# Is spring force really a conservative force?

Let us consider this picture. $$\Rightarrow$$ The first picture shows the initial position of the block when the spring is in its natural length and is kept on a smooth horizontal table.

$$\Rightarrow$$ The second figure corresponds to the situation in which the block is pulled straight on the smooth horizontal table from its natural length (indicated by the first dashed line).

$$\Rightarrow$$ The third figure corresponds to the situation in which the spring is first pulled and is then turned around a hinged nail and is then brought to the same final position as it is in the second figure (indicated by the second dashed line).

The block is moved on the horizontal table only and is not pulled vertically in any of the figures.

So in picture (b) and (c) the initial and final positions of the block is same but I don't think the spring potential energy is equal in both the cases.

Surely the spring elongated more in the third case so it stores more potential energy but we also know that the potential energy is equal to the negative of the work done by conservative forces so doesn't this mean that the work done by the spring depends on the path on which the block travels?

Can someone explain what is happening here? Where am I wrong?

• When you move the spring to the "hinged" position, you are applying forces to the spring that do work and increase its potential energy. That process is reversible. Jul 23, 2021 at 10:13

The position of the block is the same in both cases, but the position of the block does not wholey define the potential energy in the spring. The length the spring is stretched defines the potential energy (along with the spring's coefficient).

If you don't change the setup of the system, you can say that the position of the block defines the length of the spring, because you can find the length of the spring given any arbitrary block position. However, in (c) you change the system by dragging it around a nail. Now, instead of being a straight line between the wall and the block, the spring follows the two diagonal lines. The sum of the length of those two diagonals is longer than the length of the straight line, so the spring has more potential energy stored in it.

Another way you can see the same effect is if you were to start with case (b), and then use a rod to stretch the spring upwards, until it is just as in (c). Intuitively you will have to put work into that spring, so the system must be gaining energy.

• but shouldn't the work done by a conservative force be independent of the path ? Jul 22, 2021 at 14:33
• what do you mean by stretching it upwards ? Jul 22, 2021 at 14:36
• @Ankit As for the work done by a conservative force, it is independent of "path," but there's a nuance there which we sometimes miss at first. We have to talk about the entire system here. Thus the state of the system includes not just the position of the block, but how long the spring is (which might be different depending on how many times you wrap it around nails and such). Jul 22, 2021 at 14:41
• Intuitively, if you are crouched down with a bunch of books in your hands, it takes more work to lift a set of books from the ground to chest-level than it takes to drop the books and then stand up. If you only consider the position of your own body, it goes from "crouched" to "standing" in both cases, so one might argue that the work done should be the same because of path independence. But that only included the state of one part of the system: your body. If you include the state of the whole system, the books and yourself, you find that it isn't at the same state... Jul 22, 2021 at 14:43
• In the first case, where you pick the books up, your body ends up "standing" and the books end up at chest level. In the second case, where you leave the books on the ground, your body ends up "standing" and the books end up at ground level. We would call these two different states, so they don't have to have the same energy. However, what we can say is that, given any one final state (say, standing with books at chest level), it doesn't matter whether you stood up quickly or slowly. The work done will be the same. That's what they mean by path independence. Jul 22, 2021 at 14:44

You have to remember that the energy is "stored" in the spring. The block doesn't have the potential energy. The energy of the third system isn't only dependent on the position of the block (or, more specifically, the right end of the spring). You had to do work to move the middle of the spring to the hinge. You need to take this energy into account as well.

• so we know that potential energy is equal to the negative sof the work done by conservative force. So does this mean that the spring did more work in the third case ? If yes doesn't this mean that the work done by a conservative force depends on the path taken ? Jul 22, 2021 at 14:44
• @Ankit The work is still independent of the path. You could wiggle the block around, take the hinge point and move it around, etc. But if you end up at the state in the figure the work done by the spring is the same as if you did all that wiggling vs. if you had just moved the system directly to that state. Just because you have different states of the system with different amounts of energy doesn't mean the work is path dependent. Jul 22, 2021 at 14:46
• so if I got you right the state of the figure (b) is not the same as in figure (c) . Right ? If it was the same state then the work would have been same ?? Jul 22, 2021 at 14:50
• @Ankit Right. The system is not only determined by the position of the block. You have to do work in moving the middle of the spring to the hinge. But no matter how you get the middle of the spring to the hinge or how you get the block to the final position the work done by the spring will always be the same. Jul 22, 2021 at 14:54

I think that you are confused about what "independent of path" means. For conservative forces the work done between two STATES of the system is independent of path. We also can define a potential energy which is a function of the state of the system. In your case, elastic PE is a function of the state of the spring. This is the system with elastic PE. It is obvious that your (b) and (c) represent two distinct states of the spring so the work done going from (a) to (b) has no reason to be equal to the work done from (a) to (c). The path independence just refers to one pair of states. Not that going from an initial state to any other state you do the same work.

• thanks for your answer . Btw can you please tell me how do we define states ? How do we differentiate one state from other ? Jul 23, 2021 at 14:56
• This will depend on the system considered. For an ideal spring the energy depends on the deformation of the spring (extension or compression). But I understand your concern, the definition of state will depend on what are you looking for. A general approach would be the one mentioned by Dale where you use some generalized coordinates.
– nasu
Jul 23, 2021 at 16:17
• Give you my approvation to your answer. Sep 14, 2021 at 12:33

The issue that you are running into is that the system in the third case is more complicated than in the first and second cases. In the first and second cases the position of the block is sufficient to determine the amount of energy stored in the spring, so that would be two generalized coordinates.

In the third case more generalized coordinates are required to specify the state of the system. I would use two additional coordinates identifying the position of the pin on the plane and then another to identify what part of the spring is pinned. However, it is certainly possible to use different coordinates if you like.

In this larger parameter space it is clear that the work done by the spring is independent of the path taken to get there. There is a potential energy that depends only on these generalized coordinates, and not the velocity nor the history. The force is conservative.

It is conservative, it depends on the path, although the dependence of the force with $$x$$ is different just because the configuration is different. in case (3) the horizontal force is $$F=-kx\sqrt{H^2+x^2}/H$$, if H is the height of the hinge and the string is always straight (we also assume that the mass does not move along $$y$$, also assumed was rest lenght=0).