Physics Work, Energy, Power Overview

Normal kids make mistakes here. And in the same way, we have completed the second section. Now let's come to the third section, which is work, energy, power. What is work, energy and power? Come, let's go to the world of work, energy, power and understand. No break will be found. If I keep giving break, then by combining basic math and vector, Total chapter becomes 16. If I keep giving breaks, then I will have to give 8 breaks. If I keep giving breaks, then how many breaks will I have to give? Calculate and tell me. If more than 8, then I will have to give breaks. If more than 8, then I will have to give breaks. Then only breaks will work. Then when will we study? Then we will play breaks. Every two hours break. So we will play break-break only. When will we study, brother? Oh brother, if we play break-break, then when will we study, my dear? My blue, my yellow, green, black, eggplant-like face. Tell me. If we play break-break, then when will we study? When will we study? He will do the dinner again. Just eat this rice. We are going. We are going. When I talk to you. Any force applied anywhere. So remember, I am telling you this very good trick. Whenever force is applied on any object, you will not directly know the effect of that force. You will only know the effect of force when you have given time along with force or how long the force has been applied. Sorry, how long the force has been applied. Whether you have given time or displacement. If you have given force and time, then remember. by multiplying force and time, we get change in momentum and if we know change in momentum then we can calculate the velocity what will be the kinetic energy? kinetic energy will be P square by 2m then we will have a lot of stories and if you have given force and displacement then you can calculate work done from here which is F vector dot S vector ok so here we will calculate work ok, if you know work then you know change in kinetic energy is work done and kinetic energy is known, then the momentum is root under 2m. So, if time is given with force, listen carefully, if time is given with force, then where will the approach go? Change in momentum will be shown in the laws of motion. If displacement is given with force, then what will come out? Work, work can remove kinetic energy, kinetic energy can remove momentum. Then whatever you have to do, mint, coriander, ginger, keep doing from here. Okay, work done. It is F dot S. Remember, S is nothing but displacement of point of object where force is acting. Here S is not a normal S. It is the displacement of that point where force is acting. Work done depends upon frame. Yes, work done depends upon frame. Work is scalar. Work may be positive, negative or zero. If your angle theta is equal to zero degree or greater than zero degree but smaller than 90 degree, then work done will be positive. and if your angle is exactly 90 degree then work done will be zero if angle is greater than 90 degree but is equal to 180 degree or smaller than that then work done will be negative this is angle when work done is positive and negative if work done is positive then what is its physical significance kinetic energy of object increases and speed increases if work done is zero then kinetic energy is constant and speed is constant if work done is negative then remember kinetic energy will decrease means When we do work positively, we inject kinetic energy. And when we do work negatively, we take out the kinetic energy from there. Understood? Understood? Do you understand this? Do you understand? He is saying, there are only 10k students. There are only 10k seats. Understood? Sir, sir, see. There are only 10k students. The time is also going to be 12. And the seat is also 10k. Oh, sir, good. Read it, son. Small and big. Now look carefully. Neat is a question from 2016. I have already done this. But this is in a different variety. Your question is, a particle moves from a point this much to this much. This much to this much. Where a force of this much Newton is applied. How much work has been done? Now listen carefully. Sir, work done is the dot product of force and displacement. Now, 4 is given to us, 4i cap plus 3j cap. But what is the displacement? So, listen, this is the final position minus the initial position will be the displacement. So if you subtract this from 4j, then it will be 5j, which means minus j. Now 0i is not there, if you subtract this from it, then it will be plus 2i. And if there is no k from 3k, then it will be plus 3k only. So this will be the displacement. I with i dot product, 4 times 8. J with j, 3 with this, then minus 3. And if there is no k, then k dot product is 0. 8 minus 3, 5 joule work. What will you do? Final position minus initial position. What will you do, son? Final position minus initial position. Clear? Next. A body moves from the distance 10 meters along the straight line under the action of a 5-Newton force. And work done is 25 joules. Then angle bit. Work done is Fs cost theta. You were given work, force and S. So, won't the cost theta come out of here? It will come out very easily, sir. It will come out in a smokey way. Neat 2017 question. Read the question. Consider a drop of rain, water falling, mass 1 gram falling from the height 1 kilometer. Means 1000 meter. So 1 gram of an object from the height of 1 kilometer hits the ground with a speed 50 meter per second. Take g constant with a value of 10 meter per second squared and work done by the gravitational force. So how much work of the gravitational force will be done? Work done by the gravitational force will be Mgh. M value is 1 divided by 1000. We will change it in gram and kg. G value is 10. H value is 1000. If you cut 1000 from 1000, you will get 10 joule work. So 10 joule will be your positive work. This is wrong. This is wrong. This is wrong. This is right. Oh brother, so third answer. But there is more part in this. How much work will be of resistive force? This is what smart work is. Why do we have to do more politics in this? When I... Work done by gravitational force is showing that only one option is right. Work done by gravitational force is showing that only one option is right. So do I need to do more politics in this? No sir, no politics will work in this. No politics will work in this. Yes sir, look at the next. This child, you don't have to do. This is a question of NEET 2019. You just have to see. Force is equal to 20 palace 10y. X on the particle means force is applied in the y direction. in y direction. Then force F is in Newton and y is in meter. Work done by this force to move the particle from y equal to 0 to 1 meter. So we know that if there is variable force, then work into displacement, we remove it like this. F dot dx works equally. Clear? Is it a neat 2019 question? Have you done it in class? Do this and show. Neat 2023 question. Do this. And see if the same question is asked or not. Same question is asked. Means this happened. that the questions asked by NEET, the first questions which were in JEE main are now coming. And the kids say, Sir, will the question of JEE come in NEET? Do the question. And try typing it. This is the question of JEE main. And now the question of NEET has been copied completely. Do it. Sir, there was y in it, so we just made y square. So, f is equal to 5 plus 3y, 1, 0, particle in the y direction, f is in Newton. And y is in meter, then work done by the force during the y equal to 2 plus, y equal to 2, 2, 5. Do it. How will we get the work of resistive force? That is easy. We will have to use the work energy theorem in that. Where is it? I will tell you. In this we will get it like this. Work done by gravitation plus work done by friction equal to change in kinetic energy. Now initial velocity is zero and final velocity is given. So change in kinetic energy will be obtained. We know this. So work done by friction will be obtained. Is it clear? If you are interested then do it. I will teach you to do it. By the way, take it out. I will teach you. Come on, do this. 132, 117. Now decide, brother, whether it will be 132 or 117. Decide, brother, whether it will be 132 or 117. Then talk. Decide, brother. A force of 5 plus 3y square x on a particle in the y direction where f is in Newton and y is in meter. Then work done by the force during the displacement from 2 meter to 5 meter. 132, 132, 132, what will you do? F into dy equal to dw Because force is only in y What is the value of force? 5 plus 3y square into dy equal to dw Do the integration, this work will come 5 into dy will be 5y plus 3y power 3 by 3 This will be cut You will keep the limit from 2 to 5 So if you keep 5 here, then 5 first 5 then 2 So here comes 3 plus y power 3, 125 minus 2 power 3, 8. 5 times 3 is 15, plus 8 out of 125 will go, 117. 117 plus 15, so 5, 2, 7, sorry, 7, 5, 12, 2. 1 goes here, so 2, 10, 1. Is 122 coming? No, how will 122 come? 117 minus 15 is 0. Is 132 coming? 2, 1, 3. Yes, 132 is coming. So what were you telling earlier, 117 and this? you calculated this term, did you shoot it? 132 joules. So, will the dot product of work also be a question? Will the dot product of work also be a question? Did you enjoy? Did you enjoy, brother? Now, see, work is asked in this. So, you know the area of force and displacement graph is work. So, directly removing its area will tell you work. Directly removing its area will tell you work. Next, see. Next, see. If kinetic energy of the body, is increases by 300% then percentage change in momentum will be Now see carefully, what is the question? If kinetic energy of the body is increases by 300% then percentage change in momentum If kinetic energy is increased by 300% then how much change will come in momentum? So first of all p is equal to root under 2mke Is it clear? Okay Now p is proportional to root k will come? Yes sir If you increase kinetic by 300% So finally how much will be 400%? 400% means? Final kinetic energy or I can write 400% of initial kinetic energy If kinetic energy is something ke, if you subtract 300% from it, then finally how much will be? 400% If you subtract the percentage, then you will divide it by 100, which means 4 ke So the final kinetic energy will be 4 times of initial So final momentum is proportional to final kinetic energy So final momentum is equal to Final kinetic energy will be 4 ke So 4 2 will go out, root under k. Root under k initial can be called initial momentum. So final momentum is 2 times of initial momentum. Can you find this relation? Meaning, if it doubles, then it will be 100% bigger. If it doubles, then how much percent will it be bigger? Tell me. If it doubles, then it will be 100% bigger. Yes or no? Yes or no? Next, see. 90, 98, 97, 90, 89, we have asked this 3 times. 2 body of mass m and 4m moving with equal kinetic energy. Then ratio of their linear momentum. Momentum is equal to, it happens also. If its mass is m and 4m, then it will not be 1 is to 4. Why? Because momentum is equal to root under 2mke. If their kinetic energy is same, then it is same. So momentum is proportional to root m. First mass m, second mass 4m. 4 times mass. So if you keep 4 in root, then it will be 1 is to 2m. What will be the answer? The first answer will be 1 is to 2. Do you understand this well? If you understand this well, then ask this question. It is the question of JEE MIN 2022. And you will be sure that such questions which have come in JEE MIN, have come in the first NEET. They have come in the first NEET, sir. And they have come three times. That is being copied in JEE MIN. It is the question of JEE MIN 2022. You type it and check it. I am not joking. I am not joking. You type and check whether there is a question of J-E-Main or not. You type and check whether there is a question of J-E-Main or not. Tell me. Body of mass 8 kg and another mass of 2 kg are moving with equal kinetic energy. Then ratio of their momentum is 8 and 2. Equal kinetic energy. So what will be the ratio of momentum? Root M. So, root 8 divided by root 2, so 4. If it goes out of 4 roots, then 2 is 2 1. The second answer will come. The second answer will come. He said, why didn't you do it with P square by 2m? That's what I did, dude. Because kinetic energy is P square by 2m. Kinetic energy is the same. So, P square is proportional to m, means P proportional to root m. Idiot, biology. Ramlal Ekdam. Look, what is work energy theorem? Work energy theorem is that work done by all the forces is equal to change in kinetic energy. And this is the father of all the theorems. This is always valid. Always valid. Who is the father of all the theorems? Everyone's father is the same. Everyone's father is the same. Who is everyone's father? Everyone's father is your work energy theorem. Work energy theorem says, Work done by all the forces is equal to change in kinetic energy. Work done by all the forces is equal to change in kinetic energy. Second, work done by conservative force is equal to minus change in potential energy and this is also always valid. This is the definition of potential energy. Second, work done by non-conservative force is equal to plus delta U and this is not always valid. This is valid only when change in kinetic energy is zero and non-conservative force is acting against Conservative force. Non-conservative force acting against the conservative force. That's when it's valid. That's when it's valid. Clear? It's wrong. The question is from 2016. Okay. Where is it? Where did 2016 go? Which question is from 2016? This one. Let's check. I will do it according to the information given. It may be misprinted, I don't know. If you subtract this from 4j cap plus 3k cap, 4j cap plus 3k cap, this is the final position. Particle moves from a point this much to this much. So this is the final position minus this initial position. If you subtract this, it will be plus 2i and minus 5j. So this will be, from 4j, j will go minus j plus 2i plus 3k. minus z 2i plus 3k so it is correct this is the question of 2016 so remember these things I will write it again that work done is equal to change in kinetic energy and if you know work done formula is force into displacement If you apply the same force F at the same distance on any object in the world, then the kinetic energy that will develop in it will not depend on the mass. Listen carefully to what I am saying. Can I make this graph of kinetic energy versus mass for any condition? Yes. If the work is the same on any object in the world, if the work is the same on any object in the world, the developed kinetic energy in the object will not depend on the mass. Tell me. The developed kinetic energy in that will not depend on the mass. Is it clear, my brothers? Is it clear, my brothers? Now see. Body of mass 3 kg under the constant force causes the displacement S in meter and given the relation S is equal to 1 by 3 T square. Where T is the second, then work done by the force in 2 seconds is work done by the force in 2 seconds is listen carefully now here work done if you go to remove f.s then you will not be able to remove work done remove from change in kinetic energy sir change in kinetic energy i can write final minus initial final kinetic energy half m v final square minus half m v initial square so from here half m will be removed from your common v final square minus v initial square okay this displacement is given here velocity and differentiate between them. So here 1 by 3 outside, the differentiation of t square will be 2t. So this will be 2t by 3. Is it clear? Yes sir. Now I asked you, work done by in 2 seconds. So what will be the initial velocity? If you keep t 0, then 0. And what will be the final velocity? Sir, if you keep 2 time, then 2 times 4, times 3 will come. When you take out the work here, then half, what is the value of mass? 3 kg. 3 into, final velocity 4 times 3 whole square. 