What I'm going to try and do is um first of all, let's talk about work done. Okay? And then what I'm going to do later on is that we're going to talk about the energy and introduce other aspects of the lesson. But let's start with work done. So by definition, what is work done? Okay. Well, we say that work is done. Well, it's the product of a force and a displacement. we usually say in the direction of the force or it's the product of a force and a displacement parallel to the force. Okay. And that is why we usually include this um if we go back to Newton's laws you will understand to some extent what this cos of theta actually means. All right. So um but I'm not going to talk about that today. I just want us to go ahead and and and uh discuss what work done is. All right. Now, you'd appreciate the fact that there are different types of forces means that there are different types of work done. Okay. Now, what I'm going to try and do here is I want us to quickly discuss how to use this because there are quite a lot of misconceptions around this theta here, the angle. uh when it comes to the application. Okay. So essentially we say that uh if let's say for argument sake you are looking for the work done by the applied force. Let's say there's an applied force there's a body with an applied force. Then of course the force that we're going to use here would be the applied force. Okay. Right now I want us to take a quick example with that. Suppose we've got a um let's say 10 kg block. We're going to be pulling this block with a force let's say just for argument sake let's say 40 ntons. Okay. And I tell you we are pulling it across a rough horizontal surface. We're pulling it towards the right. So obviously there would be frictional force in this case. If I tell you that the value of kinetic friction I'm going to denote it as FK there. So that's kinetic friction. If I tell you that kinetic friction is 20 newtons. Okay. And first of all I want us to calculate the work done. So calculate the work done. Okay. So let's say uh first of all uh the work done by friction uh let let's just say by the applied force first of all uh by the applied force okay and then secondly I want us to calculate the work done by friction in some instances they say um the work done to overcome friction okay Um and thirdly uh the work done uh by uh gravitational force. All right. Uh if you want to let's calculate also the work done uh by the normal force. Okay. Uh and then fifthly, let's find out what is the word the net work done. The work done by the net force by the net force. I'm going to refer to this a little bit later on um as the net work done. All right. But uh what I want us to quickly do is to look at all of these forces. I'm just going to use a different color because there's something I want to illustrate to you. Right now if you remember in Newton's laws uh we can draw a free body diagram illustrating all the forces that are acting on the system. Okay. So I want us to do that quickly. So suppose here we are we've got a free body diagram. What are all the forces that are acting? So first of all we've got an applied force. So here is our applied force. That's force applied. And we know that we also have kinetic friction. Okay, it's a force that opposes uh the motion of the object. So this is going to be I'm just going to say that's FK there. And then we know definitely there is gravitational force. Okay, that's force of gravity and then as well as the normal force. All right, this is what we'd expect from Newton's laws. Now, we know what all the forces are. By the way, we also know in this particular case the applied force is greater than the frictional force. And as a result, we know that therefore it means our uh net force will be in the direction of the greater force which is in this case to the right. Okay. So quickly I want us to uh uh look at um how we can uh uh just calculate that. All right. So first of all they said to us we should calculate the work done by the applied force. So we know according to our formula work done is force multiplied by displacement. Oops uh sorry I didn't include displacement there. Okay the cos of theta. So suppose we say that uh our block has been displaced. Let's say uh just for argument sake let's just say it's been displaced for 5 m. Okay. So let's say it moved from point A to point B. It's been displaced 5 m to the right. Okay. So we want to calculate that work done by the applied force. So what I'm going to do is I'm going to put some subscripts there so that I know which uh work done I I'm actually trying to get. So the work done by the applied force which force am I going to use there? It means that I'm going to use the applied force. Okay. So do we know the magnitude of our force? Now the beautiful thing about energy is that energy is a scalar quantity. So we don't concern ourselves too much about direction. However, we know that force does have direction. Okay. So in this particular case, I don't want you when you substitute here to worry about whether a force is positive or negative. Just all we are going to focus on is essentially the the magnitude of that force. You'll see how the direction works itself out when we are using this. Now so how do we use this? We say the applied force, what is it? That's going to be 40 multiplied by the displacement. What is our displacement? It's been displaced 5 m. Okay, so that's multiplied by five. Now, notice the cost of. Now, here's