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Why is the work done on a spring independent of the amount of applied force acting on it? Why does it only depend on the spring constant and the amount of stretch or compression?

For example: $k=2$ N/m, $x(\mathrm{stretch})=10$ m.

Case-1: Applied force $F=50$ N. $$\mathrm{Total \ work \ on \ the \ spring} = 50 x - \frac{1}{2} k x^2 = 500 - 100 = 400 \ \mathrm{J}$$

Case-2: Applied force $F=100$ N.

$$\mathrm{Total \ work \ on \ the \ spring} = 100 x - \frac{1}{2} k x^2 = 1000 - 100 = 900 \ \mathrm{J}$$

It's apparent that the work done is different in both cases. However, I've witnessed in many questions, that ask for the work done in stretching a spring to a certain length, without ever mentioning the amount of applied force. What is it that I've misunderstood?

Furthermore, in problems such as the work required to raise an object to a certain height or elevate the center of mass of an object, only the work done by the gravitational force of Earth is taken into account, and the applied force is neglected. For example:

Let's consider a uniform cylinder with a radius of 'r', height 'h', and mass 'm', lying on the ground with the circular surface touching the ground. If we want to orient it on the ground with the flat circular end touching the ground, the work to be done is calculated as follows: since the center of mass moves up by 'h/2-r', the work done by gravity, i.e., mg[(h/2)-r] (negative work), is considered as the work to be done, and nothing about the applied force is ever mentioned. Why is that so?

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  • $\begingroup$ The force on a spring changes linearly with x from 0 N to 50 N in case 1. A similar effect occurs for case 2. I also note that for a given spring constant, twice the force causes twice the stretch or twice the compression. $\endgroup$ Commented Jul 16, 2023 at 16:19
  • $\begingroup$ physics.stackexchange.com/questions/770704/… <related question $\endgroup$
    – AXensen
    Commented Jul 18, 2023 at 8:58

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You aren't thinking about the problem in the right way.

First as Dale said, springs are often idealized. The idealization is that that the mass of the spring is much smaller than any objects pushing on it. And accelerations are relatively small. When you consider $F=ma$, you don't need to worry about $ma$ of the spring itself. Or about how forces on one part of the spring affect motion of other parts. This is done because the behavior of a spring is usually so close to ideal that it makes no difference. And it makes the problem simpler. It allows you to treat the spring as a massless gadget that connects two objects together and exerts equal and opposite forces on both.

At first glance it may sound like any connector must do this. But this isn't true. You might approximate a massive spring as two ideal springs with a mass in the middle. Connect this spring to two masses. Accelerate the spring's mass. It would push one mass ahead of it and pull the mass behind it. Both ends would experience forces in the same direction.

You ask about applying a force of $50$ or $100$ N to a spring. Typically you would want to think about how you would use an ideal spring to apply a force to two objects. Or about the force an object on one end of an ideal spring exerts on an object at the other end.


The force exerted by a spring is something like the reaction force exerted by a wall. Suppose you have a mass sitting against a wall. You exert a force on the mass toward the wall. If that was the only force, the mass would accelerate. The wall exerts a force just strong enough to keep the mass from penetrating the wall. The mass doesn't move. So the total force must be $0$. So the force exerted by the wall must equal your force. The important point here is the wall adjusts its force to match forces exerted on it.

A spring also adjusts its force. A spring with no forces on it has a length. If you press the two ends together, the spring will shorten. As it does so, it will exert outward forces. The behavior is similar if you pull the ends apart. The strength of the forces is proportional to the amount of shortening. $F = kx$, where $x$ is change in length.


Suppose you have two equall masses connected by the spring you describe. Initially their separation is just right so the spring force is $0$. The masses are moving toward each other at equal velocities, so the spring immediately start being compressed. The spring force grows as the separation shrinks. The masses decelerate. If the separation shrinks by $25$ m, the spring force on each mass is $50$ N. At $50$ m, the force is $100$ N.

Because the force is constantly changing, you cannot calculate the work done with $W = Fx$. Instead, you have to integrate.

Let us be careful with variable names. Let us call $s_0$ the separation that gives $0$ force, and $s$ the change in separation from $s_0$. Let $x$ be the final separation. We want to know how much work is done as $s$ changes from $0$ to $x$. So

$$W = \int_0^x F \space ds = \int_0^x ks \space ds = \frac{1}{2}kx^2$$

You can see that F matters, but it is gone from the final equation. You can bring it back. It usually isn't done because this form is more useful. But you can ask what is the final force when the separation reaches the final $x$ value? $F_{final} = kx$.

$$W = \frac{1}{2}kx^2 = \frac{1}{2}k\left( \frac{F_{final}}{k}\right)^2 = \frac{1}{2} \frac{F_{final}^2}{k}$$


So what does this work mean?

