 Fine. So again, we will, so whatever we have learned till now, let us try to use it and try to analyze a few scenarios and see how we can solve the questions based on whatever we have learned. Okay. All of you please write down. See, many of you might be doing this chapter for the first time and some of you might have already some knowledge about this chapter. Okay. So those who have some knowledge about this chapter already will obviously be, you know, in a hurry and they will be answering the question first. But that doesn't mean that a few of you who are learning your first time will be lagging behind all the time. Okay. So don't get influenced by, you know, somebody answering the question very quickly. You are learning from scratch and I'm also teaching here from the scratch. So focus on the learning. Okay. Don't skip steps or anything. So you are applying, suppose a force of two Newton on this block, which is 2 kg. Okay. Now, when you're applying this force on this block where this surface is smooth, this block will move. Right. So if this block moves by a distance, by a distance of let's say 10 meters, you need to find out work done by all the forces one by one and then find out what total work done on this 2 kg block work done by individual forces and then total work done. Okay. What is the work done by 2 Newton force? It's simple, right? It's an example of zero angle between force and displacement. The force is 2 Newton and displacement is 10 meters. Work done will be just 2 into 10. That is 20 joules. The SI unit of the work done is joules. Okay. Any other force acting on this block, if you see there are two other forces. One is mg and other is normal reaction. But since displacement is like this, the angle made by normal reaction and mg both are 90 degrees. Now do this particular question. This is question number 1. Now we are talking about question number 2. There's a block. You are applying a force of 2 Newton on this block of 10 kgs. Okay. Coefficient of friction between the block and the surface is 0.3. So mu is 0.3. All right. Now you need to find out work done by all the forces in 2 seconds. In 2 seconds, how much is the work done by all the forces individually and collectively? Will this block move? This block won't even move. Displacement will be 0 here. Why? Because if you see that, downward force will be mg. So this is 100 Newton. Okay. And there will be friction over here and you are applying a force of 2 Newton. Okay. So normal reaction is equal to 100 Newton only, which is mg. Fine. So the maximum value of friction force possible is 0.3 into normal reaction, which is 100, right? Which comes out to be 30 Newton. Okay. So if you apply force that is less than 30 Newton, the object won't even move. So if object doesn't move, the displacement will be 0 and work done by all the forces will be 0. Okay. Now let us say that you are applying a force of 50 Newton. Now find out what is the work done by all the forces in 2 seconds. Okay. What are the forces are there? You have a force of 50 Newton. So you need to find work done by 50 Newton. Then you have friction. So work done by friction. You have gravity force. Work done by mg is what? And work done by normal reaction. So all these 4 you have to find out. Once you are done, please type in that you are done. You need to find work done by all the forces in 2 seconds. Initially the block was at rest. Okay. See it's very easy to find the work done by gravity. Why? Because the displacement will be always perpendicular to the gravity. This will always be 90 degree. When the object moves, it will move in a forward direction and gravity is downward and normal reaction is upward. So the angle made by the displacement with these 2 forces will be 90 degree. So work done by gravity is 0 and work done by normal reaction is also 0. Okay. Now we need to first find out what is the displacement? Okay. Displacement directly is not given. But what is given? The applied force is given and the value of friction is given. So you can see that if you draw free buried diagram here, you have 50 Newton applied like this and then there will be friction force that is applied. How much will be the friction force? It is the maximum possible force, which is 30 Newton as it is sliding. So mu times normal reaction, which is mg is friction force. Okay. Then you have mg force, which is 100, 10 into G, 100 Newtons and this is normal reaction. Fine. So this is a mass of 10 kg. Net force horizontally is what? 10, sorry, 50 minus 30. This is a net force horizontally. This force should be equal to mass time acceleration. Right? So you get acceleration as 2 meters per second square. All right? So this is the acceleration and once you get acceleration, you can find the displacement, which is equal to half a t square assuming initial velocity to be 0. So this is what? Half a, which is 2 into t square, which is 4. 2 square is 4. So displacement is 4 meters. Okay. Now displacement is given. So you can find out the work done by all the forces individually one by one. Fine. So work done by 50 Newton, this one will be what? 50 into the displacement, which is 4. So that is 200. Okay. Work done by friction is what? 30 into 4. But then there will be minus sign over here because the angle made by friction force is 180 degree with the displacement. Okay. So you get minus 120 Newton, 120 joules. Minus 120 joules work done by friction and 200 joules is work done by 50 Newton force. Okay. So this is how you deal with this kind of scenario. If you have any doubts, please type in. Fine. We will take a few more scenarios. Then we will proceed for next concept. So suppose draw the inclined plane. This is inclined plane and then you are placing an object. Okay. This object has a mass of let us say 5 kgs. Okay. 5 kgs is its mass. This angle is, let me take 37 degrees. Why I am taking 37 degrees? Because sign of 37 degrees is 3 by 5 and cos of 37 is 4 by 5. Okay. Now, this distance is given as is 4 meters. Okay. You need to find out how much is the work done by all the forces when this block starting from this point reaches to the lower most point. So point number one, it goes to point number two. How much is the work done by all the forces acting on the block? There is a friction whose coefficient of friction is 0.1. Okay. Try to find out work done by all the forces. So let me here mention all of it as in you need to find work done by gravity, work done by normal reaction, work done by friction. That's it. Only three forces are applied over here. How much? Tell me what is the work done by gravity first? How much it is? Work done by gravity. This you can find very easily. Work done by gravity is 30. That's not correct. 200. Niranjan is saying 200. See, gravity force always acts downward. Fine. So it's a constant force that is acting downward. Whose magnitude is what? 5 into g, which is around 10. So this is 50 Newton. So 50 Newton is the gravity force that is acting down. And along the direction of this force, how much is that displacement? 