What is the force exerted by a spring when pulled by a force $F$? 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.
 A: 
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:
$$F=ma$$
More generally (for $x>0$)
$$\Sigma F_i=ma$$
So:
$$F-kx=m\ddot{x}$$
So for $x=\frac{F}{k}$:
$$F=kx \Rightarrow \ddot{x}=a=0$$
A: The reaction force will be $F$. If the displacement of the spring is linear up to the magnitude of that force, $k = \frac{F}{x}$ 
