So whenever you're pushing or pulling against a spring with some applied force, that spring pushes back on you in the opposite direction, and it forms an action-reaction pair. So, for instance, in this example, I'm going to take this spring and push against it with some applied force. That's the action. And then the spring pushes back on me with an equal but opposite direction force. That's the reaction force. And so we say that the spring force is equal to the negative of your applied force, and that's equal to negative k times x, where x represents the object's deformation here, and k is just a constant. It's just a force constant. The only thing you need to know about this negative sign here is that it just means that that force points in the opposite direction. And because that negative sign often just goes away and we're only just asked for the magnitude of these forces, we're going to use this more powerful equation. The absolute value of all these forces are equal, and that's just equal to k times x. The negative sign just goes away.

So to see how all this stuff works, let's just go ahead and take a look at an example. So I've got, in this first example here, I'm pushing the spring to the left, so that means that my applied force is equal to 120. And so that means that the spring force is in the opposite direction also equal to 120. Now I'm told that the k constant, which is that force constant, is equal to 20. And that k constant really just means it's a measure of how stiff the spring is, how hard it is to push or pull it. And so what I'm asked for is how much it compresses by. So I've got these two forces and I'm pushing on the spring and I've compressed it by some distance x, and that is the object's deformation. That deformation is always measured relative to this dotted line here, which is basically the relaxed position. By the way, the other word that you might hear for that often is the equilibrium position, where x is equal to 0.

So let's go ahead and just set up my equation here. So I've got the magnitude of these forces, right, are equal to each other, and that's just equal to k times x. I'm looking for x and I've got k, right, and I've also got the force. So my applied force, which is the spring's force, was equal to 120. My k is equal to 20, and then I've got x, which is my variable here. So you go ahead and solve for that, and you get a deformation of 6 meters. That's how much I compressed it by.

So let's take a look at a different example here because here I was pushing against the spring. Now what I'm going to do is I'm going to pull against it. So now what happens is I'm taking this spring, pulling it out so my applied force is here, which means that the spring force is equal and opposite to that. Now I'm looking for how much force I need to pull it out to some distance here. So let's just write out my formula. F_{s}=F_{a}=k⋅x. Now I'm looking for the actual forces here. I know what the k is. Now I just need to know what my deformation is. I've got 10 meters, 16 meters. Which one of those things represents my distance, my deformation? Well, before, we had a compression distance that was given to us. But here, all we're told is that the spring was originally 10 meters, and then once you've pulled on it, the spring has now become 16 meters long. So, the amount that you've pulled on it or deformed it is 6 meters. That's equal to 6 meters.

So now I've got my x, that's just equal to 6 meters. Now just go ahead and solve the problem. So I've got F_{s}=k⋅x, and I've got k equal to 40, and x is 6 here. So I've got the spring force equal to 240 newtons. In this first example, I had a deformation of 6, a k constant of 20, and then the force was equal to 120. Well, in this example, I've got the same exact deformation, 6. I've got the k constant, which is 40 (double that), and the spring force was equal to 240. So what happens is I doubled the spring constant, the force constant, and then I doubled the force. This shows that when the x is the same, and k gets doubled, then the force doubles as well. This clearly demonstrates that k represents how hard it is to deform a spring. You have to use more force to deform it the same amount. The units for k are in newtons per meter, which measures how many newtons you must use per meter of displacement. Now you should remember that, but if you ever forget, it's just F_{s}=k⋅x. You can get that. The units for F_{s} are in newtons, k is in newtons per meter, and the units for x are in meters.

So in both of these examples here, whether I've pushed or pulled against the spring, the force that acts in the opposite direction always wants to pull this system back to the center. Therefore, F_{s} is known as a restoring force, and it always opposes the deformation, whether you push or pull. That's everything we need to know about Hooke's law. Let's go ahead and use some examples.