9. Work & Energy
Work By Springs
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Guys. So up until now, all of the works we've calculated had been done by constant forces like applied forces or tensions or even gravity like mg. But not all forces are gonna be constant. Some will be variable. In fact, one of the most common variable forces that you'll see is the spring force, which remember is governed by Hooke's law, and that's just this equation which we've seen before. So in this video, I'm gonna show you how to calculate the work that's done by springs. Let's go ahead and check this out. We're gonna come back and fill the rest of this information out. So if we take a look at the problem here, we're pushing a box attached to a spring. We're gonna push this box to the right with some applied force. And because of that and because of action reaction, there's a spring force that pushes back on us. So we have the spring constant which is k and we also have the deformation. We're compressing the spring at a distance x and we know that's equal to 2 meters. Now in the first part of the problem, we wanna write an expression which means we're just gonna be working with letters here not numbers for the work that's done by our applied force and then the work that's done by the spring force. So how do we do this? Well, if we wanna calculate work, then do we just use f d cosine theta? So if we wanna calculate the work that's done by applied force, we're just gonna use the applied force times d cosine theta here. But there's a problem with this f a because remember that this f a from Hooke's law actually is equal to negative k x. And if you can tell tell from this equation, what happens is this force is not going to be constant because it depends on your deformation. The more you push on the spring, the more the spring force is gonna push back against you. So to kind of show you how this works, I'm gonna go ahead and plot sort of a graph of the strength of the applied force versus the deformation. So what happens? When you're at equilibrium position like this, your force is just equal to 0. And then what happens is the more you push on the spring as you push it to the right, your spring force and your applied force has to increase as well. So with the way that this graph looks is it kinda just looks like a straight diagonal line like this or eventually at some distance x, if I'm just looking at letters here, this f is just gonna be equal to k x. So if this force is constantly changing, what is the value that I actually plug into my FA? And that's the whole point of this video, guys. When you have works done by constant forces, you can always calculate them by using fdcosine theta. But when you have variable forces like the spring force here, then we're gonna have to use the average of the force instead. So we're gonna use f d cosine theta, but we're gonna have to figure out the average of that force. So to figure out the work done by my applied force in a spring, I'm just gonna have to figure out the average of my applied force, then I can use fd cosine theta. So how do I figure out an expression for this average force? I can actually go come back and use use this diagram here, this chart that, that I've been working with here. So if my force constantly increases diagonally like this, then the average is gonna be actually right in the middle. And so basically, the average displacement is gonna be my x over 2, and therefore the average force, my f average, if I'm going from 0 to k x, the halfway point is just gonna be 1 half k x. So this is the value that I plug in for my f a average. So really, the work done is gonna be 1 half k x. Now I have to figure out the distance. Remember the distance and the displacement are the same thing. And really this is just this x variable that I already have. Right? So you're pushing up against the spring some compression x, and that's also the distance that you move this box. Right? So this is my x as well, and then we have cosine. So what's the angle between this force and the dis displacement or distance? Well, because my force points to the right and the distance or displacement also points to the right, then our theta is just equal to 0. So we have cosine of 0, and we know this just evaluates to 1. So basically, what our expression becomes is 1 half kx squared. So that's the first part of this. Remember that the second part of part a is we also wanna calculate the work that's done by the spring. So this is w f s. And it's the same idea here except now we're just gonna calculate the average of the spring force and then use f d cosine theta. But this actually we don't have to repeat this whole process again because remember that your spring force and your applied force are the same magnitude. They just have opposite directions. So So what this means is that the average of my spring force is also equal to 1 half k x. The distance is still x. X. And now the cosine of the angle is gonna be different because the angle between my f s and my d is not gonna be 0 degrees because remember, my f s points to the left and my distance points to the right. So this angle just ends up being a 180. And remember, this just evaluates to negative one. So really, this just becomes negative one half k x squared. So those are our two expressions here. So basically, what we can see is that the work that's done by the spring when you are compressing is gonna be the negative of your applied force. So this in the same way that your Hooke's law, these two forces are just negative of each other, and they have the same value. And so the work that's done by the spring is gonna be negative one half k x squared. This is gonna be the formula that you use in your problems. Alright? So let's take let's finish off this problem here, and now we're actually gonna calculate plugging all the numbers. So the work that's done by our applied force is gonna be 1 half k x squared, which is 1 half times 500 times 2, and then this is gonna be squared. So you end up working out to a 1,000 joules. This makes sense. You get a positive number. Basically, you have to do a 1,000 joules of work. You actually have to put in a 1,000 joules of energy in order to compress the spring. Alright. And so the work that's done by the spring force is just gonna be the opposite of this. So this is gonna be negative 1 half k x squared, and you're just gonna get negative 1,000 joules when you work it all out. So whatever you have for one, answer, the other one is just gonna be the negative of whatever number you get. Alright. So that's it for this one, guys. Let me know if you have any questions.
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