4 Newton. When it comes down till here, along the direction of force, which is downward, the displacement is 4. So work done by mg force will be gravity force, which is 50 into 4. This is 200 Newton, sorry, 200 joules. 200 joules is the work done by gravity. I hope all of you have understood this. Another way to find the work done by gravity is to resolve the gravity force into two components. One is mg sin theta, mg sin theta along the incline and mg cos theta perpendicular to incline. So work done by mg sin theta is mg sin theta into this distance, which is what? 4 divided by sin theta. So it ultimately comes out to be mg into 4 only, which is 200 joule. And work done by mg cos theta is 0 because perpendicular to the incline, the displacement is 0. So this work done will be 0. So work done by mg, you can find out directly like this or you can take components, find out their individual work done and add it up. Now work done by normal direction, how much it is? It should be 0, isn't it? Because normal force is acting perpendicular to displacement all the way. The displacement is in this direction, along the incline. This is the displacement. And normal force is acting perpendicular to it. So the work done by the normal force will be 0. But then there is another force, friction. And as it is sliding, friction force should be equal to mu times normal reaction. And I think you have done this many times that normal reaction will come out to be mg cos theta. Fine. So mu is 0.1, m is 5, g is 10 and cos theta 4 by 5. This will come out to be 4 Newtons. 4 Newton is frictional force. So the work done by the friction is 4 Newton, which is backside of the displacement. That you should remember. This into 4 divided by sin theta, which is this length. This length is 4 by sin theta. That is a displacement. 4 divided by sin theta and sin theta is 3 by 5. So this is the work done by the friction, negative of that. This will come out to be equal to 4 by 16, 16 into 5. That is, I think 80. So 80 by 3 joules is a work done by the friction. So this is how you deal with the constant force scenario. And you will have to find out work done in different kinds of scenarios. So I hope you guys are comfortable in finding the work done. Now, we are going to understand this work done in a vector form. If you draw a coordinate axis like this, this is x and that is y. Initially, the particle is suppose here. This is point number 1 and then the particle goes there. This is point number 2. The position vector of point number 1 is this. Let us call it as R1 position vector. And position vector of point number 2 is that. Let us call it as R2. Can you tell me what is a displacement vector? This is a displacement vector 1 to 2. Can I write this displacement vector in terms of R1 and R2? You can see that there is a triangle over here. Since there is a triangle, let us say this displacement vector is S. So I can say that R1 vector plus S vector is the R2 vector. So S, which is a displacement vector, can be written as R2 minus R1. So this is the displacement vector. Remember this. This you can also write down as change in the position vector R. This is a displacement vector and the differential element is dr. So small change in the displacement, sorry, small change in the position vector represents what is a small displacement. When you find the average velocity, it is change in the position vector divided by what time it has taken. Or instantaneous is the average velocity for a very small interval. So V is equal to dr by dt. So all of this we have already discussed in kinematics. But the point here is that the force could be acting in let us say this direction. So vectorially you can see that this is the angle the force makes with the displacement. And let us say this is theta then f into this magnitude, the length of this into cos of theta will be the work done. So many times the questions will be not based out of some physical scenario, but it will be some mathematical type of scenario might be given and then you have to analyze it like this. Now I will briefly touch work done by the variable force. Write down work done by a variable force. Now force is changing at every moment. So force is a function of something. Every moment the force is changing. So how will we deal with such scenario? How you will deal with such scenario? If force changes with let us say displacement, every moment like if some small displacement happens force magnitude of direction changes immediately. So in order to cater to such kind of scenario what we assume that for a very small down for a very small displacement continuous variable force can be treated as constant. So a continuous function will not suddenly change its magnitude. So we can say that for a very very small displacement for a very small displacement the force has a constant magnitude and direction. It does not get time to suddenly change its magnitude and direction. So for very small displacement I can say that small work done is equal to for a very small work done and say that this f dot dr. Because for a small displacement I can say that dr is a constant vector and even though the force is changing I can treat it as a constant. So I can say that f dot dr is the small work done for that small displacement. Total work done will be what total work done has to be an integral of dw. So this is how you find out work done by a variable force f dot dr. Now when you open up the dot product you will get what magnitude of f into magnitude of dr into cos theta. Will you be able to integrate it right away? No, we will not be able to integrate it right away because there are there are three variables over here. So force is a variable, r is a variable and theta is a variable. So you cannot integrate this thing which has three variables. So you have to either write f and r as a function of theta. Basically what I am trying to say here is that you need to write down all three variables as a function of a single variable. Then only you will be able to integrate this. This is just a formula. You cannot integrate it right now and depending on with respect to what variable you are integrating your limits will be accordingly lower limit and upper limit. Getting it? All of you understood any doubts? Please message yes or no? See yeah that integration part fine. So I think you have understood that the formula for total work done is f dr cos theta integral fine. Now this integration you cannot do with three variables. For example you will not be able to integrate this x dy. Right now there are two variables x and y until as you write x as a function of y or y as a function of x you will not be able to integrate this. There are two variables. Similarly here there are three variables f r and theta fine. You will be able to integrate this only when you write all these three variables as a function of single variable. Are you getting it now? Now how you will write in terms of single variable that depends on what scenario you are dealing with. That is not generic. It is unique to the situation that you are dealing with.