2
$\begingroup$

I have learned that work done by conservative forces is independent of the path followed between initial and final position. But that's not the case for non-conservative forces, they depend on the path followed to reach the final point (eg. Friction).

My question is:

  1. Why is that so? Why do some forces' work depend on path and others not?

When I hold a thing in my hand and make it follow a short and long random path in different cases and come to some position $x$, I feel I have done a different amount of work in both cases. But gravitation being a conservative force says I have done an equal amount of work in both cases.

  1. Where am I mistaken?

  2. Also, why do 2 forces exist? Forces are forces, they must have the same nature.

Lastly,

  1. Can I state all unidirectional forces are conservative?

  2. Are there other classifications of forces?

$\endgroup$
  • 1
    $\begingroup$ I think I see at least one of your sources of confusion. If you lift an object to shoulder height and hold it there, how much work are you doing (holding it stationary)? $\endgroup$ – lemon Jul 6 '16 at 7:26
  • 2
    $\begingroup$ Have a look at this physics.stackexchange.com/q/1984 $\endgroup$ – Farcher Jul 6 '16 at 8:05
  • 2
    $\begingroup$ @lemon I feel holding it stationary also I am doing work against gravity, cause I am applying force against it. but according to definition of work, I am doing zero work cause zero displacement. I don't understand why displacement has been given the power to make work zero when force is non zero? Is it that work is just defined that way? Then what's the reason for defining it that way? $\endgroup$ – user282856 Jul 6 '16 at 15:29
3
$\begingroup$

Why do some forces' work depends on path and others not.

  • Nonconservative forces cause energy loss during displacement. For example, friction when an objects moves over a surface converts the stored energy to heat that disappears and is wasted. Therefore the final state depends on how long the path was, because that determines how much energy is lost along the way.

  • Conservative forces cause no loss in energy. Therefore, energy associated with such forces can only be converted into other stored forms in the object (kinetic energy) or system (potential energy). In fact the work done by a conservative force is what we describe as potential energy. The word "potential" gives the feeling that it is stored; it is merely a name for the work that the conservative force will do when released. And when released, that potential energy will be work done on the object and it turns into kinetic energy, which is still stored in the body. If you are told what the start and end speeds are, you therefore know that the difference in kinetic energy must be stored. Regardless of the path.

We can consider the conservation in terms of energy like here or entropy and maybe others as well. I personally find the energy approach the most intuitive.

When I hold a thing in my hand and make it follow a short and long random path in different cases and come to the some position 'x' , I feel I have done different amount of work in both cases. But gravitation being a conservative force says I have done equal amount of work in both cases. Where am I mistaken?

Gravity might be a conservative force, but the force you exert on the object is not.

Also why do 2 forces exist? Forces are forces they must be same nature.

Which two are you thinking of?

In any case, yes, forces are "the same thing" so to speak. It doesn't matter what "kind" of force or what created the force - forces are forces and they can be added, for example in Newton's laws, where we don't care about the "type" of force.

Lastly, -can I state all unidirectional forces are conservative?

What do you mean by unidirectional force?

If gravity pulls downwards, so a box slides down an incline, there can still be a friction in only one direction on the incline. The directionality is not a measure of if a force is conservative or not.

Instead think about what kind of energy that force causes. Is it potential or kinetic, then the force is conservative. Is it heat or alike, then not.

-Are there other classifications of forces?

There are many "types" of forces: Electric, magnetic, chemical, gravitational, elastic etc. Those are just names that tell us the origin of them. As stated above, the "type" or origin is of no importance; all forces can cause acceleration in the same manner.

$\endgroup$
  • $\begingroup$ you are wrong in saying that directionality is not a measure of if a force is conservative or not because if conservative forces are not unidirectional, work done by them would be different for different paths taken. $\endgroup$ – MrAP Oct 14 '16 at 15:11
  • $\begingroup$ In your incline example, if you move the box upwards friction will have another direction;its direction is opposite to the direction of motion of the box always. $\endgroup$ – MrAP Oct 14 '16 at 15:26
  • $\begingroup$ @MrAP Naturally the direction can change if the situation changes. And so can the gravitational force; just think of planets orbiting each other. $\endgroup$ – Steeven Oct 14 '16 at 18:18
  • $\begingroup$ I think that in your example, the work done by gravitational force on the planet would be different for reaching a particular point clockwise and anticlockwise $\endgroup$ – MrAP Oct 18 '16 at 17:28
  • $\begingroup$ @MrAP I'm sorry, I don't understand what you mean by clockwise and anti-clockwise. My point above was that your statement that a conservative force must be unidirectional is incorrect. The gravitational force is an example of a force which is conservative but not unidirectional. $\endgroup$ – Steeven Oct 18 '16 at 22:59
1
$\begingroup$

Work is defined as the path differential form associated with a force vector field, i. e., $dW = F_x dx + F_y dy + F_z dz$; the finite work is the integral thereof on a finite line $\gamma$. Once you integrate the variables over, the only variable left is exactly the path you are integrating upon, hence, by definition, integrals of the differential form must indeed be a function of the path, in principle.

One can show that in the very special case of forces derived by a potential function (namely conservative forces) $\textbf{F} = -\textrm{grad}\,V$ the integration over any path happens to not depend on the shape of the path, but only on its initial and end points (because of Stokes theorem on the boundaries of integration).

Forces are forces they must have the same nature.

In the universe, there are four different types of interactions and their form strongly depends on the case at hand and the distribution of masses and charges generating the force.