the part I want us to actually kind of focus on. How do we get the angle theta? The NL theta will always be the angle between the direction of motion. Okay, so where is this block going? It's going to the right. The direction of motion and the direction of the force in question. So in this case, look at this. Where's the direction of motion? Our object is moving to the right. Okay? Right? If I can just uh take these two pens uh to just illustrate. Where's the if we take this red one as the direction of motion. So we're saying the direction of motion is in that direction. Okay. But where is the direction of the force that is in question? Where is it going? Okay. So we're saying both of them are heading in the same direction. Now in this case when we refer to the angle theta we say it is the angle between direction of motion and the direction of the force in question. Now if I may to I were to ask you what is the direction of I mean what is the angle between these two at the moment? They're both going in the same direction. So the angle between them is actually zero. there's nothing or there's no uh uh there's no gap between them. So both of them are actually in the same direction. So we will say in this particular case it's the cos of 0°. Okay. Why? Because our applied force is in the same direction as the direction of the motion. So in this case that's uh what we are going to do. So we'll say 40 * 5. Okay that's 200. the cos of zero. Okay, that is a a one. All right, so you'll see that if we said 40 multiplied by 5 multiplied by the cos of 0, but we know cos of 0 is actually equal to one. And so that's why uh our answer is still zero. I mean uh 200 rather uh in this case. So remember that work done is energy and energy is always measured in jewels. Okay. So we know that the work done by the applied force is equal to 200 jewles. Okay. Right. Now let's go to the second one very quickly. Right. I want you the the the whole crux of it is that I want you to learn how to use uh that theta that angle theta there. And I'm going to use different scenarios to try and illustrate that to you. All right. Now let's talk about it very quickly. So now we want the work done by friction. Okay. So now we want the work done by friction. Okay. So work done by friction. Okay. Is equal to now which force are we going to use here? It's you've guessed right. It's going to be the frictional force. Right? So that's force of friction. um multiplied by delta x the displacement the cos of angl theta. Now please let's try that again. So now do we know the value of our kinetic friction? Absolutely. We said that's 20 and you don't need to worry yourself about its direction at the moment. Right? All that you are concerned about is it its magnitude. That's what we substitute there. So that's 20 multiplied by what is our displacement? it's five. Okay. Now, please don't concern yourself about uh the direction of each. All of all that we're concerned about is the magnitude. So, we say 20 * 5. Now, let's get to the cost of the angle. Now, once again, we are simply going to say here are my pens. Again, let's use the the the the two pointed sides as our direction. We're saying the direction of motion is to the right. Okay. But what is the direction of the frictional force? The direction of the frictional force is in the opposite direction to the direction of motion. So look at that. So we say theta is the angle between direction of motion and the direction of the force in question. So what's the direction of our force in question? It's in the opposite direction. So what is the angle between the two as we look at them here? So you guessed right that's going to be 180. Remember in the first instance we said the direction between the force okay and I mean the displacement and the force they were both in the same direction. So as a result the angle between them was zero. But this time here's the direction of the uh uh um uh direction of motion and here's the direction of our force. So in this case that's going to be 180. So then we say okay right so if you take that and put it in your calculator quickly right cos of 180 we know that that's -1. So you're saying 20 * 5 that's 100 multiplied by cos of 180 which is -1. Just a warning please make it a point that your your calculator is always in degrees. Okay. So, uh you must see that d there otherwise uh it will cause us a little bit of problems if it's in radians. Okay. So, that's 20 * 5 that's cos of 180. So, therefore we get minus 100 jewles. Now, you might ask me a question and say well but um what does it mean when I get work done to be negative? Well, it doesn't mean first of all uh what it does not mean is that it does not show us direction. Remember that work done is a scalar quantity. Okay. But what it's actually showing us in relation to this is that this is the amount of energy that has been lost in the system. Now if you think about it here this force actually did work done that is positive in a sense that it's energy that is put in the system whereas here with negative 100 it's simply telling us that it is energy that is lost if you think about it friction does cause heat. So that is energy that has been lost uh in your system there. All right so let's go to the third one. Okay, so just keep in mind there whenever we have friction and by the way whenever we have friction we know that it's going to give us that negative work done because friction always opposes motion. Okay, right