Spring force on both sides of spring I am a little confused about springs. I just wanted to know that if I pull an ideal spring of spring constant $k$ such that the spring has been symmetrically pulled and its elongation (total) comes out to be $x$ then would the force on one side by $$F=kx$$ or $$F=kx/2$$  I am a little bit confused and hence I resorted to ask it here.
 A: 
An ideal spring - Three loading cases : free,compressed,stretched.

A: There is no such thing as a non-symmetrical pull of a string or some non-"total" elongation. $x$ is elongation, and that's it. $F=kx$ is the spring force, and that's it. 
An object tied to one end of the spring experiences this spring force. The whole force. Something tied to the other end experiences by Newton's 3rd law the same force (in the opposite direction). 
No need to half it. 
Seen from each object at the ends of the spring, it would be:


*

*You draw the free body diagram of the object at one end, and it shows a spring force. This force is the push, the spring exerts on the object because it is compressed a bit and tries to return to the uncompressed state. And it is experimentally found to be proportional to the compression as $F=kx$. Which also feels intuitive: doubling the compression doubles the tendency of it to return to the relaxed state. 

*You then draw the free body diagram of the object at the other end. It also feels a spring force. Same argument holds; the force from the spring is present, because the spring is compressed a bit, and the force appears to be proportional to the compression as $F=kx$. 
We cannot talk about the "elongation $x$ caused at each end" and then say that "the total elongation is the sum of these". Because the elongation $x$ that happens is already caused by the two ends both being pulled - if only one was pulled (so only one end feels a force), there would be no elongation at all! (Assuming ideal mass less spring).
Force on one end does not contribute alone with one half of the total elongation; the forces on both ends cause it in collaboration. 
A: When you state $F = k x$ the variable $x$ represents the extension of the spring and not the position of one of the ends.
In a symmetric pull each end moves by $ \frac{\delta}{2} $ then the total displacement is $\delta$. The spring force is then $F = k \delta$ on each end of the spring.
A: The other answers simply quote Hookes law, the static relationship between displacement and force. But if you consider the question more deeply, the spring has distributed mass and distributed compliance, and so a spring all by itself is dynamic and therefore does not propagate force instantaneously as Hookes law implies all by itself.
So if you push (or pull) on a spring you will displace the coils (or leafs or whatever), and the same force will be seen on the opposite side of the spring - however delayed by the dynamic response of the spring.