4 divided by 3 whole square initial velocity 0. So, what will be 4 whole square? 16 divided by 9. So, this is 3 multiplied by 16 divided by 9 times 2. 3 divided by 9 will be 3 and 2 divided by 16 will be 8 divided by 3 joules. 8 divided by 3 joules is equal to correct answer. So, remember that many times we have to get work done from change in kinetic energy. How did we get it? We had given displacement, we differentiated and got velocity. Final and initial velocity is known, then work done is equal to change in kinetic energy. Work done is equal to change in kinetic energy. Here the question is asked, the kinetic energy acquired by the mass m, in travelling a certain distance d, starting from the rest under the action of a constant force f is directly proportional to. Meaning, kinetic energy, how will it depend on the mass? So kinetic energy does not depend on the mass. When the force and distance applied on any object is same, dia hoto. Because Carnatic energy is equal to work done. and work done F dot S will be done, so kinetic energy does not depend on mass. Next see, 3 joules of work is done in the sliding 2 kg block of the inclined plane of height 10 meter. Then work done against the friction is? Okay, okay, okay, okay, okay, okay, okay. Understand the question properly. I have asked it three times, I have asked it twice, it is an important question. 300 joules of work is done in sliding a 2 kg block up an inclined plane of height 10 meter. data. means the height of this inclined plane is 10 meters. Now to take it from here to here, the work done is 300 joules. So I have asked you, the work done against friction is? Work done by gravity plus work done by friction is equal to change in kinetic energy. So kinetic energy has not changed in this. So total work done should be zero. work done by gravity Okay, okay, okay, okay, okay, okay. The total work done in this is against gravity and against friction. Okay, okay, okay, okay. So the total work done is 300 joules. And the work done against gravity will be MGH. Say M. Say G. Your height is 10. Sir, 2 kg. M is 2 kg. So 10 times 100 is 200. So 200 joules which is the work What did we do against gravity to take it up? And 300 out of 100 joules would have been against friction. If there was no friction, if there was no friction, then I would have to work against gravity to take this block from here to here. And work done by gravity does not depend on path. Work done by gravity does not depend on path. Listen to this question carefully. Work done by gravity does not depend on path. Does not depend on path. If there was no friction, So I have to work against the gravity of the work. But gravity has worked so much against gravity. And it will be a negative work obviously because the body is going up. And the total work done is 300 joules. So I have to work against gravity of 200 joules. I have worked against friction of 100 joules extra. A body of mass 1 kg thrown upward with a velocity 20 mps. It momentarily comes after the... comes to rest after attaining a height of 18 meter. How much energy is lost due to friction? It's a good question. It's a question of work energy theorem. It's a question of work energy theorem. See, body 1 kg is thrown upward with a velocity 20 meter per second. Means, I threw a body from 20 meter per second. Now, sir, it has stopped after 18 meter. It has stopped after 18 meter. So, I have asked you, it. work done by air friction, listen total work done by gravity I have thrown it, I am not taking it gravity plus friction, equal to change in kinetic energy what will happen? if object is going up, work done by gravity will be minus mg plus work done by friction, we have to remove it, so we will not write plus minus is equal to final kinetic energy is zero, initial kinetic energy will be half m is also the whole square of initial did I write the statement correctly? yes sir so work done by friction will be minus Half, M's value is 1 kg. Velocity is 20, means 20 squared is 400. Plus, this will come here. So 1, G's value is 10 and height is 18. So if we cut this from 2, minus 200 plus 180. So how much will this be? Minus 20 joules. Work done by friction, you have got 20 joules. Just ask in magnitude. Is it clear? Work energy theorem is a good question, a good example. Next. Look, if you understand this carefully, then we have two types of force. One is our conservative force and the other is our non-conservative force. One is our conservative force and the other is our non-conservative force. Remember in this that work done does not depend on path. Conservative force's work does not depend on path. What is its work on the closed path? Zero. The concept of potential energy is only defined for the conservative force. Its work is reversible, i.e. work is not lost. Its example is gravitational force, electrostatic force, spring force. But the work of non-conservative force depends on the path. In the closed path, its work can be zero or not. The concept of potential energy is not defined for it. Work is not reversible. For which the term of potential energy is not defined. They are not. non-conservative hota hai. Jaise friction force, tension force, viscous force, normal force. Yeh sab non-conservative ke category mein aata hai. Is it clear? Here you can write work done by conservative force is equal to minus change in potential energy. Work done by conservative force is equal to minus change in potential energy. Is it clear? Next, the potential energy of the system increases if work is done upon the system by the conservative force. Look, if the conservative force works, Like the gravitational force is conservative. If its work is positive, then kinetic energy will increase. If its work is positive, then kinetic energy will increase. Potential energy of the system increases if the work is done upon the system by the conservative force. If the conservative force works, then kinetic energy will increase and potential energy will decrease. This will not happen. By the system against our conservative force. By the system. against the conservative force, then the potential energy will increase. This is correct. By the system against the non-conservative force, if you work against the non-conservative force, then the potential energy will not come. Upon the system by conservative force. On the system by conservative force. Not upon the system by non-conservative force. We will not talk about the non-conservative force. According to me, there should be a second answer. What do you say? What do you say? Okay. One is saying that 148 slides have been done. Brother, I have studied 250 slides and kept them there. Right? Look ahead. A spring of force constant K is cut into two pieces such that one piece is double of its length of the other. Then the long pieces will have force constant. Look, there is a spring whose length is L and its spring is constant K. If two pieces such that one piece is double the length, then the longer piece will have the force constant is equal to. Do it. It's a good question. Here, look at this. spring constant which is inversely proportional to length. So, what will be the spring constant of the longer piece? This is what I have asked. Let's see if you can do this or not. It is a good question. It will seem like a good concept. It is a good question. It will seem like a good concept. This question will not be solved by normal concept. Is this clear? Oh, brother! How will you do it? Tell me. How will you do it? Tell me. So see, two pieces double of its length. If I take its length L dash, its length is 2L dash. Okay, sir. Now here, the one whose length is less, its force constant will be more. So if I consider its spring constant to be K0, then its will be 2K0. Is this clear? Yes. Okay, tell me one thing. I have assumed this. Length L was of spring constant K. If length L dash is of length 12 dash, I have to divide this into two parts. So this will be 2K0, this will be K0. Sir, done, understood. Now what he is saying is, tell the value of the spring constant of the longer piece. So tell the value of K0. Now listen carefully here. If I add these two again, then spring constant will be K. And how do I add both? I will have to add them in series. And the value of spring constant in series is 2K0 multiply K0 divide. 3k0 is equal to k. In series, parallel is similar to spring constant. Oh wow sir, so k0 will be cut from k0? Yes. So k0 is equal to 3ky2. 3ky2 will be the answer. Means second answer. And till now no one is giving second answer. No one is giving second answer. No one is giving second answer. And first, there is a strong reason for being wrong. What? Jab mein I am seeing L length's spring constant K. If I reduce the length, then the spring constant will increase. And this is 2 by 3K. So it is wrong. This is less than K. Suppose this is L length's spring. If I break this, then both the lengths are less than this. So both the spring constant will increase. In this, spring constant is telling 2 by 3K. So it is telling less than K. So this cannot be the same. Tell me, did you understand it well or not? Did you understand it well or not? Did you get a good question or not? Did you get a good question or not? Tell me, next. See this question, son, J-Man 2022. And literally the language of this question is this only. You can solve it without a pen. Without pen use, solve it and show it, son, it will be fun. Means continuously studying for 9 hours will be beneficial. Continuously studying for 9 hours will be beneficial. Continuously studying for 9 hours will be beneficial. Tell me. Solve it without pain. Read the question and solve it without pain. The question of 2-1-1 is wrong. No problem. If the question is wrong, then leave it. Our sole inner figure, particle of mass 10 kg, placed at point A, when the particle is slightly displaced along the right, it starts moving and reaches the point B. The speed of the particle B is x m per second. Then value of x is, see, smooth surface, no mention of friction, 10 m height, how much it fell, 5. How it fell, no matter. Velocity will be root under 2gh. 2, G is 10, H is 5. So, root 50. Sorry, root 5 times 10 is 10. And root 100. Root 100 means 10 m per second. How many people got 10 answers? Brother, tell me. This is a question of J.E. Mendoza. This is a question of J.E. Mendoza. This is a question of J.E. Mendoza. This is a question of J.E. Mendoza. This is a question of J.E. Mendoza. Tell me, did you solve it without pain or not? Tell me. It's a 6-month question. It's a 2022 question. Tell me, did you solve it without a pen? Even such pendulum-like people ask questions. You'll blow their heads off. Brother, what is this? See what it is. What is this? The more times I have asked this question, it seems that this time it should be asked again. IIT, NMR, AIEEE, MPPMT, 94, 97, 2000, JIPMER, 2000 and IIT have asked it many times recently. Recently IIT JEE main me kai baar pucha hai. JEE main me kai baar pucha hai. Kya hai yeh sawal? Aav parthe hai. Aav parthe hai. Agar tumne work energy power kia aur yeh sawal nahi kia matlab kuch galat kia. Aav parthe hai. A uniform chain of length L and mass M. Uniform chain hai. Jiska mass M hai length L hai. Is lying on a smooth horizontal table. And one third of its length is hanging vertically down. Means, from M mass, M by 3 mass is hanging down. L by 3 will be hanging down. Over the edge of the table. If G is the oscillation due to gravity, then work required to pull the hanging part on the table is. So, how much work will be required to keep this hanging part on the table? So, the first question that develops here is, who will work? Who will work? and work energy power I am teaching so if I don't say a very famous statement then there is no fun in it where we are there is no less how many people know this where we are there is no less where less where less there there we are not what is this less less is nothing but conservation of mechanical energy meaning meaning Initially kinetic plus initial potential. Equally. Final kinetic plus final potential. This is the beta of work energy theorem. Is it valid everywhere? No, sir. It is valid only when work done by non-conservative force is zero. Do you know well? Now tell me, who will bring the chain up here? Sir, we will bring the chain. And when our work done is against gravity, can we write it directly as change in potential energy? Yes, sir. We can write external force's work done as change in potential energy. Let's write change in potential energy. Tell me. Final potential minus initial potential. Now if I put the chain on the table and consider this as the potential energy reference. Then what will be the final potential? Zero. What will be the initial potential? So the part on the table has a potential of 0. What is the lower part? m by 3. So m by 3 into g. And the center of mass of the extended part is its center. Sorry, the potential is at the center of mass. So how much is lower? ly 6. And lower that's why minus will come. Minus minus plus mgl by 18. mgl by 18. Fourth answer will come. What will be the answer? mgl by 18. NGLY 18. Is everyone's correct? Very good. Next, see. Such questions come in the lead. The potential energy of the long spring when stretched by 2 cm is U. Then the spring is stretched by 8 cm, then potential energy is stored in it. See, here the first thing to be read in the question is whether the change in potential is asked or the final potential. The potential energy stored is half kx square. He said that 2 cm is u, so we will keep x as 2. So k divided by 2, x is 2, so we will get the factor of 4. And if we cancel 4 by 2, then it is 2k. So 2k is called u. Now if the spring is stretched by 8 cm, then potential energy is. So the final potential energy will be half k, whole square of 8. So 32. 