# Doesn't the work $W = \int F \, dx$ count only the work done by the outermost point of a spring?

When we use the integral to calculate the work done by the spring force, then according to my interpretation, we are only calculating the work done by the outermost point on the spring. Why don't we sum (integrate) the work done by all the points? Am I missing something? Help me out please.

• Can you explain what you mean by the "outermost point"? – Bob D Sep 14 at 16:35
• When we are integrating kxdx, we are calculating the workdone on displacement dx of the point (the point which is in the natural length position, i.e the point we begin applying force in)with force kx (x varies from 0 to x for each dx).This means we are calculating the total work by spring force of the outermost point only. But all the points in the spring are displaced. Why don't we add all the workdones by them. – NightKruger Sep 14 at 16:37
• Now I think I understand what you are getting at, and will submit an answer. – Bob D Sep 14 at 16:38
• This question may help: physics.stackexchange.com/q/226279 – Thomas Lee Abshier ND Sep 14 at 16:49
• No offense but im asking regarding a different question. – NightKruger Sep 14 at 16:54

When we are integrating $$(-kx)dx$$, we are calculating the work done on displacement $$dx$$ off the point which is in the natural length position, (i.e. the point we begin applying force in) with force $$kx$$ ($$x$$ varies from $$0$$ to $$x$$ for each $$dx$$). This means we are calculating the total work by the spring force of the outermost point only. But all the points in the spring are displaced. Why don't we add all the work done by them?

The formula $$\int(-kx)dx$$ has within it the spring constant $$k$$, which has units of N/m. This is, of course, a linear approximation for small displacements. The infinitesimal displacement $$dx$$, multiplied by the spring constant, will result in an infinitesimal amount of Work, and integration/summing of all those infinitesimals will yield a total amount of work being done over the total displacement from $$0$$ to $$x$$.

In your question, it appears you have assumed that because only the position of the end of the spring is being considered in the integration, that the rest of the spring is being ignored/not considered by the integration?

But, in fact, the spring constant $$k$$ includes/implicitly incorporates the force contributed by every increment of the entire spring. Every increment of the spring must be stretched with every displacement of the spring's end.

Obviously, the spring constant $$k$$ will not be as large if the spring consisted of only an infinitesimally small increment of spring at the end of the spring. In other words, the spring constant has the value $$k$$ because of the total contribution of all the increments of force that must be overcome with each $$dx$$ of displacement of the end of the spring.

In other words, the concept/actuality of the spring constant $$k$$ includes the force applied against displacement of all the increments of the spring.

Thus, when integrating the force multiplied times each increment of displacement, the integration naturally/unavoidably includes the total force applied to displace all the spring's constituent inter-atomic bonds. Thus, the Work calculated for each increment of spring-end displacement implicitly includes the force needed to displace all of the interatomic bonds in the entire spring. And subsequently, integration of all those increments of Work yields the total work done by displacing the spring against the force exerted by all the bonds within the spring.

• So what you are saying is that the kx force itself is such ,that it includes the net force due to the displacement of each point on the spring. Thus when we integrate the force multiplied by increment dx, we are actually finding the workdone by all the points together contributing to total displacement of dx and the corresponding force is the net force they are applying. So when we start integrating from the displacement of 0 to x, we are simultaneously adding the small (total) works by all the points, as the force and dx we are considering are itself directly due to all the points? – NightKruger Sep 14 at 18:30
• @NightKruger Yes, that is exactly what is happening. Very good summary! – Thomas Lee Abshier ND Sep 14 at 19:52

Suppose that there is a spring fixed at one end and a force $$F$$ is applied to the spring.
The important thing is that force $$F$$ can be thought of as the "tension" in the spring and is the same throughout the spring.

Now consider the whole spring as three equal lengths.

Each length is subjected to a force $$F$$ at each end as shown in the diagram.