let's talk about work done by gravit uh gravitational force. That's the third one that we said we're going to do. So work done by gravity. Okay, so we say this is the force of gravity. Okay, I'm just going to use FG there multiplied by delta X the cos of angle theta. Now have a look at this once again. So we know we're going to use force of gravity, right? I know we didn't calculate it there. If you remember from Newton's laws, we know whenever we calculate gravity, we always say mass time gravitational acceleration. So this gives us that's going to be 10 * 9a 8. That gives us 98 ntons. Please remember 9a 8 is the um uh gravitational acceleration um on the surface of the earth. Okay. Right. So in this case we simply say well what is our gravitational force? That's 98 multiplied by our displacement is five. Let's go to the angle. Now what is our angle in this case? Now look at this once again. If we use that is the direction of motion. Okay. Where is the direction of the force? All right. So there's the direction of the force. So look at this. What would be the angle between them? Direction of motion and force in question. Well, you can say that's 90°. But you know to be quite accurate um think about it. We started with zero there and we said it's 180 on the other side. So if you you follow the cartisian plane it means that you started with zero that's 90 there okay that's 180 so this has to be 270 I hope that makes sense to you okay you can also use minus 90 okay because you're going in the opposite direction of your normal uh uh um uh revolution or your cartisian plane so in this case you can simply say our direction is 270. Either way, whether you had used minus 90 or 270, you'll realize that the work done there gives you 0 jewles. Now, let's discuss this a little bit. Why would gravity? We know that it exists. It is a force. It is actually exerting uh or or working on this particular block. But why does it give us zero work h done? Now I want you to think about it. Gravity in this case contributes nothing in that dimension in the horizontal dimension that even though it was pulling this system down but it actually did not contribute anything in causing it to be displaced horizontally. And so as a result because there was no displacement caused by gravity that is why the amount of work that it has done is actually zero. So we know that work done by gravity would be zero jewels. Now let's talk about the normal force quickly. I'm sure you're getting the hang of it by now. Okay. Right. So um now what do I say? Work done by the normal force. We know our normal force always acts perpendicular to the uh surface. Okay, in this particular case because you don't have any other horizontal I mean vertical forces we know that this uh block would not be displaced anywhere vertically. So what does that mean? It means that these two are actually balanced. All right, it means that your gravitational force is actually equal to your normal force in that case which also means that this would be 98 newtons. As I said to you, please don't worry yourself about uh uh whether positive or negative. So the work done by normal force that would be a normal force. So you can just use n multiplied by the displacement the cos of angle theta right a normal force we already know that's 98 newtons okay multiplied by the displacement which is five okay now once again let's do that again so direction of motion what is the angle between direction of motion and the force in question can you see that right here's our direction of motion. Here's the force in question. What's the angle between them? You guessed it right. It has to be 90°. And what you will also realize is that this will also give us 0 Jew. You can actually use your calculator. But we know cos of 90 is actually zero which makes this whole thing to be 0°. Now I want you to please note this once again. The same explanation that we had for gravitational h force would be the same explanation for the normal force. Even though uh the ground is exerting a force upwards but this object would not be displaced upwards and as a result uh uh the normal force does no work. Okay. Right now I want us to take the last one. Just going to make some space this side. Okay. Right. Let me pull this uh to the side here. Right. So um now number five we said we want to calculate the work done by the net force. All right. So in this case I want you to please note there are two ways in which I can do this. All right. If I remember from Newton's second law I want you to note we're looking at the horizontal motion. So we're going to actually negate those two forces because um in this case we are just focused on the horizontal motion of the object. So what are the two forces that are acting horizontally? It's going to be our applied force and frictional force. So if we wanted to find out what our net force is, okay, considering that force is an a vector quantity, right? So we know that you would probably need to choose a positive direction. Let's choose right as positive. So we know it would be what? Force applied plus our frictional force. But our friction is acting in the negative direction, isn't it? So think about this. This would be 40. All right. Minus that 20 newtons there. And so what are you left with for your net force? It would be 20 newtons. Now think about this where what is the direction of your net force? Your net force would also be to the right which would be in the