32 means 8 is 64 by 2, 32k. 32k will come. If you can write 32k, then 16 into 2k. And if you write 2k as u, then 16u will be your answer. Now 16u is fine, there is no problem. 16u will be there. But many times the question is, if you stretch 2cm, then the potential energy u comes. If I stretch 8cm, then what will be the change in potential energy? Tell me, what will be the change in potential energy? 15u. Change in potential energy will be 15U. Is this clear? One more thing. One more thing. One more thing. What? Look. Suppose there are two springs. It is very important. It is a question of 2015. One spring constant is K and the other spring constant is KB. Okay. Now we know that KB's value is more than KA. And we applied the same force on both. Tell me in which potential energy will be stored more? In A or B? This is given to us. If Kb is greater than K, then in which potential energy will be stored more? In A or B? Tell me. In which potential energy will be stored more? In A or B? Tell me carefully. In A, will potential energy be stored more or in B? Now half of the children are saying A, half of the children are saying B. Why? Because potential energy is directly proportional to K. See, there is a spring whose spring constant is k, so I applied force on it. There is a spring whose spring constant is k, so I applied force on it. If kb is greater than k, then in which case more energy will be stored? We have to compare. I applied force on both of them, so see the solution. The formula of potential energy in a spring is half kx squared. Now both have different spring constant. I applied force on them. If you put the force same, then both the elongations are same, so it is not necessary. No sir. So what is the relation of elongation? If equal to kx. So from here we can write x as if by k. So this will be half k. If by k. If by k's whole square. So this will be half if's square, if's square. This k's square, if k is cut, then k will come down. Matlab Potential energy is inversely proportional to K. So the one whose spring constant is more, energy will be stored less in that. Everyone's Lankan has been set. Is it done? Oh my God! Everyone's Lankan has been set. Paltu people, Paltu people. See, I repeated it again and again. Same force is being applied. This is the question of AIPMT 2015. And we were worried that potential energy is square from k. Yes, it happens when we write it in terms of x. In terms of force, it will be inversely proportional to k. And we have taught you this thing. Whenever we taught you in class, we taught you this thing. This is the question of J.E. Main, 2023. Brother, I remember, by mistake 95 is stuck here. Or it is asked in 95, it is also asked in NEET. Or it must be written somewhere again. I will find it and bring it. It is not written. It is not written. This is a question of J.E.Main 2023. Or 2022. Something like that. What are you talking about? Yes. Now tell me honestly. This question which you are seeing. Is it asked in NEET or not? This question which you are seeing. Is it asked in NEET or not? Tell me this. This question which you are seeing. Is it asked in NEET or not? Yes sir. This question is asked in NEET. Now see. It is asked in J.E.Main. How will 15U come? When you ask for change, then 15U will come. Now it will not come like this. When you ask for change, then 15U will come. Like this, 15U will not come. Come on, ask this question. Come on. So when is stable equilibrium? When your force is zero, then stable equilibrium happens. So what do we write the value of this force? dU by dx. So do 0 to dU by dx. Do 0 to dU by dx here. So this will be your 12, 12, 12, 12, 12a divided. x to the power of 11 is equal to 6b divided by... x to the power 5. I will subtract, right? 5 will come. Okay? I have kept this equal. So, I will subtract 12 from 6. 2. 2a by b is equal to x to the power 11 divided by x to the power 5. 11 to the power 5 will be x to the power 6. 2a divided by b is equal to x. What will come here? 2a divided by b is equal to x to the power 1 by 6. First, you don't have to do anything. You have to put 0 by doing differentiation. By doing differentiation, you will not get 13 by 7. You are doing wrong. Look, what he is doing, in differentiation, one minute, you are right, you are right, you are right, you are right, you are right, you are right, you are right, you are right, you are right, you are right, you are right, you are right, you are right, you are right, you are right, you are right, you are right, When we take it up, then if we subtract minus 12 from x power, then minus 13 is there. Oh sorry Shaktimaan. I am Ghanghor Biology, you are Ghanghor Math. So now here will come cut XY6. You are saying right. You are saying right. I am wrong. You are saying absolutely right. I did wrong. I did wrong. You did right. Sometimes you will also do right. 14 will come. Hey, 14 will come. I will take out my shoe and hit you now. You are getting excited. Once you do the right thing, you will be 14. Or else what? Once a question gets right in centuries, then it goes inside. Then you will be 14. How will you be 14? How will you be 14? First, water your mouth. Half of you are watching the lecture in sleep. Come on, come on. Water. If you don't like the lecture, then like it. 50,000 likes should be done live. 50,000 likes should be done live. A particle with total energy E is moving with a potential energy Ux. Motion of the particle restricted to the region. See, where motion cannot be done. Generally, what we know is that total mechanical energy E is kinetic plus potential. Yes sir, kinetic plus potential. So what can we write as kinetic energy? E minus U, E minus U. If you think E minus U. If it happens that U is big and E is small, then sir, if it is small, it will be negative. Whereas kinetic energy is not possible. Kinetic energy cannot be negative. Kinetic energy cannot be negative. So if kinetic energy cannot be negative, then U cannot be bigger than E. U cannot be greater than E. It can be equal. But if U becomes greater than E, then kinetic energy will be like this. I will explain again. Look, son, kinetic energy plus potential energy is equal to E. Yes, sir, it is. What do we write as kinetic? Sir, E minus U. If U is big according to this, if U is big and E is small, if U is big and E is small, then kinetic energy would have been negative. And kinetic energy is not negative. So this condition is not possible. So he is saying, which is restricted to the region. Meaning where motion is restricted. So brother, kinetic energy is negative, this is not possible. And that will happen when U is big. Meaning U is big, this is not possible. Tell me, tell me. 14 will be 16 minus 2. Where is 16 minus 2 14 happening? Where? Okay, 16 minus 2. If the spring is a stage where 8% potential energy is stored in. I didn't ask about change. I didn't ask about change. I didn't ask about change. If I had asked about change, I would have been 14. I will take off my shoes and beat you. You fool. This is U, not 2U. This is 16U, this is U. So how much change will there be? 15U. You fool, biology. You fool, biology. You are a fool. Biology. See next. Mass of 0.5 kg moving with speed 1.5 m per second on the horizontal smooth surface. Collide with a nearly weightless spring of force constant, this much. then maximum compression of the spring is, listen carefully, is there friction here? No sir. If there is no friction, then you can directly apply initial kinetic plus initial potential, equal to final kinetic plus final potential. Say, if I consider the reference of gravitational potential energy on this horizontal surface as 0, if I consider the reference of gravitational potential energy on this horizontal surface as 0, then its gravitational potential is 0, its gravitational potential is 0, and its spring potential is also 0. So, half mb squared will be saved from kinetic energy. Equal to, at maximum compression, its kinetic is zero and potential energy will be half kx squared. Son, k is also given, mass is also given, its velocity is also given. Will x be removed from here? Don't waste time in small calculations. Now, we understand vertical circular motion. Brother, my vertical circular motion, suppose, this is the length string, a mass is attached, v velocity is given. If given velocity is V0, then after any theta angle is taken, what will be its velocity? Can you find it? Yes sir, you can find it very easily. How? How can you find it? With the help of the conservation of mechanical energy. Come on, find it. I am not joking. Assume, with the help of the L length string, I have given given velocity V0 by tying the object of M mass. So when it rotates at theta angle, then how much will be its velocity here? Tell me. Come on, tell me. How much will be its velocity here? Tell me. Everyone will tell you. Okay. I don't have any happy birthday. Everyone will tell you. Tell me. So let's do one thing. Here let's assume potential energy reference is zero. So its initial potential energy will be minus mgl. Kinetic energy will be half mba2. Equal. Final potential energy will be, length is L, so vertical minus MGL cos theta. Kinetic energy will be half M V square. M, M, M, M will be cut. Minus G cos theta will go there, so plus G cos theta will be there. So G cos theta, L, L, L, L, G cos theta minus GL plus V square by 2 is equal to V square by 2. If you cut 2 from 2, then your common will be 2. This was your initial velocity V0 square. So final velocity which will come is V whole square equal to V0 whole square plus 2GL cos theta minus 1. This is the answer. This is the answer. You should know to apply conservation of mechanical energy. What is the answer? V square is equal to V0 whole square plus 2GL cos theta minus 1. This is the answer. How many people are getting this answer? Tell me please. How many people are getting this answer? Is it happening? Let it be 12 o'clock then say happy birthday. It is not 12 o'clock but he is saying happy birthday. It is not 12 o'clock but he is saying happy birthday. He is a strange person. It is not 12 o'clock but he is saying happy birthday. Let it be 12 o'clock. Now listen carefully. When it moves at a straight angle, at that time, how much tension will be there in this rope? So T will be equal. Listen carefully. mv0 given velocity of the hole is square by L. mv square by L plus minus 2mg. I will put in the terms of v0. 2mg plus 3mg cos theta. What will happen? If I keep theta 0, then cos 0 is 1. So, mb0 square by L plus mg. Right. If I keep theta 90, then cos 90 becomes 0. So, mb0 square minus 2mg will come. What will happen when theta 90 is kept? Yes, tension will come. mb0 is always square by l minus 2 mb. Right. And if I keep theta as 180, then it is absolutely right. How many people have verified this from MRSTART that the tension I had written on the generalized angle is right. Brother, I will have to agree. I remember things. In given velocity terms, the tension at any angle, at any theta angle, will be there. mv naught whole square by L minus 2mg plus 3mg cos theta. It is correct. It is correct. Did I write the velocity correctly at any theta angle or not? Tell me. Yes, so it should be minus. It is correct. Both the velocities are correct. This is also correct. Correct, thank you, Nathie. Printing mistake. 1 minus cos theta. No. Son, here 1 minus cos theta will not happen. Why? Because this is negative and this term will go there. So, cos theta term will come in positive. This is also correct. And I will kick. If I solve again, then I will kick. Idiot. I am taking it right. The velocity at any theta angle will be given velocity of whole square. I will kick, see. Given velocity of whole square plus 2mg cos theta minus 1, this will come. And this is correct. I will explain the reason for this. I will explain the reason for this. When I gave velocity V0 at the lowest point, then we have to get velocity at any theta angle. So first of all think that I am getting velocity at theta angle of this mass. What would have happened if theta was 0? If theta was 0, then the velocity would have been equal to v0 If theta was 0, then the velocity would have been equal to v0 Is this logic being verified here? That if I keep theta, then the velocity will be equal to v0 Keep theta 0 first Because if you are looking at theta angle, then if theta was 0, then the velocity would have been equal to v0 Come here and tell me If I keep theta 0 here Then what will be the cos 0? 1, 1, minus 1, 0, this will end See v theta equal to v0 will come, first thing Second thing If I keep theta 90, then the particle will reach here. And the velocity here will be less than V0. Yes sir. So when I keep theta 90, then cos 90 is 0 and this will be minus 1. So V is equal to V0 whole square. If this is 0, then minus 2 mg. Minus 2 mg. This will be the answer. And the right thing is, if you say here also minus and here also minus, then the velocity will increase. Whereas as the particle goes up, the velocity will decrease. I have done the right thing. If I don't have any happy birthday, don't spam. You are still waiting, means you are students. So don't spam, I don't have any happy birthday. Birthday comes very late. Tell me, did I write this velocity right or wrong? Tell me, if I have given velocity v0 at the lowest point, then the velocity I have written at theta angle is right or wrong? Tell me once. If you don't understand, then open the notes and tell me. I don't have any happy birthday. Our happy birthday comes on 27th September. It's right or wrong, open the notes and tell me. Secondly, if I write theta at any angle, then the retention will be mba2bl given velocity, lowest point velocity I am giving, mba2bl minus 2mg plus 3mg cos theta. It is possible that I will put cos theta here by mistake, it is possible that I will take this plus and this minus, but I know that when I take theta 0, then the value of tension at this point is t up, mg down. and below centrifugal force, if I am making FBT here then Mb0 whole square plus L, so what will be the value of tension? The value of tension will be Mb0 whole square plus L plus Mg when I will keep theta 0 Keep theta 0 here, keep theta 0 here, sub cos 0 is 1 2mg will be deducted from 3mg, plus Mg, Mb square plus Mg, tell me if you enjoyed it or not Second thing, the minimum, ok whatever tension is there here 3mg is less here, yes, 3mg is less here 3 mg is more than that then 3 mg is more maximum tension is here minimum tension is here do you know this? do you know this? if I play the game of tension tension here if I play the game of tension tension here listen to me play the game of tension tension so the tension will be here 3 mg will be less than that yes or no? 3 mg will be less than that 3mg will be less the tension will be less here the