The work done on spring 1 by the forces stretching it by a small amount $$\delta x$$ is $$F\,\delta x$$.
The work done on spring 2 by the forces stretching it by $$\delta x$$ is $$F\,2\delta x - F\,\delta x = F\,\delta x$$.
The work done on spring 3 by the forces stretching it by $$\delta x$$ is $$F\,3\delta x - F\,2\delta x = F\,\delta x$$.

So the total work done in stretching the three springs is $$F\,\delta x +F\,\delta x+F\,\delta x = F\,3\delta x = F\Delta x$$ where $$\Delta x (=3\delta x)$$ is the total extension of the whole spring.

The whole spring can be cut up into as many segments as you like and then the work done extending the spring can be written as $$\displaystyle \int_0^{x_{\rm final}} F\,dx$$ where $$x_{\rm final}$$ is the final extension of the whole spring.

• That's a good way to explain it without describing the spring constant. Upvote from me. – NightKruger Sep 14 at 18:45
• Why would part 2 strech by 2dx and part 3 strech by 3dx??? – Schwarz Kugelblitz Sep 15 at 21:43
• @SchwarzKugelblitz Look at elemental spring 2. The left end moves $\delta x$ to the right and the right end moves $2\delta x$ to the right so spring 2 is stretched by $2\delta x -\delta x = \delta x$. – Farcher Sep 15 at 21:47
• How does the right side move 2dx? – Schwarz Kugelblitz Sep 15 at 21:48
• @SchwarzKugelblitz Because each elemental spring is stretched by $\delta x$. – Farcher Sep 15 at 21:49

Why don't we sum (integrate) the work done by all the points?

In essence we are doing just that. When we take the integral of $$(kx)dx$$ between zero (the relaxed position of the end of the spring) to $$x$$, the stretched position, we are macroscopically summing up the work done on the microscopic level to pull the molecules apart at each point along the length of the spring in opposition to the electrostatic forces between them.

Envision the spring being comprised of molecules connected together by microscopic springs between them. The spring constant, $$k$$, is the macroscopic parameter equivalent to the microscopic spring constant. The integral of $$(kx)dx$$ sums up all the work being done on the microscopic level in stretching the microscopic springs.

But then are summing work done on each point only for distance dx

What I mean is the sum of all work at the molecular level equals the total work at the macroscopic level, and the sum of all the intermolecular displacements, $$dx_i$$, add up to the final displacement, $$x$$ as @Farcher showed pictorially. Or

$$\sum_{i=0}^{n}k_{i}dx_{i}=\frac{kx^2}{2}$$

Where $$\sum dx_{i}$$ = $$x$$.

Hope this helps.

• But then are summing work done on each point only for distance dx. – NightKruger Sep 14 at 16:56
• @NightKruger Although I see you already accepted another answer, I updated my answer to respond to your comment. Hope it helps. – Bob D Sep 14 at 19:33
• You all are equally responsible for my crystal clear understanding now. Thomas Lee Abshier ND made me understand theoretically. Farcher made it pictorial and You my good sir with the basic idea. I accepted Thomas's because he explained the little point I was missing clearly. – NightKruger Sep 14 at 19:38
• @NightKruger And that's exactly what you should do. As they say on this site, acceptance of an answer does not necessarily mean it is the best answer nor that the other answers are necessarily wrong. It means it is the answer that worked best for the person. – Bob D Sep 14 at 19:43

The restoring force doesn't depend on the relaxed length of the spring or number of particles in it. It just depends on how far it is stretched or compressed with respect to the mean position and the spring constant.

When finding the work done (potential energy stored) we just need the force which we must apply to stop the spring from relaxing, at a certain distance and hence we just need the force at that point and the small displacement $$dx$$ over which it acts. Finally we integrate it all to get the total work done.