positive direction. So you can calculate net force this way. You can say well net force I mean net work rather okay would be equal to fnet multiplied by the displacement the cos of angle theta. Sorry I have to squeeze for space there uh a little. Right. So the network what is our net force the value of our net force that's 20 multiplied by delta x we know that our displacement is 5 m okay by the way our displacement has to always be meters multiplied by the co now once again what is the direction of motion I mean what is the angle rather between the direction of motion and the force in question we said the force in question here it's the net force So where's our net force going? It's actually going in the same direction as our our displacement. So as a result that will also be zero. I'm sure you are getting the hang of this. So what does that give us? It will give us 100 jewles. Now just um you know I always kind of want to uh uh think uh ahead because we're going to need this as we go along. All right. So what I want us to do quickly is I want to show you another way in which we could have calculated that network done. Okay. Right. So what I could have done is if I said to you that the net force is the sum of forces then it would actually still hold as well that the net work done is the sum of work done. Okay. Right. So in this case it would be now think about uh in this case what what would it be? It would be the work done by the applied force. Okay. Plus the work done by friction plus the work done by gravity plus sorry because I'm squeezing for space there. Work done by the normal uh the normal force. And please remember in this particular case um I'm not too concerned about h direction because it is scalar quantities. So in this case look at this. We know both of these are zero isn't it? the normal force, the work done by the normal force, the work done by gravity. But remember, we got a value for the work done by the applied force. Okay, what was that? This was 200 newtons uh uh jewels, sorry, uh 200 juwles plus in this case the work done by friction that's 100 jewles. But remember that that was work done h that was work that was lost in our system. So that's negative work done. Okay. So in this case 200 minus 100 and guess what? We get to the very same answer as we did before uh using that method. I hope that makes sense. All right. So what we'll do is we'll move on to the next thing. Right? I just want that to linger a little bit so that uh if you want to pause it and just reflect and think about it, there it is in front of us. Okay. Right. So um I thought I should just uh include this question uh for now because I know that there are some issues um when it comes to uh inclined plane whether you are looking at it from the side of Newton's laws or work energy and power uh you know this is where we really experience tend to experience problems uh when it comes to the application of uh work done. So what I want us to include is this here. Right? And I want you to please follow me. Okay? Right? So here it is. We've got an inclined plane. All right? So now on this incin plane we are applying a force of 80 newtons. All right. Suppose let's just say just for argument sake we've got a frictional force. I'm just going to include that there. Let's say we've got a frictional force. let's say of uh 12 newtons okay acting on the block all right and um we are pulling this up an inclined plane right now I want us to calculate the different types of work done on this particular system so the first thing I want you to actually learn to do is please all the time start with your free body diagram okay start with your free body diagram because it will be a an actual guideline as to what to do. Okay. Right now, what I'm going to do is I'm going to draw it over here. Right. So, let's draw our free body diagram. So, what are the forces that are acting here? So, I actually t like to actually show the incline on a dotted line so that you can uh be able to see the forces acting clearly. All right. Now, first of all, what forces do we have? We've got the applied force that is acting parallel to our incline. Can you see that those two lines are parallel? So, this is going to be a force applied. And then uh we also have um another force there which is friction. Okay, we did say this would be a rough horizontal I mean a incline plane. So we know we've got a frictional force there. Uh so that's friction acting also parallel to the incline. Now think about it. Friction always opposes motion. If I'm pulling up the incline, okay, friction is acting down uh that incline plane there. All right. What other forces acting? All right. So another force that is acting is our normal force. Now please I want you to note this clearly because that's this is another area where we experience problems. Please note that the normal force must always be perpendicular to the incline plane uh uh to the plane uh to the surface. So in this case perpendicular simply means that it is at 90°. Okay. Now please note all right our plane is kind of skewed okay 30° in this case so uh the perpendicular cannot be straight up like that so our perpendicular has to also be skewed so that it makes a right angle with that 90° h I think that is uh you can see that clearly all right so this is our normal force okay And then what other force do we have? We've got the gravitational force. Now please I want you to also note gravity always works or acts vertically