# Is the restoring force of a spring an action-reaction pair?

Let's imagine a spring is attached to a rigid wall. One end is free to move. Now I exert force on that end and make some elongation there. According to Newton's third law, there should be a action reaction pair. So a restoring force will occur in the spring. But when I release that end there shouldn't be any restoring force (because I no longer exert any force) in the spring and the spring should remain in such an elongated state. But reality differs. Why?

Second question is that, if restoring force and external force are an action-reaction pair, then they are equal in magnitude, which means that we cannot exert a constant force. I mean external force can't be constant because restoring force being $$-kx$$ depends on $$x$$ which is a variable. But common sense says it's easily possible to exert constant force.

However I am literally stuck in these two conceptual questions and will be glad to see detailed answer.

• You exert a force on the spring. The spring exerts a force on you. When you let go, the force exerted by spring on you is zero as you would expect. The restoring force still exists and hence the springs goes back.
– sku
Commented Mar 14, 2022 at 6:19
• @sku why does restoring force exist then?When and why is actually it produced in the spring ?
– user325381
Commented Mar 14, 2022 at 6:26
• A spring wants to be in the relaxed state. If you compress or pull, you are creating a restoring force which is working against you.
– sku
Commented Mar 14, 2022 at 6:40
• If it was an ideal spring, and you exclude gravity, the spring will continue to oscillate with frequency $\frac{1}{2\pi}\sqrt{\frac{k}{m}}$ But real springs are not ideal. That is why reality differs from theory. And although the force varies with $x$, for each value of $x$, $F_{\text{exerted}}=kx$ In other words, you have to keep increasing the exerted force for increasing values of $x$ and Newton's third law always hold. Commented Mar 14, 2022 at 6:48
• @josephh your frequency is for a massless spring with a mass m on the end. That is not what we have here. The only mass is the spring. If only it were that simple. Commented Mar 14, 2022 at 7:22

But when I loose that end there shouldn't be any restoring force(because I no longer exert any force) in the spring and the spring should remain in such an elongated state. But reality differs . Why?

You can imagine a spring as a chain (collection) of particles. First particle in the chain pushes the second, the second pushes the third and so on.. You are pushing the last particle in the chain, and the same particle is pushing you back (action-reaction). Once you stop pushing, the last particle also stops pushing you back, but is still being pushed by other particles in the chain until spring restores to its original position.

if restoring force and external force are action reaction pair then they are equal in magnitude.

restoring force being -kx depends on x which is a variable.

Correct and correct.

But common sense says it's easily possible to exert constant force.

Of course it is possible to exert a constant force. For a spring to compress, there must be two external forces acting on the spring - one from the left and one from the right. If one of these two forces is larger than the other, the spring will compress but it will also move (accelerate) in the direction of the larger force.

If one end of the spring is attached to a fixed wall and you apply some force while spring is relaxed, the wall will (almost) immediately apply the same force to the other end. The pushback that you feel from the spring at $$\Delta x = 0$$ is not restoring force by the spring but the normal force by the wall propagated through the spring.

• Can we say this? As I exert force the 'x' gets larger . Does it mean that I am applying variable force? Or it says if I apply a particular force there will be a particular amount of elongation or say value of 'x' . Again we apply another particular force which gives another particular elongation. And doing this for different constant forces everytime (not at a time) gives some values. Now if I graph those numbers it's linear. Hence we get F=Kx. But this relationship doesn't mean that as I continue to pull, x gets larger and F becomes variable.
– user325381
Commented Mar 14, 2022 at 8:42
• @VedflewquestaPercycloxirieta What it means is that, in equilibrium, for a certain constant force you exert on the spring it will elongate $\Delta x_0$ from its relaxed state. If you want to elongate it more than $\Delta x_0$ you need to provide (exert) more force. Commented Mar 14, 2022 at 8:54
• So basically,the restoring force of spring and the force of the man are NOT the action reaction pair?? Commented Apr 28, 2023 at 2:00
• @Aakash From my answer: The pushback that you feel from the spring at $Δx_0$ is not restoring force by the spring but the normal force by the wall propagated through the spring. Commented Apr 28, 2023 at 5:44

But common sense says it's easily possible to exert constant force.

Constant forces are nice and easy to calculate in physics problems. But it is difficult to exert a constant force in some situations.

You might easily be able to exert 100N against a wall. But you would have tremendous difficulty exerting 10N against a feather. We can see that 10N on a feather would produce immense acceleration. And your hands/arms cannot deliver such.

Similarly, if you had a goal of exerting a constant 10N on a light spring, you would almost certainly fail. The end of the spring would simply accelerate away rather than push with any significant force until the displacement is greater.

If you really could push much, much faster so that you really did produce the target force immediately, then it would be because the end of the spring has a non-zero mass, and the acceleration of that mass requires a force. We are no longer considering light, ideal springs in that case.