 All right, let's look at another example of using integration to calculate work done by a variable force So imagine in this situation we have a We have a spring. Okay, so the there's the classic spring problem So a force of 40 newtons is required to hold a spring That has been stretched from its natural length of 10 centimeters to its length of 15 centimeters Let's talk about that for a second. So what I mean by spring, right? We have this coil and there's some type of tension in the coil that it keeps it at a certain length, right? And so we refer to this as its natural length if no outside forces are Acting on the spring. How long does it want to be? Now in this example? We say that its natural length is 10 centimeters All right, but there is a force being exerted on our spring and so it's been stretched out So it's been elongated because it's been stretched out and So it's been stretched out to this factor of 15 centimeters So it's a little bit long and if you've ever pulled a string out before so it's been stretched here If you've ever stretched out a string a spring of some kind, excuse me You can often feel that there's this force that's pulling the two sides of the string back together It wants to return to its equilibrium state And this how strong that is depends a lot on the spring if you pull a slinky out There's not a lot of force going on there. If you take like a trampoline spring, holy cow There's a lot of force pulling that thing back together If you get bored in class you take apart your pen and start squeezing it and pulling it apart the spring there again You get a different type of force, right? And so according to Hooke's law from physics It turns out the force required to hold a spring stretched Any distance from its natural length is actually going to be the the force is going to be directly Proportional to the length that you've exerbated, right? So we get this force function f of x equals kx where x here represents the distance it stretched beyond its natural length And f of x would be the force going on there So in the situation we have right now We with the information we have we know it's natural like this 10 centimeters It's been to stretch it 15 centimeters. That equates to 40 newtons of force. All right And so we're gonna make it we're gonna find the spring constant doing that now remember with newtons newtons Newtons their basic units is gonna be a kilogram by meter per second squared So if you're using newtons with centimeters, you're gonna have to switch you either have to switch to centi newtons Or you have to switch your centimeters to meters. So that actually I'm just gonna switch centimeters to meters That's gonna be so the easier approach Divide them by a hundred because there's a hundred centimeters in a meter there So you get point one five meters and this is gonna be point one meters So it's naturally this point one And it's it's been stretched to one point five But we have to also keep track of x is measuring the distance we go beyond its natural length So in terms of our x-coordinate, we have a measurement at the x-coordinate zero point zero five Okay, so we go point zero five units past the natural length. That's where this 40 comes into play so by Hooke's law f F of point zero five will equal 40 So that's the observation we have been in the Hooke's law tells us this week will k times point zero five So divide both sides by point zero five. We're gonna figure out that the spring constant What did I say 40? Over point zero five That's gonna turn out to be one eight hundred right and you don't really have to worry about the units of this spring constant because with these type of Direct variation problems you often see in science the units are just for the constant just whatever makes the units work right this distance and the force are related to each other and so we get the spring constant now be aware This is not actually the question that was asked. I haven't even read the question What we've done so far is if we know the force to hold a Spring at some distance from its natural length. We can compute this spring constant. We're gonna need that What we're asked to do is find the work Necessary and stretching the spring from 15 centimeters to 18 centimeters now be aware The first situation was not we weren't moving the spring out We're saying it's already stretched and it took 40 newtons to hold it stretched out But what if it's 15 centimeters? We want to stretch it to 18 centimeters. All right. Well, this is where the This is where integration comes into play here because what we know is that the work the work to stretch it from 15 centimeters or I'm gonna write this in meters still point one five To one point eight here the work to stretch it from point one five. I'm sorry We need to do this from the natural length. This is a common mistake here so 15 centimeters actually is point zero five in our units for X and then the 18 centimeters will be point eight Meters beyond the natural length. So we want to integrate the force function To get work right here and the force functions will be figured out a moment ago What was it it was 800 Times X DX and so what we've seen before is if we can integrate this function with respect to X That'll give us the the work to do and so this this is it's not a variable The stretching out the spring is a variable force because the farther and farther and farther you get past the natural length The harder and harder and harder it is to stretch that spring So we can't just take oh we stretched out three centimeters times by some force or something We have to integrate this thing, but the integral itself is not too hard to do Just by the power rule. We're gonna get four hundred X squared going from point zero five to point zero eight Do make sure your bounds are the right numbers and there's a little bit of arithmetic chore going on here But that's all there's going to be here. We plug in point zero eight and point zero five in for X So we're going to get point zero eight squared minus point zero five squared point zero eight squared That's going to be point six four and then we subtract from it point two five Which was point zero five squared their difference Is just going to be point three nine and if you times that by four hundred you get about 1.56 joules Of work