downwards. Okay, vertically downwards. So think about it. Uh this would be horizontal. So in this case vertical would be in that direction. So notice the normal force and the force of gravity this time are not acting in a straight line. Can you see that? Okay. They are not actually h in a linear to each other. They don't form a linear pattern. Okay. So in this case you can see that force of gravity there is vertically downwards and the normal force is perpendicular to the incline. Now this is where this becomes very interesting and by the way all the time your mark allocations when they ask you to draw a free body diagram your mark allocations will always tell you how many forces you need to include. Okay. Right. So typically this would would have been four marks. Okay. Right now in this case I want us to do something quite peculiar because we're going to need it. Okay. for the videos or or or for for the questions going forward. Okay, we might not do it in this particular video, but uh we're going to actually see that as we proceed. Now, look at this. I can actually redraw, okay, this drawing. And it's up to you really which drawing or or which free body diagram you'd prefer. And I'll show you why I'm I'm I'm drawing the other one. Okay. So, nothing changes. We know we've got force applied. Okay. We've got frictional force there. There's a friction acting downwards. Okay. We've got the normal force acting in that direction. Okay. Remember, we said it's at 90° and to our incline. I know that that apploid force is pep is pep is parallel rather. So therefore it has to be at 90° to that. So uh there are the forces. Now the problem with this sto with this drawing is that what it does is that your forces are not necessarily in the same dimension if I may put it as such. Okay. So what we want to do is let's rather get this guy to comply and be in the same dimensions as the other two guys. So what normally tends to happen this is what um uh this would be kind of a revision of what we did in Newton's laws. Okay. So in this case when we have something which is on an incline plane we know that we can now break gravity into components. Okay. So I'm messing up that uh drawing. And by the way when I ask you to draw this please don't put the components together with force of gravity in the same drawing. Okay. So it's either you'll draw a drawing with without those. Okay. or you will draw the one that I'm going to show you in in just a little while. Okay. So, what happens there? We know we've got force of gravity. Here it is. There. All right. Now, these two components, this component of gravity is actually acting perpendicular. So, I'm going to call it FG perpendicular. All right. And here's another component of gravity which is parallel. FG parallel. Now both these components acting together. Okay. In terms of vector sums, they will give us that h um uh that force of gravity there. So it means this would be the result of these acting to uh these acting together. So what I'm going to do is I'm going to draw them there. So there's our FG perpendicular and I'm going to state something very important about them. Okay, FG that's the perpendicular component of gravity. Now notice I want to include the force of gravity parallel in my drawing but I can't just have it lingering somewhere there. So what I'm going to do is I'm not changing anything about it. I'm just moving it to that dot there so that you can see that it's a force that is acting on that dot there. So here's my force of gravity parallel. Okay. Now one thing also that you need to note is that to calculate that uh those two okay please I won't go h uh through it as you know much as I would like to but uh because this would be kind of revision in a sense. Now to get those FG parallel and FG perpendicular, what we then use is that angle there. Now this is where our learners tend to make a lot of errors. Right? This angle here, we will only use it to calculate those components of gravity. I'm going to show you that because I want to make it easier. You know, there is another way of calculating this, but I tend to think it uh it it just kind of complicates things. Okay. So now let's talk about FG uh parallel. We say to calculate FG parallel, you simply are going to say it's going to be the force of gravity multiplied by the sign. Now the reason why it's the sign. Now look at this. This angle here would be the same angle theta as there. Okay? So the angle there is simply the same angle there. All right? uh I don't want to go through the mathematics of it right so in this case because those two are equal so the force of gravity is actually opposite that so that's opposite over hypotenuse so as a result we end up with fg parallel is simply force of gravity multiplied by the s of theta please if you didn't understand what I just said right now please don't worry too much about it all I want you to remember is simply remember this Okay. Right. The same thing for our FG perpendicular. Right. Our perpendicular component of gravity. This is going to be the force of gravity multiplied by the cos of the angle theta. Now the confusion arises here because we we actually are using the same symbol theta. You know, if I had it my way, I'd actually use a different symbol. Okay, because then it tends learners tend to think that uh because we're talking about theta here then they can just simply use h even when we talk about work done then they simply use the same angle