If we consider an experiment of pulling a spring with a constant force $F$, then by Newton's Third Law of Motion we should experience an equal reaction force $F$ in the opposite direction. But by Hooke's Law, a stretched spring should exert a force proportional to the amount of stretching ($x$), that is, force should be $k\times x$. How can the two possible at the same time? I am sure that I must be missing something, maybe something to do with the internal of spring, but I can't figure out what.

  • 1
    $\begingroup$ When you pull, one end of the spring by a distance $x$, you apply force of $kx$ in one direction. The spring then pulls back with a force $-kx$ (by Newton's Third Law) to keep the spring in equilibrium. What is the confusion here? $\endgroup$
    – Sam
    Jan 26, 2020 at 13:07
  • $\begingroup$ It is not true that you can only pull with force kx, it can be anything, after all, you are the one who is pulling it. $\endgroup$ Jan 26, 2020 at 13:09
  • 1
    $\begingroup$ To pull the spring by distance $x$, you will have to apply a force of $kx$ so long as Hooke's law is still obeyed for that distance $x$. $\endgroup$
    – Sam
    Jan 26, 2020 at 13:21
  • $\begingroup$ Hooks low hold till elastic limit. And if you say force can be anything then they will be beyond elastic limit, so no law will be applied. $\endgroup$ Jan 26, 2020 at 13:25
  • 1
    $\begingroup$ This question has nothing to do with the elastic limit. NOTHING. Can we please drop that discussion? Suggestion to the original poster: if you change the first sentence to "ideal spring" this discussion goes away on its own. $\endgroup$
    – garyp
    Jan 26, 2020 at 14:12

2 Answers 2


If we consider an experiment of pulling a spring with a constant force $F$, then by Newton's Third Law of Motion we should experience an equal reaction force $F$ in the opposite direction.

The spring provides a restoring force $F=kx$, as long as it is not stretched beyond capacity.

But stretched beyond capacity it will still provide a restoring force but it will no longer be proportional to $x$.

But before the response is such that $F=kx$, that is, $x $ is less than $F/k$, what is the reaction?

We need to look at this dynamically. Assume a point mass $m$ attached to the spring, where the force $F$ will act on. The spring is kept horizontal $x$-axis (so we don't need to account for gravity)

Say that at $t=0$, $x=0$ and we start applying the constant force $F$ (assume also the spring to be of $0$ mass). The spring's restorative force is also $0$ (because at that point $x=0$).

Since there is now a net force acting on the point mass, by N2L there must be acceleration:


More generally (for $x>0$)

$$\Sigma F_i=ma$$ So: $$F-kx=m\ddot{x}$$ So for $x=\frac{F}{k}$:

$$F=kx \Rightarrow \ddot{x}=a=0$$

  • $\begingroup$ I am not at all asking if force is kx till a particular capacity or not, I am just asking why Newton's Law says that force exerted by spring should be F (which is constant) and Hooke's Law says it should be kx. $\endgroup$ Jan 26, 2020 at 13:20
  • 1
    $\begingroup$ I thought that was self-evident? As the force $F$ extends the spring, the spring's response gradually increases, until $F=kx$. $\endgroup$
    – Gert
    Jan 26, 2020 at 14:16
  • $\begingroup$ But before the response is such that F=kx, that is, x is less than F/k, what is the reaction? Newton's third law says it should be F whereas Hooke's law says it should be kx(<F). $\endgroup$ Jan 26, 2020 at 14:25
  • $\begingroup$ Please see my edit. $\endgroup$
    – Gert
    Jan 26, 2020 at 15:30
  • $\begingroup$ So there's a mass on a spring and you are pulling it with force F. Gert's post shows that initially there is an acceleration because there is a net force on the mass because kx < F. When x is large enough so that kx = F, the acceleration is 0. Newton's third law says that the mass always exerts an equal and opposite force to F throughout this process. But that reaction force is not what is considered when finding the motion of the mass as it is not acting on the mass. $\endgroup$ Jan 26, 2020 at 15:42

The reaction force will be $F$. If the displacement of the spring is linear up to the magnitude of that force, $k = \frac{F}{x}$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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