 So now let's look at the work done by spring forces. Well, our spring force is a variable force. And by Hooke's law, that force was equal to minus kx, where k was the spring constant and x was the distance you were away from your equilibrium point. If I want to find the work done by that spring force, I can do it graphically using the area under the curve. And the curve for the spring force is going to look something like this. Our minus kx, meaning at any particular value of x, the force is minus kx. Well, the area under the curve there is going to be a triangle. And the area of a triangle is 1 half base times height, where my base is x and my height is a minus kx. Rearranging this just a little bit to combine my two x's and bring my minus sign out front, I have minus 1 half kx squared. Now that assumed I was starting from my 0 point. If instead, I've got a situation where I've got a final position, and again, that gives me an area of minus 1 half kx final squared. But I have an initial position that's not at 0. That initial position defines another little area, minus 1 half kx initial squared. And the area I really care about for the work is this little trapezoid that's left over in here, which is the large red triangle minus the small green triangle. And rearranging with our minus signs and putting my positive term first, I've got 1 half kx initial squared minus 1 half kx final squared as my area of the work done by the spring. And I could do something similar using calculus. And in calculus, I'm using the integral of the fdx and possibly the fdy and the fdz if I have forces in three dimensions. In this case, my spring is just in the x direction, so I can use just the integral with respect to x. Plugging in my specific force of the minus kx, I've got this integral here. Now the minus k are constant, so they can pull out front, leaving me with the integral of xdx, which is 1 half x squared. And I still have to evaluate it over my limits. Evaluating it over the limits, I've got minus 1 half kx final squared plus a 1 half kx initial squared. And again, rearranging my terms, I get the exact same thing I saw when I looked at the area graphically. So let's look at applying that force. Let's say I've got a 10 Newton per meter spring, so that's my k value. And it's stretched from 1 meter to 2 meters. Well, my general work equation here is going to be the same one that I looked at from both the calculus and the area. And plugging in my specific values, what I find is that I end up with a minus 15 Newton meters. So what this means is that as I was stretching the spring, the spring force itself was pulling back, resisting that motion. So the work was negative, or the spring force is trying to take energy out of the system. If instead I looked at a spring which was released from 2 meters and it returned to 1 meter, well, that's going to have exactly the same equation, but now I'm going to flip my final position and my initial position. So the 2 meters is my initial, and the 1 meter is my final. And in that case, I'd get a positive 15 Newton's per meter. So as it moves from 2 meters to 1 meters, the spring force is doing positive work, adding some energy into the system. Now, if I did release it from 2 meters, yeah, it's going to go through 1 meters, but it's probably going to go back on to the equilibrium position. But here, we're just looking at one part of the path. So now let's kind of turn this around a little bit and say, what about the work needed from an outside force to stretch the spring from 1 meter to 2 meters? Well, when I was looking at the force of the spring, that was minus kx. But the force I need to apply to stretch it against that force is going to be positive kx. So while the work done by the spring has x initial and x final in those positions, I get the exact opposite when I'm looking at the work done by an applied force, which is going against the spring force. So while I had a negative 15 Newton's per meter done by the spring, I'd have a positive 15 Newton's per meter done by the applied force. I have to add energy in the system to stretch that spring out. The spring at the same time is trying to take energy out. So that kind of wraps up the work done by spring forces and how we can calculate it for various situations.