tension will be less here from here 3mg will be less from here 6mg will be less minimum tension is at the top point maximum tension is at the bottom point here 3mg will increase here 3mg will increase here 3mg will decrease here 3mg will decrease here 3mg will increase here 3mg will decrease here and here the difference of tension is 6mg if it is written in tension mba square by l plus mg what is written? mb square b L plus mg so the tension here will be mb square b L minus 5 mg why? because from here 6 mg tension would have been reduced how much tension will be there here? here tension will be there mb square b L minus 2 mg yes it will be there how? we have to reduce 3 mg from here mb square b L plus mg if we reduce 3 mg then minus 2 mg here 3 mg will be reduced 5 mg here 3 mg will increase here 3 mg will be reduced here 3 mg will increase, here 6 mg, here 0, here 3 mg will decrease, here 3 mg will decrease, here 6 mg will decrease, here 3 mg will increase, here 3 mg will increase, here 6 mg, here 0 say, say, will you remember this? take, take brother Oh brother, I am also very smart. Here is the correct attention written. My God, but I swear on physics, I did not see. You must be thinking, sir, you will become more intelligent. You see from here and write there. And you say that we are high. Oh, I did not see. I swear on physics, I really did not see. I swear on physics, I did not see. Minimum velocity should be at the minimum point to complete the circular loop. And the minimum velocity should be at the bottom. The minimum velocity at the bottom must be at the root under 5gr. And the maximum velocity at the maximum point should be at the root under gr. To complete the vertical circular motion. Where? At the maximum height. To complete. Vertical circular motion. So complete vertical circular motion. Okay, brother? What will be the maximum tension? mb square bl plus mg below. And the minimum tension will be at the top point. Now in this, let's ask such a question. Where is the probability of the rope breaking the most? The tension is wrong. The tension is wrong. Let's see. It is possible that something has gone wrong in the speed. Let's say it. No problem. MBA square VL minus 2mg. Let's reduce 3mg from this. It's right. MBA square plus mg at the lowest point. Where is it wrong? MBA square VL plus mg at the lowest point. It's right. Maximum tension at the lowest point. MBA square VL plus mg. It's right. Print one is wrong. mb square of l plus mg. Right. Up T, below mg, below. mb square of l. T max minus T minimum 6 mg. Minimum will be mb square of l minus 5 mg. Right. mb square of l minus 2 mg plus. Okay. Oh my God. Who sees this minute? This is my brother. Come on. I thought you made some conceptual mistake. means I have made a big mistake who looks like this lets ask NEET 2019 question before that lets see one more thing if we get such a question in NEET that let me give you this this is a vertical loop and here I have given a v0 velocity to an object to complete this circular path how much minimum velocity should I give remember root under 4gr is there If you take the root out of 4, then it will be 2 root gr. Why? Because this is your support from below. If you tie it with a simple string, then it becomes 5 gr. Okay, let's take a massless rod. If you take a massless rod, what will you take? Massless rod. Massless rod. And here you will tie the object of m mass and throw it with v velocity. So here also root 4 gr will come. Is it clear? Come on, ask this question. A mass m attached to the thin wire and whistle in the vertical circle. Then wire will most likely to break wind. What a joke! What kind of a joke is this? Let's move on. What? A body inclined at the wrist and sliding along the frictionless track from the height H. As shown in the figure. Just complete a vertical circular of diameter AB equal to D. This diameter is given to us as AB equal to D. So, what will be its radius? 30 by 2. Now, here the height H is equal to. So, the value of this H is asked so that When I release this ball from here, it will fall down and complete the vertical circular. The biggest thing is that when the particle reaches here, how much velocity will it get? Sir, the velocity it will get will be root under 2gh. Yes or no? Yes sir. If the vertical height is h, then the velocity will be root under 2gh. And the required velocity to complete it, what will be it? Root under 5G It is not a support loop. 5g into R. What will I write R as? D by 2. If this is equal to it, then the vertical circular motion will be complete. Now if you square the roots on both sides, then write it simply. 2gh is equal to 5g d by 2. G will be less than g. Keep it here with 2. So h will be equal to 5d by 4. How many people have solved this by themselves? How many people have solved this by themselves? Tell me. We will lift our hands and tell you. See next. If the object is here, then its velocity is u, so ui cap will be there. When it reaches here, then at this point, the velocity will come. Final minus initial velocity vector's magnitude is asked. What is asked? Magnitude of change in velocity. How much velocity is there here? We know this, right? Some u to the whole square minus 2gr will be there. Yes, it will be. Minus 2gr. Yes, yes, it will be. It will be. Where is this? In K. If it is square, then if you take it here, then the root will come. This will come. Now, change in velocity. Change in velocity happens. Final velocity, root under U square minus 2GR. J cap minus initial velocity Ui cap. This will happen. This will be your change in velocity. If you take out its magnitude, then magnitude is its whole square. Root under this whole square plus this whole square. So this whole square will be u square minus 2gr plus u square. This whole square, connect this whole square. Magnitude is being taken. So this will be root under 2u square minus 2gr. If you take 2 common, the first answer will come. Stupid kids give third answer. Stupid kids give third answer. Next is power. Now remember, I have written it very strongly. I have decorated it. Do you see how much I have decorated it? This power is of two types. One is instantaneous and the other is average. Right? What will be the average power? Sir, you had said that whoever's average integration has to be taken out, divide it by t and it will be integration of t. Sir, this will be the average power. It is not wrong. Brother, there is no mistake here. It is not wrong. It is not wrong. It is absolutely correct. Okay. By the way, average power is total work done divided by total time. We can write total work done as change in kinetic energy divided by total time taken. This will be the average power. Okay. So write average power as total work done by time. Or do average power integration of P into DT divided by integration of DT. This will also work. This will also work. There is no problem. Okay. Next. instantaneous power is nothing but rate of work done instantaneous power is nothing but rate of work done It's a formula of velocity. Oh, brother, it's written in mass. There is no mass here. Is there any mass written in the previous one? Let's see. I won't leave you wrong. You eat your life. Oh, brother, brother, brother, brother. This 2R will come here. 2gr cos theta minus 1. Where will mass come in the formula of velocity? Where will mass come in the formula of velocity? By the way, I had taken out here by COME. Where is mass? Where is mass in the formula of velocity? Where is the mass? 2G L cos theta minus 1. V0, cos2, GL, cosθ-1. You won't get mass. Brother, you made a drive twice. It was a little bit of a printing mistake. Did you make a drive twice or not? Tell me. Say sorry, Shaktimaan. You made a drive twice. Next goal. Listen to me. P is equal to dw by dt. What can I write dw? P into dt. Now listen carefully, what will be the slope of the work time graph? The slope of the work time graph will be power. The area of the work time graph will be nothing. The slope of the power time graph will be nothing. The area of the power time graph will be work done. Is this right? See, it is written here. See, it is written here. What? Area of power time graph is work. Area of power time graph is work. Slope of work time graph is slope of work. time graph is power slope of work time graph is power can you understand from here differentiation means slope integration means area you can write dw as f dot ds if constant force is out then ds by dt will be velocity so power is force into velocity dot word we can write it as fb cos theta if theta is positive sorry power is scalar but power may be positive negative and zero suppose theta is is 0 degree or is less than 90 degree then how will your power come? It will come negative. And negative power means that energy is coming out of it. If your angle is greater than 90, whose angle is it? Angle between force and velocity. Instantaneous power is every cost hitter. Second thing, remember that if you ask the power of any engine, remember that engine's power is directly proportional to the power of the pump is directly proportional to V cube. And the force that the pump is applying to flow the liquid, that force is proportional to V square. For example, let's ask you this question. Let's assume that there is a power P of an engine, from which X liter of water is coming out. If I want to take out 2X liter of water per second, then how much power will I have to do? I will have to do power 8 times. Because the amount of water will come out as soon as the velocity. Is this thing in your mind? Have you thought about this? Okay. Next, this is a question from NEET 2016. A question, pay attention. Body of mass 1 kg begins to move under the action of a time dependent force F, where I and J are the unit vector along X and Y axis. What power will be developed by the force at the time t? Now see, he is asking you whether it is force or power. No problem, listen carefully. This is force. How much? 2Ti cap plus 3T square j cap. If I get velocity, then power is the dot product of force and velocity. Power is the dot product between force and velocity. Does it happen? Yes sir, it happens. Now I know the force here. Can I calculate the velocity? Yes, because the mass is 1 kg. If the mass is force divided by acceleration, then the acceleration will be this. 2Ti cap plus 3T square j cap. And if you integrate the acceleration, then your velocity will be this. Don't do anything, integrate it. Can you write the result of direct integration without writing? Yes. It is 2T, so its integration will be 2T square by 2. This is 3T square, so its integration will be 3 times. The integration of T square is 3T cube divided by 3. 2 from 2 will be divided by 3 from 3. This is your T square I cap plus T cube J cap. This is your velocity. Sir, very good. Let's multiply this velocity with this. So 2T to T square. So this will be 2T cube plus 3T square to T cube. So it will be 3T 5. 2T cube plus 3T 5. 2T cube plus 3T 5. This will be your answer. Hi, everyone. This will be your answer. This will be your answer. Tell me yes or no. Yes. Tell me yes or no. Is it clear? Next. Next question is from Mains 2012. What? A car of mass M starts from wrist. acceleration so that instantaneous power delivered to the car has a constant magnitude P naught. Then instantaneous velocity of the car is proportional to. So initial velocity, sorry, instantaneous velocity of the car. Such questions are asked a lot. Listen carefully. Listen carefully. Is power constant? We can write power as force into velocity dot. Now, power is given as constant. It's okay. We write constant p naught. We can write force as dv by dt into m. Mass into acceleration. m into dv by dt into v take this dt here p0 into dt equal to mv into dv integrate both sides p0 out, dt's integration is t equal to m out, v into dv's integration is v square by 2 ignore the constant can I say v square is proportional to t so v is proportional to v is proportional to root t second answer See how easy it is. Power half. Finished. Next, see. A body is projected vertically from the earth, reaches the height equal to the earth radius. Before returning to the earth, the power exerted by the gravitational force is greatest. See, when you were throwing it up from here, its speed was maximum. Gradually, the speed was decreasing. Here, it became zero. Then, when it came down, it was running here. Gravitational force was falling down here. So, what will happen here? MG. V into cos 180 degree because force and velocity have the same angle here the power will be zero because the velocity is zero when it will collide here, then the velocity below and the force below here the power of gravitational force sorry I am putting the vector on the power, mg into V0, so where will be the power maximum? just instant the body is projected, no it remains constant all the time, the velocity is changing at the instant just before the body hits the earth now you, assume When the body was hitting the earth, then the power was there. And when it was thrown, then the power was there. Are the magnitudes same in both places? Yes, the magnitudes are same in both places. But one is plus and one is minus. So if you have to choose between the two options, then your second option is the best. By the way, while throwing and when hitting, the magnitude of the power is the same in both cases. Next, look, son. Ask the question. Engine pumps water through the horse pipe. Water passes through the pipe and leave it with velocity 2 meters per second. When Mars per unit length of the water of the pipe is 100 kg per meter. What is the power of the engine? See, in this kind of question, if he asks you the power of the engine, then remember the power of the engine will be mass per unit length into VQ. Mass per unit length is 100 and velocity is 2Q. So, it will be 100 multiplied by 2Q which will be 8. So, this will be 800 watt. up your answer will be fourth. Okay? But sir, A is also sometimes given, wrongly given. Yes, if this is given, rate at which, rate at which, at which, kinetic energy, supplied, to the, to the water, then this will be, half mu V cube. This will be 400 watts. Meaning, if we ask the power of the engine, So that will be µVQ. µ is mass per unit length. If you are asked to tell the rate of kinetic energy, the rate at which we are providing kinetic energy to the water, tell that rate. So that will be µVQ. Okay, sometimes you are asked to tell the average power. If you are asked to tell the average power, what are you asking? If you are asking the average power, then also you will write If you are asking the instantaneous power, then you will write µVQ. Clear? If you are asking the power of the pump, then you will write µVQ. Clear? Clear? So, we have done some questions about power from here. Now, this portion ends here.

