 All right, so we have started this chapter last class, work by energy. OK, read this down. So what we did last class was that we have just introduced the concept of work, OK? And then we have introduced kinetic energy, OK? And towards the latter part, we have got a relation between the work done on a mass, OK? And how much kinetic energy is changing, OK? So we got a very simplistic expression that total work done, total work done on a single mass is equal to change in kinetic energy. So k2 minus k1 is the work done, OK? It's a very simple straightforward relation. So this is for a single mass, OK? Now if there are multiple masses, OK? Suppose there are two or three masses, then the work done, if you consider two or three masses together, fine? So you find out the work done on all the masses, OK? So work done on all the masses will be equal to change in kinetic energy. Let's suppose there are two masses. So we'll take finite kinetic energy of both the masses and subtract it with initial kinetic energy for both the masses, OK? So we can not only use work energy theorem for a single mass, but we can use this work energy theorem for the multiple masses also, fine? So this is the total work done on both the masses, all right? So you'll see that work done by the internal force. As in the force between one and two masses, it will come out to be 0, fine? So ultimately what you will end up with is a work done by the external forces with respect to one and two both objects together, all right? So this is what we have learned towards the end of the session, OK? So we'll start the session with a numerical, OK? So that you properly recollect whatever we have done last, OK? You guys remember the work done by the spring? The work done by the spring force, we have derived it to be half kx1 square minus half kx2 square, fine? So I'm writing it here because we are going to use this again and again, fine? Where x1 is initial deviation from its natural length and x2 is the final deviation from the spring's natural length, all right? So x1 and x2 can be extension or can be compression. It doesn't matter, fine? All right, let us take up this question, all of you. So I'm starting with simplistic of all the questions. This is a spring constant k. There is a mass, right? There is a mass of mass m that is coming towards the spring, the mass is coming towards a spring with velocity v0, all right? You need to use work energy theorem to find out the maximum compression in the spring. What is the max compression on the spring? All of you, find out quickly. Welcome, everyone. You guys saw this question? OK, Bharath got an answer. Others, what is the condition for maximum compression? When the compression will be maximum? See, this mass is moving forward with velocity. When it hits the spring, its velocity will decrease, right? So its velocity will keep on decreasing. But then it is moving forward. And as it moves forward, spring becomes, you know, spring gets compressed more and more, all right? So when this mass momentarily comes to rest, OK, that is the position where the spring will encounter the maximum compression, all right? So after the mass m comes to rest, after that point onwards, it will start going backwards, right? So we need to use this work energy theorem, whatever we have derived in the last class, between this position where the mass has velocity v0 and where the mass comes to rest, right? So between these two points, I have to use work energy theorem. Now, who is doing work on the mass? There are forces like mg. OK, mg is a force. Normal reaction is a force. There is no friction, fine? mg and normal reaction, they are perpendicular to displacement. So mg and normal reaction do not do any work. But the spring is doing work on small m. When the spring gets compressed, it applies a force, fine? And we know the work done by the spring formula is this. So total work done is nothing but work done by the spring. And this is equal to half k x1 square. Now, what is x1? Initial division from its natural length. Initially, the spring is in its natural length. So x1 is 0, right? Finally, let's say spring has a compression of x. So this is the work done by the spring, OK? This comes out to be work done by the spring. This should be equal to change in the kinetic energy. Final kinetic energy is 0 minus initial kinetic energy, which is half m v square, getting it? So using this relation, you will get velocity. Sorry, we need to find x, right? So you'll get the value of x to be equal to v0 under root m by k, it will come, fine? Any doubt with respect to this question? Hello, yesh, Krishna, Shitos. Any doubts with respect to this question? Fine, so let's modify this question a little bit. Let's say that the coefficient of friction between the surface and the mass is mu, OK? So there is a friction between mass m and the surface now, which is mu, getting it? So mu is a coefficient of friction, and it starts acting as in the surface is rough from the point where the spring starts. So before the spring starts, the surface is smooth, getting it? So after this, when the spring starts, then onwards, the surface is rough and coefficient of friction is mu. Now you have to find out the maximum compression in the spring. Now can you find out what is the compression in the