $\endgroup$ – MrAP Oct 29 '16 at 19:44
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Work done by a potential field (or against it) is always path-independent because object's energy is different in different points of space. And only way to change energy is displacing it in the field. E.g. Electric field, gravitational field.

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