No break will be found. If I keep giving break, then by combining basic math and vector, Total chapter becomes 16. If I keep giving breaks, then I will have to give 8 breaks. If I keep giving breaks, then how many breaks will I have to give? Calculate and tell me.

If more than 8, then I will have to give breaks. If more than 8, then I will have to give breaks. Then only breaks will work.

Then when will we study? Then we will play breaks. Every two hours break. So we will play break-break only.

When will we study, brother? Oh brother, if we play break-break, then when will we study, my dear? My blue, my yellow, green, black, eggplant-like face. Tell me. If we play break-break, then when will we study?

When will we study? He will do the dinner again. Just eat this rice. We are going.

We are going. When I talk to you. Any force applied anywhere.

So remember, I am telling you this very good trick. Whenever force is applied on any object, you will not directly know the effect of that force. You will only know the effect of force when you have given time along with force or how long the force has been applied. Sorry, how long the force has been applied.

Whether you have given time or displacement. If you have given force and time, then remember. by multiplying force and time, we get change in momentum and if we know change in momentum then we can calculate the velocity what will be the kinetic energy?

kinetic energy will be P square by 2m then we will have a lot of stories and if you have given force and displacement then you can calculate work done from here which is F vector dot S vector ok so here we will calculate work ok, if you know work then you know change in kinetic energy is work done and kinetic energy is known, then the momentum is root under 2m. So, if time is given with force, listen carefully, if time is given with force, then where will the approach go? Change in momentum will be shown in the laws of motion. If displacement is given with force, then what will come out? Work, work can remove kinetic energy, kinetic energy can remove momentum.

Then whatever you have to do, mint, coriander, ginger, keep doing from here. Okay, work done. It is F dot S. Remember, S is nothing but displacement of point of object where force is acting. Here S is not a normal S.

It is the displacement of that point where force is acting. Work done depends upon frame. Yes, work done depends upon frame. Work is scalar. Work may be positive, negative or zero.

If your angle theta is equal to zero degree or greater than zero degree but smaller than 90 degree, then work done will be positive. and if your angle is exactly 90 degree then work done will be zero if angle is greater than 90 degree but is equal to 180 degree or smaller than that then work done will be negative this is angle when work done is positive and negative if work done is positive then what is its physical significance kinetic energy of object increases and speed increases if work done is zero then kinetic energy is constant and speed is constant if work done is negative then remember kinetic energy will decrease means When we do work positively, we inject kinetic energy. And when we do work negatively, we take out the kinetic energy from there. Understood?

Understood? Do you understand this? Do you understand?

He is saying, there are only 10k students. There are only 10k seats. Understood?

Sir, sir, see. There are only 10k students. The time is also going to be 12. And the seat is also 10k. Oh, sir, good.

Read it, son. Small and big. Now look carefully.

Neat is a question from 2016. I have already done this. But this is in a different variety. Your question is, a particle moves from a point this much to this much. This much to this much.

Where a force of this much Newton is applied. How much work has been done? Now listen carefully.

Sir, work done is the dot product of force and displacement. Now, 4 is given to us, 4i cap plus 3j cap. But what is the displacement? So, listen, this is the final position minus the initial position will be the displacement. So if you subtract this from 4j, then it will be 5j, which means minus j.

Now 0i is not there, if you subtract this from it, then it will be plus 2i. And if there is no k from 3k, then it will be plus 3k only. So this will be the displacement.

I with i dot product, 4 times 8. J with j, 3 with this, then minus 3. And if there is no k, then k dot product is 0. 8 minus 3, 5 joule work. What will you do? Final position minus initial position.

What will you do, son? Final position minus initial position. Clear? Next.

A body moves from the distance 10 meters along the straight line under the action of a 5-Newton force. And work done is 25 joules. Then angle bit.

Work done is Fs cost theta. You were given work, force and S. So, won't the cost theta come out of here?

It will come out very easily, sir. It will come out in a smokey way. Neat 2017 question. Read the question.

Consider a drop of rain, water falling, mass 1 gram falling from the height 1 kilometer. Means 1000 meter. So 1 gram of an object from the height of 1 kilometer hits the ground with a speed 50 meter per second.

Take g constant with a value of 10 meter per second squared and work done by the gravitational force. So how much work of the gravitational force will be done? Work done by the gravitational force will be Mgh. M value is 1 divided by 1000. We will change it in gram and kg.

G value is 10. H value is 1000. If you cut 1000 from 1000, you will get 10 joule work. So 10 joule will be your positive work. This is wrong.

This is wrong. This is wrong. This is right. Oh brother, so third answer. But there is more part in this.

How much work will be of resistive force? This is what smart work is. Why do we have to do more politics in this? When I... Work done by gravitational force is showing that only one option is right.

Work done by gravitational force is showing that only one option is right. So do I need to do more politics in this? No sir, no politics will work in this. No politics will work in this.

Yes sir, look at the next. This child, you don't have to do. This is a question of NEET 2019. You just have to see. Force is equal to 20 palace 10y.

X on the particle means force is applied in the y direction. in y direction. Then force F is in Newton and y is in meter.

Work done by this force to move the particle from y equal to 0 to 1 meter. So we know that if there is variable force, then work into displacement, we remove it like this. F dot dx works equally.

Clear? Is it a neat 2019 question? Have you done it in class? Do this and show. Neat 2023 question.

Do this. And see if the same question is asked or not. Same question is asked. Means this happened.

that the questions asked by NEET, the first questions which were in JEE main are now coming. And the kids say, Sir, will the question of JEE come in NEET? Do the question. And try typing it.

This is the question of JEE main. And now the question of NEET has been copied completely. Do it.

Sir, there was y in it, so we just made y square. So, f is equal to 5 plus 3y, 1, 0, particle in the y direction, f is in Newton. And y is in meter, then work done by the force during the y equal to 2 plus, y equal to 2, 2, 5. Do it. How will we get the work of resistive force? That is easy.

We will have to use the work energy theorem in that. Where is it? I will tell you.

In this we will get it like this. Work done by gravitation plus work done by friction equal to change in kinetic energy. Now initial velocity is zero and final velocity is given. So change in kinetic energy will be obtained. We know this.

So work done by friction will be obtained. Is it clear? If you are interested then do it. I will teach you to do it.

By the way, take it out. I will teach you. Come on, do this.

132, 117. Now decide, brother, whether it will be 132 or 117. Decide, brother, whether it will be 132 or 117. Then talk. Decide, brother. A force of 5 plus 3y square x on a particle in the y direction where f is in Newton and y is in meter. Then work done by the force during the displacement from 2 meter to 5 meter.

132, 132, 132, what will you do? F into dy equal to dw Because force is only in y What is the value of force? 5 plus 3y square into dy equal to dw Do the integration, this work will come 5 into dy will be 5y plus 3y power 3 by 3 This will be cut You will keep the limit from 2 to 5 So if you keep 5 here, then 5 first 5 then 2 So here comes 3 plus y power 3, 125 minus 2 power 3, 8. 5 times 3 is 15, plus 8 out of 125 will go, 117. 117 plus 15, so 5, 2, 7, sorry, 7, 5, 12, 2. 1 goes here, so 2, 10, 1. Is 122 coming?

No, how will 122 come? 117 minus 15 is 0. Is 132 coming? 2, 1, 3. Yes, 132 is coming. So what were you telling earlier, 117 and this?

you calculated this term, did you shoot it? 132 joules. So, will the dot product of work also be a question?

Will the dot product of work also be a question? Did you enjoy? Did you enjoy, brother?

Now, see, work is asked in this. So, you know the area of force and displacement graph is work. So, directly removing its area will tell you work. Directly removing its area will tell you work.

Next, see. Next, see. If kinetic energy of the body, is increases by 300% then percentage change in momentum will be Now see carefully, what is the question?

If kinetic energy of the body is increases by 300% then percentage change in momentum If kinetic energy is increased by 300% then how much change will come in momentum? So first of all p is equal to root under 2mke Is it clear? Okay Now p is proportional to root k will come? Yes sir If you increase kinetic by 300% So finally how much will be 400%? 400% means?

Final kinetic energy or I can write 400% of initial kinetic energy If kinetic energy is something ke, if you subtract 300% from it, then finally how much will be? 400% If you subtract the percentage, then you will divide it by 100, which means 4 ke So the final kinetic energy will be 4 times of initial So final momentum is proportional to final kinetic energy So final momentum is equal to Final kinetic energy will be 4 ke So 4 2 will go out, root under k. Root under k initial can be called initial momentum.

So final momentum is 2 times of initial momentum. Can you find this relation? Meaning, if it doubles, then it will be 100% bigger. If it doubles, then how much percent will it be bigger? Tell me.

If it doubles, then it will be 100% bigger. Yes or no? Yes or no?

Next, see. 90, 98, 97, 90, 89, we have asked this 3 times. 2 body of mass m and 4m moving with equal kinetic energy.