Normally in physics problems the spring is considered to be an ideal spring, with negligible mass and negligible internal energy dissipation. So although the various parts of the spring do move as it is compressed and expanded, if you to work out the $$W=Fd$$ work associated with all of those various individual parts of the spring, you would realize that the force $$F$$ associated with moving each of those parts is zero, so no work is done on those parts. You only have to concern yourself with the work done on whatever the ideal spring is pushing.

If you were to ever encounter a physics problem in which a compressed spring is, for example, embedded in some very thick medium so that there is a force associated with the movement of every element of the spring itself as it is released, then one would have to do as you suggested and figure out the $$W=Fd$$ work associated with every element of the spring and add them all up to get the total work.

• But then that too sounds unnatural. If we consider work by all the points of the spring, then that would equal infinite! – NightKruger Sep 14 at 17:01
• @NightKruger - No, an infinite summation over infinitesimally small parts does not in general result in an infinite value as you will learn when you take calculus. – Samuel Weir Sep 14 at 17:04
• Yes I know that but the net work done on each point is of course not infinitesimal. – NightKruger Sep 14 at 17:06
• @NightKruger - You would have to work out the forces on each of the small elements of the spring and then take the limit as the elements get infinitesimally small. Again, this sort of procedure is all covered in calculus courses. – Samuel Weir Sep 14 at 17:10

We're calculating the work done by the spring on some external point which isn't part of the spring. When you calculate the work done on some part of the spring, you're calculating how this contributes to the total kinetic energy of the spring itself.

The short answer to the question in the title is, "Yes, but the work done by that part of the spring on what?". In particular, if the integral is positive, that means energy was transferred 'to' an object, negative means 'from.' What you're asking is the same as asking, "Where did that energy come from before it got to the end of the spring?" As you correctly intuited, that energy gets stored all along the entire spring, but it is in a more advanced class on elastic media you'll learn how to handle that.

In the mean time, if you want to think about it on your own, explore the topic of using multiple springs to build a single "spring." For example, "If I connect two springs with constant $$k$$ end to end, what is the constant $$k_{\mathrm{effective}}$$ of the combination?" or the same question if I connect two springs side by side.

The other answers are all good, but I love this question too much to pass it up:

Let's break our spring with constant $$k$$ into $$n$$ little springs with spring constant $$nk$$ (shown here in ASCII):

|~$$nk$$~|~$$nk$$~|$$...$$|~$$nk$$~|~$$nk$$~|~$$nk$$~|

You can verify that the total spring constant is indeed $$k$$:

$$\frac 1 {\frac 1{ nk} +\ldots+\frac 1{nk} }= \frac 1 {\frac n {nk}}=k$$

since series springs are added harmonically. The surprising result is that the spring segments have a larger spring constant than the total spring.

When the total spring is stretched by $$x$$, each segment stretches by $$x/n$$.

Note that the tension in each segment is:

$$T = (nk)(x/n) = kx$$

as it must be. Moreover, each segment has done work:

$$\Delta W = \frac 1 2 (nk)\big(\frac x n\big)^2 = \frac 1 2 \frac {kx^2} n$$

so obviously:

$$W = n\times\Delta W = \frac 1 2 kx^2$$

In the limit $$n\rightarrow \infty$$ you can see that the energy is uniformly distributed along the length of the spring.

• But isn't the work supposed to be 1/2 of what you mentioned? – NightKruger Sep 14 at 19:28
• @NightKruger not anymore. – JEB Sep 24 at 16:52
• +1 - I'm puzzled about the downvote! This is a good understanding of how springs operate!. If you cut the spring in half, you double the spring constant. The grand constant is the integral of the little contributions of each coil. – Bill N Sep 24 at 17:42
• @BillN well I had dropped a factor of 2 for while, but it's there now. – JEB Sep 24 at 22:24
• Wow! Missing a factor of two certainly isn't worth a downvote, IMO. That's easily handled in a comment. Downvotes should be used to signal off-topic answers or poor physics, not some numerical factor that's easily missed. Oh, well ... – Bill N Sep 25 at 13:34