 Hi, I'm Zor. Welcome to Indizor Education. I would like to consider a couple of cases of how to calculate the work. And these cases will be just a bit more complicated than before. So it's this lecture and the next one. In this lecture, we will talk about spring elasticity. That's the work of antelasticity. And the next one will be gravitational force. So what's so peculiar about the force of elasticity about the spring? Well, because as you push it in or stretch it out, the force is variable. Force depends on how much you compress or stretch the spring. And that's why this makes a little bit more complicated case. And that's why I would like to dedicate this particular lecture, which is relatively short actually and simple. But still, it's important to understand how the work can be calculated in case the force is variable. Now, this lecture is part of the course called Physics 14, presented on Unizor.com. I suggest you to watch this lecture from the website because the website contains not only the lectures, but also the textual description basically of everything which is presented in every lecture, which makes the whole course as a combination of visual and textbook-like presentation. And it's a course. I mean, you just have everything which is built into certain logical sequence. Now, there is a prerequisite for this course. It's called Mass 14 on the same website. And I do suggest you to go through the Mass 14 first if you didn't do it yet. Because I'm using basically lots of mathematical apparatus which is presented in that course in this one, in the physics. Especially, it's important for vector algebra and calculus. So these are very much involved in almost everything, whatever we consider in the physics proteins. All right, so back to business. We are talking about the work against the variable force which is presented by the spring. So let's consider you have a spring and you would like to compress it. Let's say it's compressed. So you would like to compress it with certain force, F. If this is the neutral state, you would like to compress it to the length L. So my question is what kind of work this force performs by compressing the string by the length L? Well, let's consider we have elasticity coefficient K for the spring. Now, what is this elasticity coefficient? Well, basically it means that the force which is needed to, well, in this case, let's say, compress. But for the stretching, it's exactly the same thing. It's equal to this K times the distance you are compressing. The more you compress, the stronger the force you should actually apply to hold it in that particular position. So in the beginning, when the spring is in the neutral state, the force is equal to zero. But as we push it in by actually developing certain force, the force should increase because the resistance of the spring increases proportionally to the length by which you have compressed it. Okay. Now, how can we approach this particular thing? Well, I have two different approaches. One is simpler, another is more complex, a little bit more complex, calculation-wise. But it's kind of more traditional for the general approach in physics. The general approach in physics is when we depend on the time. So first, my approach to come up with the formula of what is the work which is supposed to be performed, I will use the time as the main parameter. And for this reason, let me just assume that we are pushing this particular end of the spring with constant speed V. Now, if it's a constant speed, that means that the force is exactly equal to the resistance which spring actually exhorts when it's compressed. Okay. Now, if that's the case, then let's have the parameter s, which is the distance by which we have compressed, in this case it's this one, as a function of t of time, all right? Now, and obviously it's equal to speed times time, right? We start from t equals zero in the neutral state, and then we are pushing the spring with constant velocity V, and that means that the length by which we have compressed the spring is equal to s of t is equal to V times t, speed times time. At that particular moment, we know what is the force, which is supposed to continue compressing our spring. It's equal to k times s of t, which is k times V times t. Okay, fine. Now, we have to calculate the work. Now, we cannot use a generalized formula because f is not constant, f is changing all the time. But what we can do is, we can use a little bit more universal and correct actually formula, which is this. So, on the infinitesimal increment of the s from s to s plus ds, plus differential, it's an infinitesimal increment of s. Now, during this infinitesimal increment, my force is considered to be constant. Now, if I will multiply this constant force by the distance, this infinitesimal distance by which we are compressing at the time t, then I will have a differential, an infinitesimal increment of work, which this force is supposed to perform. Okay, great. This is just pure calculus and very simple one. Because if we know what is s of t, then differential s, this is the derivative, the first derivative, which is V times differential of time. I'm just using elementary calculus in this particular case. If we have a function, linear function in this case of t, then differential of the function is equal to the first derivative, which is V, and times the differential