Then ratio of their linear momentum. Momentum is equal to, it happens also. If its mass is m and 4m, then it will not be 1 is to 4. Why? Because momentum is equal to root under 2mke. If their kinetic energy is same, then it is same.

So momentum is proportional to root m. First mass m, second mass 4m. 4 times mass. So if you keep 4 in root, then it will be 1 is to 2m. What will be the answer?

The first answer will be 1 is to 2. Do you understand this well? If you understand this well, then ask this question. It is the question of JEE MIN 2022. And you will be sure that such questions which have come in JEE MIN, have come in the first NEET.

They have come in the first NEET, sir. And they have come three times. That is being copied in JEE MIN. It is the question of JEE MIN 2022. You type it and check it.

I am not joking. I am not joking. You type and check whether there is a question of J-E-Main or not. You type and check whether there is a question of J-E-Main or not.

Tell me. Body of mass 8 kg and another mass of 2 kg are moving with equal kinetic energy. Then ratio of their momentum is 8 and 2. Equal kinetic energy.

So what will be the ratio of momentum? Root M. So, root 8 divided by root 2, so 4. If it goes out of 4 roots, then 2 is 2 1. The second answer will come. The second answer will come.

He said, why didn't you do it with P square by 2m? That's what I did, dude. Because kinetic energy is P square by 2m. Kinetic energy is the same.

So, P square is proportional to m, means P proportional to root m. Idiot, biology. Ramlal Ekdam. Look, what is work energy theorem?

Work energy theorem is that work done by all the forces is equal to change in kinetic energy. And this is the father of all the theorems. This is always valid.

Always valid. Who is the father of all the theorems? Everyone's father is the same.

Everyone's father is the same. Who is everyone's father? Everyone's father is your work energy theorem. Work energy theorem says, Work done by all the forces is equal to change in kinetic energy. Work done by all the forces is equal to change in kinetic energy.

Second, work done by conservative force is equal to minus change in potential energy and this is also always valid. This is the definition of potential energy. Second, work done by non-conservative force is equal to plus delta U and this is not always valid.

This is valid only when change in kinetic energy is zero and non-conservative force is acting against Conservative force. Non-conservative force acting against the conservative force. That's when it's valid. That's when it's valid. Clear?

It's wrong. The question is from 2016. Okay. Where is it?

Where did 2016 go? Which question is from 2016? This one.

Let's check. I will do it according to the information given. It may be misprinted, I don't know. If you subtract this from 4j cap plus 3k cap, 4j cap plus 3k cap, this is the final position.

Particle moves from a point this much to this much. So this is the final position minus this initial position. If you subtract this, it will be plus 2i and minus 5j. So this will be, from 4j, j will go minus j plus 2i plus 3k.

minus z 2i plus 3k so it is correct this is the question of 2016 so remember these things I will write it again that work done is equal to change in kinetic energy and if you know work done formula is force into displacement If you apply the same force F at the same distance on any object in the world, then the kinetic energy that will develop in it will not depend on the mass. Listen carefully to what I am saying. Can I make this graph of kinetic energy versus mass for any condition?

Yes. If the work is the same on any object in the world, if the work is the same on any object in the world, the developed kinetic energy in the object will not depend on the mass. Tell me.

The developed kinetic energy in that will not depend on the mass. Is it clear, my brothers? Is it clear, my brothers?

Now see. Body of mass 3 kg under the constant force causes the displacement S in meter and given the relation S is equal to 1 by 3 T square. Where T is the second, then work done by the force in 2 seconds is work done by the force in 2 seconds is listen carefully now here work done if you go to remove f.s then you will not be able to remove work done remove from change in kinetic energy sir change in kinetic energy i can write final minus initial final kinetic energy half m v final square minus half m v initial square so from here half m will be removed from your common v final square minus v initial square okay this displacement is given here velocity and differentiate between them.

So here 1 by 3 outside, the differentiation of t square will be 2t. So this will be 2t by 3. Is it clear? Yes sir. Now I asked you, work done by in 2 seconds.

So what will be the initial velocity? If you keep t 0, then 0. And what will be the final velocity? Sir, if you keep 2 time, then 2 times 4, times 3 will come. When you take out the work here, then half, what is the value of mass? 3 kg.

3 into, final velocity 4 times 3 whole square. 4 divided by 3 whole square initial velocity 0. So, what will be 4 whole square? 16 divided by 9. So, this is 3 multiplied by 16 divided by 9 times 2. 3 divided by 9 will be 3 and 2 divided by 16 will be 8 divided by 3 joules.

8 divided by 3 joules is equal to correct answer. So, remember that many times we have to get work done from change in kinetic energy. How did we get it? We had given displacement, we differentiated and got velocity. Final and initial velocity is known, then work done is equal to change in kinetic energy.

Work done is equal to change in kinetic energy. Here the question is asked, the kinetic energy acquired by the mass m, in travelling a certain distance d, starting from the rest under the action of a constant force f is directly proportional to. Meaning, kinetic energy, how will it depend on the mass? So kinetic energy does not depend on the mass. When the force and distance applied on any object is same, dia hoto.

Because Carnatic energy is equal to work done. and work done F dot S will be done, so kinetic energy does not depend on mass. Next see, 3 joules of work is done in the sliding 2 kg block of the inclined plane of height 10 meter.

Then work done against the friction is? Okay, okay, okay, okay, okay, okay, okay. Understand the question properly.

I have asked it three times, I have asked it twice, it is an important question. 300 joules of work is done in sliding a 2 kg block up an inclined plane of height 10 meter. data. means the height of this inclined plane is 10 meters.

Now to take it from here to here, the work done is 300 joules. So I have asked you, the work done against friction is? Work done by gravity plus work done by friction is equal to change in kinetic energy. So kinetic energy has not changed in this.

So total work done should be zero. work done by gravity Okay, okay, okay, okay, okay, okay. The total work done in this is against gravity and against friction. Okay, okay, okay, okay.

So the total work done is 300 joules. And the work done against gravity will be MGH. Say M. Say G. Your height is 10. Sir, 2 kg. M is 2 kg. So 10 times 100 is 200. So 200 joules which is the work What did we do against gravity to take it up?

And 300 out of 100 joules would have been against friction. If there was no friction, if there was no friction, then I would have to work against gravity to take this block from here to here. And work done by gravity does not depend on path. Work done by gravity does not depend on path.

Listen to this question carefully. Work done by gravity does not depend on path. Does not depend on path. If there was no friction, So I have to work against the gravity of the work. But gravity has worked so much against gravity.

And it will be a negative work obviously because the body is going up. And the total work done is 300 joules. So I have to work against gravity of 200 joules.

I have worked against friction of 100 joules extra. A body of mass 1 kg thrown upward with a velocity 20 mps. It momentarily comes after the...

comes to rest after attaining a height of 18 meter. How much energy is lost due to friction? It's a good question. It's a question of work energy theorem.

It's a question of work energy theorem. See, body 1 kg is thrown upward with a velocity 20 meter per second. Means, I threw a body from 20 meter per second.

Now, sir, it has stopped after 18 meter. It has stopped after 18 meter. So, I have asked you, it. work done by air friction, listen total work done by gravity I have thrown it, I am not taking it gravity plus friction, equal to change in kinetic energy what will happen?

if object is going up, work done by gravity will be minus mg plus work done by friction, we have to remove it, so we will not write plus minus is equal to final kinetic energy is zero, initial kinetic energy will be half m is also the whole square of initial did I write the statement correctly? yes sir so work done by friction will be minus Half, M's value is 1 kg. Velocity is 20, means 20 squared is 400. Plus, this will come here. So 1, G's value is 10 and height is 18. So if we cut this from 2, minus 200 plus 180. So how much will this be?

Minus 20 joules. Work done by friction, you have got 20 joules. Just ask in magnitude.

Is it clear? Work energy theorem is a good question, a good example. Next.

Look, if you understand this carefully, then we have two types of force. One is our conservative force and the other is our non-conservative force. One is our conservative force and the other is our non-conservative force.

Remember in this that work done does not depend on path. Conservative force's work does not depend on path. What is its work on the closed path?

Zero. The concept of potential energy is only defined for the conservative force. Its work is reversible, i.e. work is not lost. Its example is gravitational force, electrostatic force, spring force. But the work of non-conservative force depends on the path.

In the closed path, its work can be zero or not. The concept of potential energy is not defined for it. Work is not reversible.

For which the term of potential energy is not defined. They are not. non-conservative hota hai. Jaise friction force, tension force, viscous force, normal force.

Yeh sab non-conservative ke category mein aata hai. Is it clear? Here you can write work done by conservative force is equal to minus change in potential energy. Work done by conservative force is equal to minus change in potential energy.

Is it clear? Next, the potential energy of the system increases if work is done upon the system by the conservative force. Look, if the conservative force works, Like the gravitational force is conservative. If its work is positive, then kinetic energy will increase.

If its work is positive, then kinetic energy will increase. Potential energy of the system increases if the work is done upon the system by the conservative force. If the conservative force works, then kinetic energy will increase and potential energy will decrease. This will not happen.

By the system against our conservative force. By the system. against the conservative force, then the potential energy will increase. This is correct.

By the system against the non-conservative force, if you work against the non-conservative force, then the potential energy will not come. Upon the system by conservative force. On the system by conservative force. Not upon the system by non-conservative force. We will not talk about the non-conservative force.

According to me, there should be a second answer. What do you say? What do you say?

Okay. One is saying that 148 slides have been done. Brother, I have studied 250 slides and kept them there. Right? Look ahead.

A spring of force constant K is cut into two pieces such that one piece is double of its length of the other. Then the long pieces will have force constant. Look, there is a spring whose length is L and its spring is constant K. If two pieces such that one piece is double the length, then the longer piece will have the force constant is equal to. Do it.

It's a good question. Here, look at this. spring constant which is inversely proportional to length.

So, what will be the spring constant of the longer piece? This is what I have asked. Let's see if you can do this or not.

It is a good question. It will seem like a good concept. It is a good question.

It will seem like a good concept. This question will not be solved by normal concept. Is this clear? Oh, brother!

How will you do it? Tell me. How will you do it?

Tell me. So see, two pieces double of its length. If I take its length L dash, its length is 2L dash. Okay, sir. Now here, the one whose length is less, its force constant will be more.

So if I consider its spring constant to be K0, then its will be 2K0. Is this clear? Yes. Okay, tell me one thing.

I have assumed this. Length L was of spring constant K. If length L dash is of length 12 dash, I have to divide this into two parts. So this will be 2K0, this will be K0. Sir, done, understood.

Now what he is saying is, tell the value of the spring constant of the longer piece. So tell the value of K0. Now listen carefully here.

If I add these two again, then spring constant will be K. And how do I add both? I will have to add them in series. And the value of spring constant in series is 2K0 multiply K0 divide.

3k0 is equal to k. In series, parallel is similar to spring constant. Oh wow sir, so k0 will be cut from k0?

Yes. So k0 is equal to 3ky2. 3ky2 will be the answer.

Means second answer. And till now no one is giving second answer. No one is giving second answer.

No one is giving second answer. And first, there is a strong reason for being wrong. What?

Jab mein I am seeing L length's spring constant K. If I reduce the length, then the spring constant will increase. And this is 2 by 3K.

So it is wrong. This is less than K. Suppose this is L length's spring. If I break this, then both the lengths are less than this. So both the spring constant will increase.

In this, spring constant is telling 2 by 3K. So it is telling less than K. So this cannot be the same.

Tell me, did you understand it well or not? Did you understand it well or not? Did you get a good question or not? Did you get a good question or not? Tell me, next.