spring? Surface is no longer smooth. Now you will see that not only work done by the spring will be there, but work done by the friction will also feature. Getting it? So there will be two work done now. Earlier, friction was not there, and normal and MG were not doing any work, OK? So only work done by spring used to come. Now the friction is also there. So work done by friction will also come in this particular case. Any one of you got the expression? Sukhi, not present today? I think many others, there are at least five or six people who are not present. OK, Bharath sent me an answer, OK? Bharath, x has to be positive. So although you get a quadratic equation, you need to reject the roots of the equation where x is negative. Getting it? So mathematics do not care about a physical scenario. But when you are studying physics, you should use the mathematics, all right? And once you get the answer in to check whether it makes physical sense or not, fine? So x being negative doesn't make any sense. So work done by the spring will be equal to, again, minus of half kx square only, fine? Now this x is not equal to that x, OK? x is a variable. And work done by friction will be what? So friction will act back side, backwards, fine? And since it is sliding, friction force will be equal to mu times n, OK? And the value of n is equal to mg, fine? So the work done by friction will be minus of mu times n, which is mg, multiplied by the distance it has moved into x, fine? So total work done on this block will become how much? Minus of half kx square minus mu mgx. This is the total work done. And this is equal to change in its kinetic energy. Final kinetic energy is 0 minus initial kinetic energy, which is half m into v square, OK? So any doubt with respect to this question? I hope many of you got this equation, correct? All of you got it? Any doubts? OK, I am assuming all of you understood this particular question. In case you have any doubts, feel free to message on the chat box, or you can message me personally or WhatsApp, whatever it is, OK? So good. We'll take up a few more questions, because see, mechanics is more of problem practice, fine? So we cannot just discuss theory, and that's it. So it is 90% problem solving, and only 10% is the theory part. Anyways, draw this diagram with me. This is A, this length is given as L, OK? And this height is given as H, fine? So particle slides along a track with elevated ends, and the flat central part, as shown in the figure. So B to C is a flat central part, OK? The flat central part has a length of L is equal to 3 meters, fine? The curve portion of the track are frictionless. So there are no friction on these curved parts. So A to B and C to D, they are smooth, fine? And for the flat part, from B to C, there is a friction with mu is equal to 0.2, getting it? Now the particle is released from a height of this H value is given as 1.5 meter, OK? And also ignore any bumps it can encounter near the transition between the, let's say, a flat portion and the curved portion, so it smoothly transitions. So there is no loss of energy as such, OK? So you need to find out till what height it will reach on the other curved surface. Let's say it reaches here. You need to find out what is this height? How much is this H? Letting it, you need to find the value of H. All of you, please try this out. See, whatever is not given, you can assume, OK? You can say, OK, mass is M. If it is not given, probably the answer is independent of that. But then if you want, just take it as M. OK? People are getting answers. That's wrong. Sukirt, Bharat, Ashutosh, that's not correct. Anyways, so OK, let's see. Let's try solving this. So between what and what points I can use this work energy theorem, I'm going to use this work energy theorem between two points, OK? Work done is equal to change in kinetic energy. So these two points, I'll directly take A and D. Getting it? So point D is the second point and point A is the first point. Fine? So work done can be split into two parts. Work done by the gravity and work done by the friction. Isn't it? Now what is the work done by the gravity between point 1 and 2? So work done by the gravity can also be split into two parts. Work done by the gravity when it goes from A to B, then work done by gravity when the object goes from C to D. Getting it? Now there is no work done by the gravity when the object is moving between B and C, because gravity is acting downwards and object is moving forward. So it is 90 degrees. Fine? So between B and C, there is no work done by the gravity. So I'm just taking A to B and C to D. Fine? So work done by the gravity between A to B is mg into H. Fine? Because displacement along the direction of force, which is vertically down, is H and mg is down. Fine? So mg multiplied by displacement along the direction of force, which is H. So work done by the gravity from A to B is mg H. All right? And when the object goes from C to D, then the movement of object is upwards. OK? It is against the direction of gravity. So you need to write it as minus of mg H. Fine? So this is the work done by the gravity. And then we need to take into account work done by the friction. Fine? So that is equal to frictional force, which is mu mg into L. And there will be a minus sign coming in here