of the parameter, which is time. Great. Now, we can calculate differential of w of t. So, this is an infinitesimal increment from time t to times t plus dt. So, the work is incremented on every infinitesimal interval dt, and this is equal to f of t times differential is equal to this. But now we know what is f of t, right? So, it's k times V times t times V again, so it's V square and times dt. Now, how can we work out the work? How can we find out what is the work, actually, which is performed on the whole interval from 0, from s is equal to 0 to s is equal to capital L? Well, that's an integration, obviously. So, all we have to do is we have to take integral from time t is equal to 0 to the time when we have reached, let's say it's capital T, and it's equal to integral from 0 to t k V square t dt, which is equal to k V square. Now, integral of t is basically what is the function derivative of which is equal to t. That's t square over 2, right? So, it's t square divided by 2. We have to take it into the limits from 0 to t. That's the formula of Newton-Leibniz, right? And it's equal to k V square t square divided by 2. If you will substitute t is equal to capital T, that will be this, and if t is equal to 0, it will be 0. Now, what is this? Well, V is a speed, right? It's a constant speed we are compressing the spring. Capital T is the time it takes to compress it from 0 length 0 to length L. So, V times t is equal to basically L. So, this is equal to V times t is L. So, it's k L square divided by 2. So, this is the final formula for the work. So, this is my work. Now, this is my first variation of the calculating the work, which is performed by the force compressing the spring with the constant speed. Now, what's very remarkable about this? It does not depend on the force F. It does not depend on the speed by which we are compressing this spring, the D. It depends only on the total length of compression. There is one dependency though. We are talking about constant velocity of compressing. Well, is this really a kind of a restriction or a requirement for this formula? Well, actually not. And there are a couple of considerations. The most important is we can always divide this into any number of little intervals. And on each interval we will have exactly this. So, basically there is no difference between variable speed and the total speed. And the constant speed, sorry. But I would like actually to come up with this formula in a little bit, I would say, more natural way. And here it is. Now, F, which is the force, as we know, is equal to K times S. And this is function of T. That's true. Now, the W, which is work, which is also function of T, is equal to F of T, actual differential of the W. F of T times differential of S of T. Now, what actually is much simpler here, forget about dependency on the time. Let's consider the S as our main parameter, everything depends on. Now, if S is the main parameter, let's just forget about what the time actually here is. So, S is parameter and F of S is equal to K times S, right? So, S is a parameter how much we compressed. So, forget about the time. We are talking only about the distance by which we compress. So, this is the main formula. Great. Now, in this case, differential also can be represented as function of S. Well, it's equal to F of S times DS. So, there is no time. S is an independent variable. So, increment of the distance by which we have compressed times the force gives us the increment of the work. Now, all we have to do right now is to integrate it. Now, instead of integrating by time from 0 to capital T, we integrate by lengths from 0 to L. And that what gives us KS DS from 0 to L. Again, integral of S DS is S squared divided by 2. So, it's K S squared divided by 2. So, from 0 to L is equal to K L squared divided by 2. So, I have exactly the same formula, but I think I have derived it simpler. And what's also very important, there is absolutely no consideration about the speed, about time, etc. No matter how I do it, I can do it faster and then slower or I can do it with a uniform speed doesn't really matter. Because if I will take the displacement of the spring as a main parameter, everything much simpler actually leads us to exactly the same formula. And, okay, that's basically the completion of this little exercise. This is the formula of how much work external force should really perform to compress the spring by the lengths L. If the coefficient of elasticity of the spring is K. By the way, for compression, this is formula for compression, but it's exactly the same as stretching. Because the coefficient works in exactly the same way. It doesn't really matter. It's a displacement which is really very important. And obviously it's not for any lengths. It's probably the lengths within the region of elasticity of the spring. I mean, obviously we cannot stretch it more than some physical limitations of the spring. So, within certain interval around the neutral position of the spring, this formula is correct. And that actually completes our lecture about elasticity. So, the next I will consider the gravity. So, thanks very much. I do suggest you to read the textual description of this lecture on unizor.com. It's part of the course Physics 14. It's mechanics, mechanical work. And in there I have a chapter for work and elasticity. Thank you very much and good luck.