See this question, son, J-Man 2022. And literally the language of this question is this only. You can solve it without a pen. Without pen use, solve it and show it, son, it will be fun. Means continuously studying for 9 hours will be beneficial.

Continuously studying for 9 hours will be beneficial. Continuously studying for 9 hours will be beneficial. Tell me.

Solve it without pain. Read the question and solve it without pain. The question of 2-1-1 is wrong.

No problem. If the question is wrong, then leave it. Our sole inner figure, particle of mass 10 kg, placed at point A, when the particle is slightly displaced along the right, it starts moving and reaches the point B. The speed of the particle B is x m per second.

Then value of x is, see, smooth surface, no mention of friction, 10 m height, how much it fell, 5. How it fell, no matter. Velocity will be root under 2gh. 2, G is 10, H is 5. So, root 50. Sorry, root 5 times 10 is 10. And root 100. Root 100 means 10 m per second. How many people got 10 answers? Brother, tell me.

This is a question of J.E. Mendoza. This is a question of J.E.

Mendoza. This is a question of J.E. Mendoza.

This is a question of J.E. Mendoza. This is a question of J.E. Mendoza.

Tell me, did you solve it without pain or not? Tell me. It's a 6-month question. It's a 2022 question.

Tell me, did you solve it without a pen? Even such pendulum-like people ask questions. You'll blow their heads off. Brother, what is this?

See what it is. What is this? The more times I have asked this question, it seems that this time it should be asked again.

IIT, NMR, AIEEE, MPPMT, 94, 97, 2000, JIPMER, 2000 and IIT have asked it many times recently. Recently IIT JEE main me kai baar pucha hai. JEE main me kai baar pucha hai.

Kya hai yeh sawal? Aav parthe hai. Aav parthe hai. Agar tumne work energy power kia aur yeh sawal nahi kia matlab kuch galat kia. Aav parthe hai.

A uniform chain of length L and mass M. Uniform chain hai. Jiska mass M hai length L hai.

Is lying on a smooth horizontal table. And one third of its length is hanging vertically down. Means, from M mass, M by 3 mass is hanging down.

L by 3 will be hanging down. Over the edge of the table. If G is the oscillation due to gravity, then work required to pull the hanging part on the table is. So, how much work will be required to keep this hanging part on the table? So, the first question that develops here is, who will work?

Who will work? and work energy power I am teaching so if I don't say a very famous statement then there is no fun in it where we are there is no less how many people know this where we are there is no less where less where less there there we are not what is this less less is nothing but conservation of mechanical energy meaning meaning Initially kinetic plus initial potential. Equally. Final kinetic plus final potential. This is the beta of work energy theorem.

Is it valid everywhere? No, sir. It is valid only when work done by non-conservative force is zero. Do you know well? Now tell me, who will bring the chain up here?

Sir, we will bring the chain. And when our work done is against gravity, can we write it directly as change in potential energy? Yes, sir. We can write external force's work done as change in potential energy.

Let's write change in potential energy. Tell me. Final potential minus initial potential.

Now if I put the chain on the table and consider this as the potential energy reference. Then what will be the final potential? Zero. What will be the initial potential?

So the part on the table has a potential of 0. What is the lower part? m by 3. So m by 3 into g. And the center of mass of the extended part is its center. Sorry, the potential is at the center of mass. So how much is lower?

ly 6. And lower that's why minus will come. Minus minus plus mgl by 18. mgl by 18. Fourth answer will come. What will be the answer?

mgl by 18. NGLY 18. Is everyone's correct? Very good. Next, see.

Such questions come in the lead. The potential energy of the long spring when stretched by 2 cm is U. Then the spring is stretched by 8 cm, then potential energy is stored in it. See, here the first thing to be read in the question is whether the change in potential is asked or the final potential. The potential energy stored is half kx square.

He said that 2 cm is u, so we will keep x as 2. So k divided by 2, x is 2, so we will get the factor of 4. And if we cancel 4 by 2, then it is 2k. So 2k is called u. Now if the spring is stretched by 8 cm, then potential energy is.

So the final potential energy will be half k, whole square of 8. So 32. 32 means 8 is 64 by 2, 32k. 32k will come. If you can write 32k, then 16 into 2k.

And if you write 2k as u, then 16u will be your answer. Now 16u is fine, there is no problem. 16u will be there.

But many times the question is, if you stretch 2cm, then the potential energy u comes. If I stretch 8cm, then what will be the change in potential energy? Tell me, what will be the change in potential energy?

15u. Change in potential energy will be 15U. Is this clear?

One more thing. One more thing. One more thing.

What? Look. Suppose there are two springs. It is very important.

It is a question of 2015. One spring constant is K and the other spring constant is KB. Okay. Now we know that KB's value is more than KA.

And we applied the same force on both. Tell me in which potential energy will be stored more? In A or B?

This is given to us. If Kb is greater than K, then in which potential energy will be stored more? In A or B? Tell me. In which potential energy will be stored more?

In A or B? Tell me carefully. In A, will potential energy be stored more or in B?

Now half of the children are saying A, half of the children are saying B. Why? Because potential energy is directly proportional to K. See, there is a spring whose spring constant is k, so I applied force on it.

There is a spring whose spring constant is k, so I applied force on it. If kb is greater than k, then in which case more energy will be stored? We have to compare. I applied force on both of them, so see the solution.

The formula of potential energy in a spring is half kx squared. Now both have different spring constant. I applied force on them.

If you put the force same, then both the elongations are same, so it is not necessary. No sir. So what is the relation of elongation? If equal to kx. So from here we can write x as if by k.

So this will be half k. If by k. If by k's whole square. So this will be half if's square, if's square. This k's square, if k is cut, then k will come down.

Matlab Potential energy is inversely proportional to K. So the one whose spring constant is more, energy will be stored less in that. Everyone's Lankan has been set.

Is it done? Oh my God! Everyone's Lankan has been set.

Paltu people, Paltu people. See, I repeated it again and again. Same force is being applied.

This is the question of AIPMT 2015. And we were worried that potential energy is square from k. Yes, it happens when we write it in terms of x. In terms of force, it will be inversely proportional to k. And we have taught you this thing. Whenever we taught you in class, we taught you this thing.

This is the question of J.E. Main, 2023. Brother, I remember, by mistake 95 is stuck here. Or it is asked in 95, it is also asked in NEET.

Or it must be written somewhere again. I will find it and bring it. It is not written. It is not written. This is a question of J.E.Main 2023. Or 2022. Something like that.

What are you talking about? Yes. Now tell me honestly.

This question which you are seeing. Is it asked in NEET or not? This question which you are seeing.

Is it asked in NEET or not? Tell me this. This question which you are seeing.

Is it asked in NEET or not? Yes sir. This question is asked in NEET. Now see. It is asked in J.E.Main.

How will 15U come? When you ask for change, then 15U will come. Now it will not come like this. When you ask for change, then 15U will come. Like this, 15U will not come.

Come on, ask this question. Come on. So when is stable equilibrium?

When your force is zero, then stable equilibrium happens. So what do we write the value of this force? dU by dx. So do 0 to dU by dx.

Do 0 to dU by dx here. So this will be your 12, 12, 12, 12, 12a divided. x to the power of 11 is equal to 6b divided by... x to the power 5. I will subtract, right?

5 will come. Okay? I have kept this equal. So, I will subtract 12 from 6. 2. 2a by b is equal to x to the power 11 divided by x to the power 5. 11 to the power 5 will be x to the power 6. 2a divided by b is equal to x. What will come here?

2a divided by b is equal to x to the power 1 by 6. First, you don't have to do anything. You have to put 0 by doing differentiation. By doing differentiation, you will not get 13 by 7. You are doing wrong.

Look, what he is doing, in differentiation, one minute, you are right, you are right, you are right, you are right, you are right, you are right, you are right, you are right, you are right, you are right, you are right, you are right, you are right, you are right, you are right, you are right, you are right, When we take it up, then if we subtract minus 12 from x power, then minus 13 is there. Oh sorry Shaktimaan. I am Ghanghor Biology, you are Ghanghor Math.

So now here will come cut XY6. You are saying right. You are saying right. I am wrong. You are saying absolutely right.

I did wrong. I did wrong. You did right. Sometimes you will also do right.

14 will come. Hey, 14 will come. I will take out my shoe and hit you now.

You are getting excited. Once you do the right thing, you will be 14. Or else what? Once a question gets right in centuries, then it goes inside.

Then you will be 14. How will you be 14? How will you be 14? First, water your mouth.

Half of you are watching the lecture in sleep. Come on, come on. Water.

If you don't like the lecture, then like it. 50,000 likes should be done live. 50,000 likes should be done live.

A particle with total energy E is moving with a potential energy Ux. Motion of the particle restricted to the region. See, where motion cannot be done. Generally, what we know is that total mechanical energy E is kinetic plus potential. Yes sir, kinetic plus potential.

So what can we write as kinetic energy? E minus U, E minus U. If you think E minus U. If it happens that U is big and E is small, then sir, if it is small, it will be negative. Whereas kinetic energy is not possible.

Kinetic energy cannot be negative. Kinetic energy cannot be negative. So if kinetic energy cannot be negative, then U cannot be bigger than E. U cannot be greater than E.

It can be equal. But if U becomes greater than E, then kinetic energy will be like this. I will explain again.

Look, son, kinetic energy plus potential energy is equal to E. Yes, sir, it is. What do we write as kinetic?

Sir, E minus U. If U is big according to this, if U is big and E is small, if U is big and E is small, then kinetic energy would have been negative. And kinetic energy is not negative.

So this condition is not possible. So he is saying, which is restricted to the region. Meaning where motion is restricted. So brother, kinetic energy is negative, this is not possible. And that will happen when U is big.

Meaning U is big, this is not possible. Tell me, tell me. 14 will be 16 minus 2. Where is 16 minus 2 14 happening? Where? Okay, 16 minus 2. If the spring is a stage where 8% potential energy is stored in.

I didn't ask about change. I didn't ask about change. I didn't ask about change.

If I had asked about change, I would have been 14. I will take off my shoes and beat you. You fool. This is U, not 2U.

This is 16U, this is U. So how much change will there be? 15U. You fool, biology. You fool, biology.

You are a fool. Biology. See next. Mass of 0.5 kg moving with speed 1.5 m per second on the horizontal smooth surface. Collide with a nearly weightless spring of force constant, this much.

then maximum compression of the spring is, listen carefully, is there friction here? No sir. If there is no friction, then you can directly apply initial kinetic plus initial potential, equal to final kinetic plus final potential. Say, if I consider the reference of gravitational potential energy on this horizontal surface as 0, if I consider the reference of gravitational potential energy on this horizontal surface as 0, then its gravitational potential is 0, its gravitational potential is 0, and its spring potential is also 0. So, half mb squared will be saved from kinetic energy.

Equal to, at maximum compression, its kinetic is zero and potential energy will be half kx squared. Son, k is also given, mass is also given, its velocity is also given. Will x be removed from here? Don't waste time in small calculations. Now, we understand vertical circular motion.

Brother, my vertical circular motion, suppose, this is the length string, a mass is attached, v velocity is given. If given velocity is V0, then after any theta angle is taken, what will be its velocity? Can you find it?

Yes sir, you can find it very easily. How? How can you find it? With the help of the conservation of mechanical energy.

Come on, find it. I am not joking. Assume, with the help of the L length string, I have given given velocity V0 by tying the object of M mass. So when it rotates at theta angle, then how much will be its velocity here?

Tell me. Come on, tell me. How much will be its velocity here?

Tell me. Everyone will tell you. Okay. I don't have any happy birthday. Everyone will tell you.

