As I said, one hard part of keeping up with these energy-work theorems is the negative signs. So we'll test the question, can you have negative work? Yes. You can because Delta K, the change in kinetic energy, can be negative just not K. Kinetic energy is always positive or zero. But, of course, you can have a change in kinetic energy. So to help us think about this one, and let's say a mass at V collides with a spring. Let's think about what happens. What is the maximum deflection? So we can draw that real quick. Let's see. So we're going to think about it this way. We'll have an x-axis like that. Here's a wall that the spring is attached to, and here is its natural resting position at zero of the x-axis. The mass is going to come flying in like this at velocity V. So you can imagine what's going to happen, is the spring will push, and it'll bring the mass to a stop. That's the question, is how far does it go in to bring the mass to a stop? If you have a head for conservation of energy, and you've done these before, you know that you just set one-half mv squared equals to one-half k deflection squared, and you're done. But let's think about it carefully in terms of systems, and external and internal, and positive and negative, and see if we can do it in careful detail. To do it in careful detail, we have to define the system. In this case, the system will just be the mass. So the system is the mass, therefore, the force of the spring is external. So we've set it up that way. Let's see if everything makes sense according to the work-kinetic energy theorem. So external work equals Delta K, the change in kinetic energy. So what would be the external work in this case? Well, it is going to be the integral of F.dr, or the force dotted with the displacement. So as this thing pushes into the spring, the force is negative x. So the force is -kx, and that is times the displacement, is in the positive x-direction dx for the integral. I didn't do the full vector parts. But if we'd thought of the force as a vector this way, and the displacement is a vector that way, they're opposite directions. That's why it comes out with a negative sign. So -kx dx is that integral, and we're going to integrate from zero to, let's just call it, big D, the displacement zero to D. That's how far it gets. So that's one side. The other side is Delta K. That must be the kinetic energy minus the kinetic energy initial, final minus initial of the system. The system being just the mass. The mass is kinetic energy. Well, let's do our integral. This is minus one half kx squared evaluated from zero to big D, and this is, let's think about it. It's started out, the final kinetic energy is zero. We're figuring how far it deflects before it stops, so that's zero. Then, minus the initial kinetic energy, is what it had when it first hit the spring, and that's just one-half mv squared. So when we evaluate our limits here, this is minus one half kd squared. Minus one-half kd squared equals minus one-half mv squared. You can see the minuses become pluses, and you can see this is just that simple idea. I'll just take the kinetic energy and equate it to the energy that gets dumped into the spring. But here, we've done it, I'd say, a little more carefully. We thought carefully about our system. Everything makes sense. But notice the work was negative. You do negative work when you push on something, but it moves against the push. Before I did positive work when I pushed the mass because I pushed along the direction that it moved, here the spring is catching, and the spring is doing negative work. You do negative work because you have a lowering and negative in your change in kinetic energy. The thing lost kinetic energy. So the energy or the work-kinetic energy theorem part makes sense. I guess we should finish this. The one-halves go away, and this comes over here. No, I'm sorry. Xd is what we're solving for. The D, the displacement, becomes the square root of m over k times v definition. So there's the answer. I can also show you negative work by pushing the cart. So you won't believe how good I am at this. So I'm going to have the cart come. It's going to come and I'm going to use my finger to apply negative work. I'm going to push on it. I'm going to push this way, the cart is going to displace this way. It'll bring it exactly to zero. Practical effects.