Two things in this case. It is the loss in kinetic energy of the masses as the spring decelerates them. And it is the increase in potential energy of the spring.

In another problem, you might hang a mass from a fixed ceiling. Gravity pulls the mass downward with a constant force. The spring stretches until the spring force becomes equal to the weight. The work done as the spring stretches is equal to the loss in potential energy of the mass. It is also equal to the gain in potential energy of the spring.

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    $\begingroup$ beautifully explained $\endgroup$
    – hyportnex
    Commented Jul 16, 2023 at 17:09
  • $\begingroup$ @mmesser314 Pardon my lack of insight but I'm talking about an external constant force being applied on the spring by some other agent that does work on the spring. $\endgroup$ Commented Jul 16, 2023 at 17:52
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    $\begingroup$ @AyushNaman, based on Newton's 3rd law, it's not possible to apply a constant force to a spring. $\endgroup$ Commented Jul 17, 2023 at 1:05
  • $\begingroup$ The example above of a mass hanging from the ceiling is an example of a constant for on the two ends of a spring. Nothing moves, so no work is done. Work was done setting it up, but that was done with non-constant force on the spring. $\endgroup$
    – mmesser314
    Commented Jul 17, 2023 at 4:11
  • $\begingroup$ You cannot start with the spring at length $s_0$ and exert a constant force on the ends without violating the assumptions that make the spring ideal. If you do, you find that you have to consider the mass of the spring and the effect of one part of the spring pulling on other parts. You can get waves propagating up the spring. Advanced classes go over this in detail. One example of this is sound waves. $\endgroup$
    – mmesser314
    Commented Jul 17, 2023 at 4:15
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Why is the work done on a spring independent of the amount of applied force acting on it? Why does it only depend on the spring constant and the amount of stretch or compression?

This is because the springs are also typically assumed to be massless. With a massless spring it is not possible to exert a force on it that is different from the Hooke’s law force. Any other force would result in an infinite acceleration.

In your examples you calculated assuming a constant force. With massive springs the difference between the work done and the elastic potential energy would go into the kinetic energy. For a massless spring this cannot be done.

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    $\begingroup$ Please ignore my limited insight, but isn't the work done on an object independent of the mass of the object? Also, please share your insights on why the applied force is not taken into account when asked about the work done to move an object to a certain height. $\endgroup$ Commented Jul 16, 2023 at 13:53
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    $\begingroup$ @AyushNaman the applied force is taken into account, but you are assuming that it is possible to apply an arbitrary force. It is not. This can be seen by a free body diagram and Newton’s 2nd law. Only a force equal to the Hooke’s law force can be applied to a massless spring $\endgroup$
    – Dale
    Commented Jul 16, 2023 at 14:03
  • $\begingroup$ Where can I look this up in detail? $\endgroup$ Commented Jul 16, 2023 at 14:13
  • $\begingroup$ @Dale What if there is a massless charge on one end of the spring and a very strong static electric field over all the space? Cannot one then apply an arbitrary force-unconstrained by the Hooke's law-to the spring? $\endgroup$
    – aystack
    Commented Jul 19, 2023 at 3:14
  • $\begingroup$ @aystack I don’t think so, but I would have to see a free body diagram and the math to go with it to be sure either way $\endgroup$
    – Dale
    Commented Jul 19, 2023 at 3:31
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In both your cases it seems you are treating the externally applied force as a constant force applied over the stretching distance. It is not. The force is a function of the displacement and is equal and opposite to the opposing force of the spring at each position per Newton's 3rd law. That force is proportional to the stretching distance times the spring constant. Therefore

$$F(x)=kx$$

A differential amount of work done is then $$dW=F(x)dx=kxdx$$

And the work done stretching the spring by an amount $x$ is $$W=\int_0^xkxdx=\frac{1}{2}kx^2$$

For $k=2N/m$ and $x=10$ the work done by the external force is 100 J. The spring does an equal amount of negative work of 100 J, for a net work of zero. The negative work done by the spring takes the energy given it by the positive work of the external force and stores it as elastic potential energy.

Hope this helps.