there all right I'm going to emphasize this again please just note okay right please just note here we are talking about the angle that is actually the incline angle okay right please remember fg is equals to m * g right so that's m * g the s of uh sorry that's the s of theta here that's mg the cos of theta right now let's just do that uh calculate these uh these two so if you wanted to get that fg parallel this will give you the mass is 10 * 9a 8 * the sign now what is our angle between the horizontal. Okay. And that incline there. What is our angle? Our angle is 30°. Okay. Right. If you note that there, that will give us uh that's 10 * 9a 8 uh time s of uh 30. Okay, that gives us 49. So we know our FG parallel is 49 mton and FG perpendicular is 10 * 9A 8 the cos of 30° okay 10 * 9.8 8 time the cos of 30° and that is what we get that gives us 84.87 newtons. All right. Now what do we do in that particular case? We simply are going to say to ourselves all right now let's calculate what our right. So we've just concluded with these uh FG parallel and FG perpendicular. This had nothing to do by the way with the question. Okay, this had everything to do uh with just what we need to do in preparation for what for for what we're about to answer. Now let's just calculate our uh our forces our work done rather. So how do we get the work done by the applied force? So we're simply going to say all right I'm going to do it over here. The work done by the applied force. This is equals to force applied times delta x the cos of uh theta. Oops. You see there again I forgot to put the displacement. Let's just say for argument sake our block has been displaced ah um let's just say uh 5 mters. Let's stick with 5 mters. say there's been displaced between A and B okay for 5 mters going up the incline so now look at this once again let's see how to apply this please don't make that error okay right that's why I'm teaching you this so what is our force applied our force applied is 80 ntons what is our displacement displacement is five okay the cost of now please let's do that again. So we said how do you get that angle? We said it's the angle between the direction of motion. Where would the block be going? If you notice our 80 Newton force is much bigger than the frictional force. It's much bigger than the component of gravity parallel the the parallel component of gravity. So definitely our block would be moving up. So now think about this. What would be the direction between what would be the angle between direction of motion? Here's our block moving up and the force in question. So there it is. Our applied force is actually acting upwards as well in the same direction as the motion. So look at this. What's the angle between them? All right. Please don't be confused. We're not talking about the incline angle. We are talking about the angle between direction of motion and the force in question. And what is that angle in this particular case? You guessed right. It is zero. So the angle there would be angle zero. So all that we need to do simply going to say well that's 80 * 5 the cos of uh 0. All right I know you already have the answer in front of you. That's going to be 400 juwles. So that is the work done by our applied force. Let's go to the second one. All right, very quickly. Right. So what is the work done by friction? So I'm just going to denote it as FK there. So that's going to be friction multiplied by delta x the cos of theta. All right. So once again, what is our frictional force? All right. We said the value of our friction. Let's take that to be 12 multiplied by the displacement. Okay, which is five the cos of right let's note the angle once again what is our direction of motion direction of motion is upwards and what is the direction of the force in question right note this friction is opposing motion the uh the object is going upwards and friction is going down the incline so what is that angle there it has to be you guessed it right it's 180 okay so in this case we're going to simply Say this is 12 * 5 okay multiplied by the cost of 180 and you'll see that that gives us negative work done and that's -60 JW okay right number three so what we want to find out now is the work done by gravity okay right now there are two ways of actually doing this but I actually like to make things much easier for myself. Okay, if you go with that first diagram, what you would need to do is find out what is the angle between that direction of motion, right? And the force in question. Okay? Right now, please uh uh because I don't like to make it complicated. I can show you this in so many other ways. All right? But what I want us to do is now you remember that gravity has two components. Okay? Right? Now note this. If you used the perpendicular component of gravity. All right? What would you have? Okay. I want to show you this for the first one. Okay? And then you'll see why I make the decision that I I make all the time. So the perpendicular component of gravity it's fg perpendicular delta x the cos of angle theta. Now note this what is my perpendicular uh um uh uh component? That's going to be 84a 87. My displacement is 5. What is the angle? All right. So direction of motion there it is there. and the force in question. You still remember that FG perpendicular. Those two are actually at 90°. That's why we said it's perpendicular. It's at 90° to the incline. So in this case, what would it be? If you remember my