$\endgroup$
  • $\begingroup$ To whoever downvoted this... why? $\endgroup$ – anon01 Mar 21 '17 at 4:26
-1
$\begingroup$

I look at [non]conservative forces in terms of what they turn energy into. When moving against a force, you do work; when moving with a force, work is done on the object. In a conservative forces, these are both "efficient" energy conversions: 100% of the work you do is turned into usable potential energy, and then 100% of that is turned back into kinetic energy of the object. This is the case with, say, an object falling under gravity in a vacuum: its energy is exactly the same at the bottom as the top.

Friction does not do this. Friction, by definition, converts all of the work done into heat. The force opposing movement in a certain direction $x$ is the sum of the work required to get the potential energy, and the energy that will be lost. In friction there is no potential energy stored, so you are always doing work to move.

The force of your arm on an object is also not conservative. Perhaps, with very efficient muscles, you can lift a weight 100% efficiently. But when you lower it again, the potential energy is not converted back into energy in your arm. In fact, you'll have to start firing your muscle cells to slow it down as it falls, which will waste heat, so you do work even when it's coming down. Again, all the work is turned into heat.

As an example of something in between, an object flying through air under gravity: going up, most of its energy is turned into potential energy, and a bit into turbulence in the air (and eventually heat); and going down, most is turned back into kinetic energy. So perhaps, upon reaching its original height, it has 90% of its original speed. This force was mostly conservative.

Why are there "2 kinds of forces"? In the same sense that there "are two kinds of English sentences": true and false. Conservatism is just a property I can talk about, indicating 0 loss. This is the same as "[perfectly] elastic" and "inelastic".

Why are the 4 fundamental forces conservative? -- because they have nowhere else to put the energy! On a microscale, all energy is "tracked", including heat, as a the kinetic energy of particles creating heat. By definition, energy is the thing that is conserved by all the forces. So we built energy such that the fundamental forces would be conservative, and this determined it. Then, if you choose to ignore certain kinds of energy (such as heat), you can get nonconservative forces.

$\endgroup$
  • $\begingroup$ That's just not what conservative and non-conservative forces are. Also, energy conservation is a different point and need not always hold. "we built energy such that the fundamental forces would be conservative, and this determined it" that's not entirely true at all, as nor is the very last statement. $\endgroup$ – gented Jul 6 '16 at 9:28
  • $\begingroup$ I'm confused by your objections. It's not perhaps the standard phrasing of defining a conservative force (either "path independence" or the as a relation between some derivatives), but it's clearly equivalent in the case of a force that depends only on position. The question's poster doesn't seem to be asking for a mathematical derivation -- they seem to understand that -- as much as an elaboration on the reasons/examples/intuition for these forces, which I think I provide. Re: "energy conservation .. need not always hold." I'm very confused what you mean by this part. $\endgroup$ – Alex Meiburg Jul 6 '16 at 10:58
  • $\begingroup$ Certainly energy conservation is true on a global scale. As a result, any nonconservative forces are a result of neglecting to count some form of energy. Do you disagree with this? $\endgroup$ – Alex Meiburg Jul 6 '16 at 10:59
  • $\begingroup$ They may be equivalent in same cases, but they are different concepts in general (this was my remark). "energy conservation needs not always hold" means exactly what it says, namely that given any general system, energy conservation might or might not hold according to the case at hand. As for the last remark "any nonconservative forces are a result of neglecting to count some form of energy" I'm not particularly convinced this is the case: it is true for thermodynamic heat for sure, but I'm not sure it holds in general (it might though). $\endgroup$ – gented Jul 6 '16 at 11:23
  • $\begingroup$ For example the Lorentz force acting on a charged particle is non-conservative, but I don't see any instance of your examples holding for electromagnetism. $\endgroup$ – gented Jul 6 '16 at 15:18
-1
$\begingroup$

What makes a force conservative is not the nature of the force, but the configuration, or setup in which you feel it. For instance, a static configuration of charges or masses can create a force field that is conservative, as the force you feel at a given place is always the same. But if the charges were moving, this would be no longer true, the force that a probe charge will feel at a given position will be a function of time, and the field no longer conservative.

Example 1: conservative. The electric force field created by a charge at rest.

Example 2: non-conservative. The electric field by the same charge if the charge is attached to a spring that oscillates periodically. At a given point in space the electric field will change with the position of the charge, so it is not conservative.

$\endgroup$
  • $\begingroup$ Could you please just help me visualise your explanation with an example of each (conservative and nonconservative) case? $\endgroup$ – user282856 Jul 6 '16 at 18:15
  • $\begingroup$ I added the examples to the answer, hope it helps $\endgroup$ – Wolphram jonny Jul 6 '16 at 18:20
  • $\begingroup$ But my point is this: how can you have a static system in any case cause there is going to be displacement in case of work being done. So all along the path the field is going to vary. $\endgroup$ – user282856 Jul 6 '16 at 18:41
  • $\begingroup$ yes, of course, the object feeling the force will feel a different force across space, but will feel the same force if he comes back to the same place. This is a necessary condition for the field to be conservative (although not sufficient). When the charge that generates the electric field is moving, the electric field at a given position changes, so when the object that feels the field returns to a given position, he will feel a different force. This is enough to make it nonconservative. $\endgroup$ – Wolphram jonny Jul 6 '16 at 19:17
  • $\begingroup$ The exact condition for a force field to be conservative is explained in more detail here en.wikipedia.org/wiki/Conservative_force $\endgroup$ – Wolphram jonny Jul 6 '16 at 19:18

protected by Qmechanic Oct 14 '16 at 16:32

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.