because friction is acting backwards, but the displacement of object is forward. So minus sign will come. Fine? So total work done will be equal to minus of mu mg L plus mg H minus mg capital H. This will be equal to change in the kinetic energy. And how much is the change in kinetic energy? Initial kinetic energy is 0 because velocity is 0. And final is also 0 because at point D, velocity is 0. It comes to rest. OK? So 0 minus 0. So from here, you can get the value of H. You can see that M get cancelled out. Fine? So H, you will get here as G times H minus mu GL, OK? So divided by G also, sorry. So you will get H minus mu L. Fine? So H is what? H is 1.5, mu is 0.2, and L is 3. So it comes out to be 0.9 meters. Fine? So like this, you have to do this particular question. OK? So there is a tendency that you will break up the entire motion into two or three components. Like you can say that, OK, first you'll apply work energy theorem between A and B. Then you apply between B and C and then between C and D. It doesn't matter, OK? You can directly take initial point and the final point. If you're able to calculate the total work done, it does not matter. Getting it? So this is how you have to solve this particular question. In case you have any doubts, please message immediately. Else I'm moving to the next question. Any doubts? Niranjan says no doubts on behalf of everyone. Others also please, you know, you should not feel shy. Just communicate, OK? Anyways, we'll take up a couple of more questions. All of you draw this. Fine. So two blocks are connected by the string as shown in the figure, OK? They are released from the rest. Show that after they have moved a distance L or after they have moved a distance of L, what is the velocity of M1 and M2, OK? Where coefficient of friction between the horizontal surface and M1 is mu, OK? Pulley is massless and frictionless. The string is inextensible and massless, OK? So get the value of the velocity of M1 and velocity of M2 when both the block move by a distance of L. They both move the equal distance, right? M1, if it moves by L, M2 also moves by L. Now you have to apply a vergenge theorem on both blocks together, M1 and M2, right? Velocity of M1 and M2. Bharat, the answer you sent me, it's difficult to understand what you've written. Sorry. Can I send it again? Those who get the answer, you can WhatsApp me. Just take a pic of whatever you're doing and send it across. It's very easy. OK, Sukirt has sent one answer which is independent of coefficient of friction looks like. He's saying it doesn't depend on coefficient of friction, is it? Yes, now it is correct. I'm getting very different answers. I got multiple WhatsApp messages. Anyways, so I'll try to solve it. I hope everybody have tried this out. OK, so see, at times you will get the answer by many different means, but then you need to understand the kind of intricacies that are involved in a particular question, OK? So focus on how it is solved. So I am going to apply this work energy theorem on both the masses together, fine? So kinetic energy for M1, finite kinetic energy for M1, finite kinetic energy for M2 minus initial kinetic energy for M1 plus initial kinetic energy for M2. This is a total work done, fine? And this work done is a work done on M1 and M2 together, fine? So the work done will be some of the work done on M1 and M2. So who are doing work? Friction on M1 is doing work, isn't it? So if this block moves a distance of L, so how much the friction will do the work? Minus of mu M1g into L, because friction is acting opposite of the displacement. So minus mu M1g into L is a work done by the friction on M1, fine? And who else is doing work? You may say that tension, which is acting on M1, is doing work, all right? So let's include that for a while, plus tension force into L, even tension is doing work here, right? Now this is the work done. This is total work done on M1. Gravity is not doing any work on M1 because gravity is acting down, which is perpendicular to displacement. And normal reaction also, same case. Now let's add up the work done on M2. So if M1 moves forward, M2 will go down by the same amount, let's say L, fine? So if it goes down, that is the direction of gravity. So gravity will do the work on M2. So that is M2g into L, right? And also there is a tension acting, T. Now tension is acting upward and displacement is downward. So minus T into L will come, fine? So you'll see that the work done by tension get cancelled off, OK? So whenever you find the work done in a system, you have to only consider the external forces, which is, in this case, friction and the gravity, fine? So this is the total work done. This is equal to change in kinetic energy. So final kinetic energy for M1 is half M1 into V square, OK? Plus for M2, it is half M2 into V square, right? Because velocity of both the masses will be same. They're connected with an inextensible string. So this is the final kinetic energy of the system minus initial kinetic energy of the system is 0 for both the masses, fine? Just you equate these two, you'll get the answer for velocity, OK? So I hope you have got this. Let's move to next question. OK, what is the final answer? What Sukrit has sent is correct. Just multiply M1 with mu. Then the Sukrit answer is correct.