cartisian plane, right? I took it from that direction. So in that case, it means that it's actually 270. and you'll get a work done of zero with the gravity perpendicular component of gravity. But remember we also have now a FG parallel. So with FG parallel so work done by gravity parallel this is going to be FG parallel delta x the cos of theta. Note this we found that to be 49 multiplied by 5 * the cos of angle theta. Now note this here is our parallel component. You can see it's in the same direction as friction. So there's our direction of motion and our parallel component is actually opposing that it's going down. So therefore it's going to be 180. So in this case all right. So that's going to be 49 * 5 cos of 180 cos 180 and then we get - 245 juwles. Now what I would advise is that when they ask you h in the case of an incline when they asking for the work done by gravity you might as well just forget about that perpendicular component. Why? Because it will give you a work done of zero. So it means the work done by gravity is actually now being done by the parallel component of gravity. And then uh uh almost the last one. So now we want to find out what is the work done by the normal force. I'm sure you can you you kind of guessed it by now. So that's going to be our force normal multiplied by the displacement multiplied by the cos of theta. What is our normal force? Now in this case, notice this that your force of gravity perpendicular and your normal force are actually in the same plane or you can say they're in the same dimension so to speak. So what's happening? This is acting downwards that's acting upwards. So they have the same magnitude. They are actually opposing each other. Okay. So in this case our normal force would be the same as the uh u um FG perpendicular. So that's going to be 84.87 87 times our displacement which is 5 the course of now please I want you to note once again direction of motion there it is upwards normal force we said it's normal it's actually perpendicular to the uh uh to the um to the surface so that must be at 90° so this would be 90 and so that is why you'll get zero jewels of work it contributes absolutely nothing to the motion of the object. Now, very lastly, uh I know I didn't include this in as part of my questions. Remember we said you must now calculate the network done. Okay, so what we could do is to simply say it's going to be fnet multiplied by delta x the co of angle theta. Now let's go back to our drawing. What is our net uh force? Okay, we know these two uh oppose each other. We are focusing now on the forces that are parallel to the incline. Right? These oppose each other. So that's going to be that applied force minus gravity minus the frictional force. Right? So um don't have a lot of space here. Um so what I'm going to do is I'm going to write it over here. Right? So it means that if I wanted to get that net force, it's going to be F applied plus minus F gravity parallel plus a negative. Why am I saying it's negative? I'm actually choosing direction upwards as positive. Rule of thumb is that choose the direction of motion as positive so that it always works out uh much better for you. So uh minus force uh kinetic friction. So what is our applied force? That was 80. That's - 49. That's - U 12. Is that the case with that? Yes. So that's 80 minus that 49 - 12 and we get a force of 19 newtons. That's our net h our net force. So where would that net force be acting? Note once again the biggest force is actually the applied force. So it means that your net force would be in that direction as well uh uh um when you compare it to those two. Okay. So what are we saying? Our net force is simply going to be 19 multiplied by 5 the co. Now once again notice what is the angle between direction of motion and the force in question both of them going upwards. So that is going to be uh 0° there. So as a result that's going to be 19 * 5 and that would give us 95 jewels of work done. Note once again just the last thing that I want us to note right as much as we said Fnet is the sum of forces the net work done would also be the sum of work done. Okay. So if I if I said to you okay so it means that it's going to be the work done by the applied force plus the work done by friction plus the work done by gravity. Okay. Right. Uh you can include normal if you want to but uh you can already see that uh uh it's not going to actually contribute. Notice uh I'm not going to calculate it to make an actual calculation there. But look at this. Work done by uh applied force that's 400. Okay. But remember work done by friction was actually -60, isn't it? So that's -60. All right. Work done by gravity parallel that's also minus 245. Okay. And then plus zero if you want to. Okay. And what do you get? You get that same value of 95 jewles. So you can either take the net force or you can simply take the sum of the work done. Um I'm just going to leave that there so that you can see it. All right. And um I hope that kind of makes sense for you. We're going to explore this even more because in our next session we are going to be talking about the principle of conservation of mechanical energy. I want us to explore that uh as we explore the subject of work energy and power. Hope that's been worthwhile. Please remember you can email me if you didn't understand anything. Please just make a comment and then I will try to actually explain that better in our next session. Thank you.