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  • $\begingroup$ why can't external force be a constant force? Rather, it should be the opposing spring force that depends on the deformation of the spring $\endgroup$ Commented Jul 16, 2023 at 17:49
  • $\begingroup$ When you apply an external force at some point on the spring the spring must exert and equal and opposite force on that point. That's Newton's 3rd law. And the force the spring applies is not constant, it varies with the displacement. $\endgroup$
    – Bob D
    Commented Jul 16, 2023 at 17:59
  • $\begingroup$ @AyushNaman I am assuming your spring is massless. Is that a correct assumption? $\endgroup$
    – Bob D
    Commented Jul 16, 2023 at 18:29
  • $\begingroup$ Yes all conditions are ideal. Oh, I get it. You're implying that since the mass of the spring is zero, the net force on the spring must be zero. $\endgroup$ Commented Jul 16, 2023 at 18:45
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    $\begingroup$ @AyushNaman, do you understand the implications of Hooke's Law? $F=kx$. This means that the force that is required to change the length of the spring depends on how much the spring is stretched or compressed. It also means that it's impossible to apply a constant force to a spring, but it is possible to apply a maximum force to a spring. $\endgroup$ Commented Jul 17, 2023 at 1:11
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Did you ever do experiments with springs? Say you have a spring with $k=100\,$N/m. Now, to get constant force, the easiest way is via some weight $mg$. Say, you have a weight of $100\,$g, i.e. about $1\,$N, if you hang it on the spring it extends $1\,$cm. To achieve $2\,$cm you have to add the same weight again, so you have $2\,$cm, and so on, until you have $10\,$kg or $100\,$N and your spring is extended by $1\,$m. So the force you need is proportional to the elongation. That is why you can either calculate the force or the elongation in an exercise; the work will be the same.

Also if you lift a weight up you need the force $mg$ to hold it and a little bit more to lift it slowly. So again the work done is $mgh$. You can not apply more force if you do not want to accelerate the mass.

With little kids, I demonstrate lifting an empty box and pretend that I need all my possible force, then I ask why they do not believe me.

So you must exactly state how you would calculate the force to extend a spring or lift a mass, and following the formula for work.

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  • $\begingroup$ @naturallyInconsistent Thank you very much for editing my bad text! $\endgroup$
    – trula
    Commented Jul 18, 2023 at 10:38
  • $\begingroup$ Of the stuff I have edited, yours is rather ok. It is definitely not bad text $\endgroup$ Commented Jul 18, 2023 at 11:24
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I had a similar doubt about the work done against gravity.

The key point is this: such problems assume you start at rest and end at rest. This is a constraint on your applied force.

When an object is initially and finally at rest, then the net work done on it must be precisely the change in its potential energy. It is of course impossible for the work to be less than the change in potential energy. If the work is greater than the change in potential energy, then the excess will become kinetic energy, contrary to our assumption that it is finally at rest. This is why we don't need to know anything about the exact nature of the applied force in such problems (and is precisely why reasoning about physical systems through energy conservation is so useful -- it lets us "black box" the applied forces so long as we know the initial and final states).

I believe this is so confusing when you first learn about it because everything in our everyday experience seems to speak against it. If I lift a heavy weight and set it back down, surely I've done work. And yet the work-energy theorem will say the work done on the weight is $0$. Am I wasting my time in the gym? As far as the weight is concerned, yes. You really did no work on the weight. But it turns out that doing no work on a weight in this manner is an excellent way to get your body to do work on itself, which is how you signal to your body that it should build more muscle, improve its cardiovascular system, etc.

But wait, if I did work on myself, why didn't my kinetic energy change? It seems we've identified a violation of the work-energy theorem. On the contrary, the heat you feel in your muscles after a workout is (essentially) microscopic kinetic energy. A lot of the work you did on yourself went into heating up your body (and the rest more or less went into heating up other things).

To avoid this kind of confusion in the future, it might be helpful to remember in physics, work is not something you do that makes you tired, it's a force acting over a distance to increase the kinetic and/or potential energy of a system.

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this is your system

system.

you pull the mass with constant force $~F~$ slowly to the left (acceleration equal zero) thus

$$F-k\,x=m\,\ddot x=0\tag 1 $$

from here the work

$$W=\int_{x_i}^{x_f} (F-k\,x)\,dx=F\,(x_f-x_i)-\frac k2\,(x_f^2-x_i^2)$$

and accordioning to equation (1) $~x_f=\frac Fk~$

thus if you move the mass with distance $~\Delta x~$ the spring also move with $~\Delta x~$

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I think the other answers are skipping a step. It looks like what you're missing is that you can only apply as much force on the spring as the spring will apply back on you. i.e., every action has an equal and opposite reaction.

Try to apply 10lbs of force to a feather floating in the air, for example. You cannot, because it won't resist you.

When you stretch a spring, the force you apply is determined by the spring constant, because that determines how hard the spring resists you. That is why the "applied force" doesn't appear separately in the equations. This is not entirely accurate, because a spring has mass which resists acceleration, but it's close enough. The other answers explain well starting here.

Similarly, when we talk about the work required to lift an object, only the force of gravity is considered. Obviously, the object is not considered massless in this case, because then it wouldn't be heavy. If you apply a force greater than the object's weight, however, then it will accelerate. If, when you get to the top, you have put in more work that was required to lift the object against gravity, then you will be throwing it, not lifting it.

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