>> Let's do the following experiment. Let's say we take a block of mass m and we slide it along on a frictionless table and we hit a spring. It compresses that spring at distance x1. But now let's send it in three times as fast. Let's let this be 3v and let's calculate the new distance x3 that it would be compressed. Okay. All right. So we've got two different cases here. Let's tackle the first case. First we just have m at v -- Compressing to x1. Okay. All right. Let's think about energy. When that block is sliding, what sort of energy do we have? Kinetic energy. When it has fully compressed the spring, what sort of energy do we have? Spring. Okay. Now that equation is going to hold for different v's and x's. So for instance, if we go back and change this to 3v and we look at x3 then this equation becomes what? 1/2m times 3v quantity squared equals 1/2k times x3 squared. Okay. And now what we're looking for is x3. So let's take this last equation and let's solve it for x3. Okay. We can do that pretty easily. If I multiply it by 2 on each side, I get rid of those 1/2s. If I divide by k then I get x3 squared equals m over k times 3v quantity squared. And then I want to square out that 3v so I get 9mv squared all over k or x3 equals the square root of 9mv squared all over k. That doesn't look that helpful except some of these things are in this first equation right here. In fact, if I look at this first equation, I can rewrite it as the following: mv squared over k is equal to x1 squared. Right. I just cross out the 1/2s. I divide by a k. And I get this equation right here. And now look, all these terms are right there. And so this becomes the square root of 9 times mv squared over k which is x1 squared. So how much bigger is x3 in this case? It is root 9 of x1. Okay. Kind of cool. All right. Hopefully that's clear. Cheers.