Tell me. So let's do one thing. Here let's assume potential energy reference is zero.

So its initial potential energy will be minus mgl. Kinetic energy will be half mba2. Equal.

Final potential energy will be, length is L, so vertical minus MGL cos theta. Kinetic energy will be half M V square. M, M, M, M will be cut. Minus G cos theta will go there, so plus G cos theta will be there. So G cos theta, L, L, L, L, G cos theta minus GL plus V square by 2 is equal to V square by 2. If you cut 2 from 2, then your common will be 2. This was your initial velocity V0 square.

So final velocity which will come is V whole square equal to V0 whole square plus 2GL cos theta minus 1. This is the answer. This is the answer. You should know to apply conservation of mechanical energy.

What is the answer? V square is equal to V0 whole square plus 2GL cos theta minus 1. This is the answer. How many people are getting this answer?

Tell me please. How many people are getting this answer? Is it happening? Let it be 12 o'clock then say happy birthday. It is not 12 o'clock but he is saying happy birthday.

It is not 12 o'clock but he is saying happy birthday. He is a strange person. It is not 12 o'clock but he is saying happy birthday. Let it be 12 o'clock.

Now listen carefully. When it moves at a straight angle, at that time, how much tension will be there in this rope? So T will be equal.

Listen carefully. mv0 given velocity of the hole is square by L. mv square by L plus minus 2mg. I will put in the terms of v0. 2mg plus 3mg cos theta.

What will happen? If I keep theta 0, then cos 0 is 1. So, mb0 square by L plus mg. Right.

If I keep theta 90, then cos 90 becomes 0. So, mb0 square minus 2mg will come. What will happen when theta 90 is kept? Yes, tension will come. mb0 is always square by l minus 2 mb. Right.

And if I keep theta as 180, then it is absolutely right. How many people have verified this from MRSTART that the tension I had written on the generalized angle is right. Brother, I will have to agree. I remember things.

In given velocity terms, the tension at any angle, at any theta angle, will be there. mv naught whole square by L minus 2mg plus 3mg cos theta. It is correct.

It is correct. Did I write the velocity correctly at any theta angle or not? Tell me. Yes, so it should be minus.

It is correct. Both the velocities are correct. This is also correct. Correct, thank you, Nathie.

Printing mistake. 1 minus cos theta. No. Son, here 1 minus cos theta will not happen. Why?

Because this is negative and this term will go there. So, cos theta term will come in positive. This is also correct.

And I will kick. If I solve again, then I will kick. Idiot.

I am taking it right. The velocity at any theta angle will be given velocity of whole square. I will kick, see.

Given velocity of whole square plus 2mg cos theta minus 1, this will come. And this is correct. I will explain the reason for this. I will explain the reason for this. When I gave velocity V0 at the lowest point, then we have to get velocity at any theta angle.

So first of all think that I am getting velocity at theta angle of this mass. What would have happened if theta was 0? If theta was 0, then the velocity would have been equal to v0 If theta was 0, then the velocity would have been equal to v0 Is this logic being verified here? That if I keep theta, then the velocity will be equal to v0 Keep theta 0 first Because if you are looking at theta angle, then if theta was 0, then the velocity would have been equal to v0 Come here and tell me If I keep theta 0 here Then what will be the cos 0?

1, 1, minus 1, 0, this will end See v theta equal to v0 will come, first thing Second thing If I keep theta 90, then the particle will reach here. And the velocity here will be less than V0. Yes sir. So when I keep theta 90, then cos 90 is 0 and this will be minus 1. So V is equal to V0 whole square. If this is 0, then minus 2 mg.

Minus 2 mg. This will be the answer. And the right thing is, if you say here also minus and here also minus, then the velocity will increase. Whereas as the particle goes up, the velocity will decrease.

I have done the right thing. If I don't have any happy birthday, don't spam. You are still waiting, means you are students. So don't spam, I don't have any happy birthday.

Birthday comes very late. Tell me, did I write this velocity right or wrong? Tell me, if I have given velocity v0 at the lowest point, then the velocity I have written at theta angle is right or wrong?

Tell me once. If you don't understand, then open the notes and tell me. I don't have any happy birthday. Our happy birthday comes on 27th September. It's right or wrong, open the notes and tell me.

Secondly, if I write theta at any angle, then the retention will be mba2bl given velocity, lowest point velocity I am giving, mba2bl minus 2mg plus 3mg cos theta. It is possible that I will put cos theta here by mistake, it is possible that I will take this plus and this minus, but I know that when I take theta 0, then the value of tension at this point is t up, mg down. and below centrifugal force, if I am making FBT here then Mb0 whole square plus L, so what will be the value of tension? The value of tension will be Mb0 whole square plus L plus Mg when I will keep theta 0 Keep theta 0 here, keep theta 0 here, sub cos 0 is 1 2mg will be deducted from 3mg, plus Mg, Mb square plus Mg, tell me if you enjoyed it or not Second thing, the minimum, ok whatever tension is there here 3mg is less here, yes, 3mg is less here 3 mg is more than that then 3 mg is more maximum tension is here minimum tension is here do you know this?

do you know this? if I play the game of tension tension here if I play the game of tension tension here listen to me play the game of tension tension so the tension will be here 3 mg will be less than that yes or no? 3 mg will be less than that 3mg will be less the tension will be less here the tension will be less here from here 3mg will be less from here 6mg will be less minimum tension is at the top point maximum tension is at the bottom point here 3mg will increase here 3mg will increase here 3mg will decrease here 3mg will decrease here 3mg will increase here 3mg will decrease here and here the difference of tension is 6mg if it is written in tension mba square by l plus mg what is written?

mb square b L plus mg so the tension here will be mb square b L minus 5 mg why? because from here 6 mg tension would have been reduced how much tension will be there here? here tension will be there mb square b L minus 2 mg yes it will be there how? we have to reduce 3 mg from here mb square b L plus mg if we reduce 3 mg then minus 2 mg here 3 mg will be reduced 5 mg here 3 mg will increase here 3 mg will be reduced here 3 mg will increase, here 6 mg, here 0, here 3 mg will decrease, here 3 mg will decrease, here 6 mg will decrease, here 3 mg will increase, here 3 mg will increase, here 6 mg, here 0 say, say, will you remember this?

take, take brother Oh brother, I am also very smart. Here is the correct attention written. My God, but I swear on physics, I did not see.

You must be thinking, sir, you will become more intelligent. You see from here and write there. And you say that we are high. Oh, I did not see.

I swear on physics, I really did not see. I swear on physics, I did not see. Minimum velocity should be at the minimum point to complete the circular loop. And the minimum velocity should be at the bottom. The minimum velocity at the bottom must be at the root under 5gr.

And the maximum velocity at the maximum point should be at the root under gr. To complete the vertical circular motion. Where? At the maximum height. To complete.

Vertical circular motion. So complete vertical circular motion. Okay, brother? What will be the maximum tension?

mb square bl plus mg below. And the minimum tension will be at the top point. Now in this, let's ask such a question.

Where is the probability of the rope breaking the most? The tension is wrong. The tension is wrong. Let's see.

It is possible that something has gone wrong in the speed. Let's say it. No problem.

MBA square VL minus 2mg. Let's reduce 3mg from this. It's right. MBA square plus mg at the lowest point.

Where is it wrong? MBA square VL plus mg at the lowest point. It's right.

Maximum tension at the lowest point. MBA square VL plus mg. It's right. Print one is wrong. mb square of l plus mg.

Right. Up T, below mg, below. mb square of l.

T max minus T minimum 6 mg. Minimum will be mb square of l minus 5 mg. Right.

mb square of l minus 2 mg plus. Okay. Oh my God.

Who sees this minute? This is my brother. Come on. I thought you made some conceptual mistake. means I have made a big mistake who looks like this lets ask NEET 2019 question before that lets see one more thing if we get such a question in NEET that let me give you this this is a vertical loop and here I have given a v0 velocity to an object to complete this circular path how much minimum velocity should I give remember root under 4gr is there If you take the root out of 4, then it will be 2 root gr.

Why? Because this is your support from below. If you tie it with a simple string, then it becomes 5 gr.

Okay, let's take a massless rod. If you take a massless rod, what will you take? Massless rod.

Massless rod. And here you will tie the object of m mass and throw it with v velocity. So here also root 4 gr will come. Is it clear?

Come on, ask this question. A mass m attached to the thin wire and whistle in the vertical circle. Then wire will most likely to break wind.

What a joke! What kind of a joke is this? Let's move on. What? A body inclined at the wrist and sliding along the frictionless track from the height H.

As shown in the figure. Just complete a vertical circular of diameter AB equal to D. This diameter is given to us as AB equal to D.

So, what will be its radius? 30 by 2. Now, here the height H is equal to. So, the value of this H is asked so that When I release this ball from here, it will fall down and complete the vertical circular. The biggest thing is that when the particle reaches here, how much velocity will it get? Sir, the velocity it will get will be root under 2gh.

Yes or no? Yes sir. If the vertical height is h, then the velocity will be root under 2gh.

And the required velocity to complete it, what will be it? Root under 5G It is not a support loop. 5g into R.

What will I write R as? D by 2. If this is equal to it, then the vertical circular motion will be complete. Now if you square the roots on both sides, then write it simply.

2gh is equal to 5g d by 2. G will be less than g. Keep it here with 2. So h will be equal to 5d by 4. How many people have solved this by themselves? How many people have solved this by themselves?

Tell me. We will lift our hands and tell you. See next. If the object is here, then its velocity is u, so ui cap will be there.

When it reaches here, then at this point, the velocity will come. Final minus initial velocity vector's magnitude is asked. What is asked? Magnitude of change in velocity. How much velocity is there here?

We know this, right? Some u to the whole square minus 2gr will be there. Yes, it will be. Minus 2gr.

Yes, yes, it will be. It will be. Where is this? In K. If it is square, then if you take it here, then the root will come. This will come.

Now, change in velocity. Change in velocity happens. Final velocity, root under U square minus 2GR.

J cap minus initial velocity Ui cap. This will happen. This will be your change in velocity. If you take out its magnitude, then magnitude is its whole square.

Root under this whole square plus this whole square. So this whole square will be u square minus 2gr plus u square. This whole square, connect this whole square. Magnitude is being taken. So this will be root under 2u square minus 2gr.

If you take 2 common, the first answer will come. Stupid kids give third answer. Stupid kids give third answer.

Next is power. Now remember, I have written it very strongly. I have decorated it. Do you see how much I have decorated it? This power is of two types.

One is instantaneous and the other is average. Right? What will be the average power? Sir, you had said that whoever's average integration has to be taken out, divide it by t and it will be integration of t.

Sir, this will be the average power. It is not wrong. Brother, there is no mistake here.

It is not wrong. It is not wrong. It is absolutely correct.

Okay. By the way, average power is total work done divided by total time. We can write total work done as change in kinetic energy divided by total time taken. This will be the average power.

Okay. So write average power as total work done by time. Or do average power integration of P into DT divided by integration of DT.

This will also work. This will also work. There is no problem.

Okay. Next. instantaneous power is nothing but rate of work done instantaneous power is nothing but rate of work done It's a formula of velocity.

Oh, brother, it's written in mass. There is no mass here. Is there any mass written in the previous one? Let's see. I won't leave you wrong.

You eat your life. Oh, brother, brother, brother, brother. This 2R will come here. 2gr cos theta minus 1. Where will mass come in the formula of velocity? Where will mass come in the formula of velocity?

By the way, I had taken out here by COME. Where is mass? Where is mass in the formula of velocity?

Where is the mass? 2G L cos theta minus 1. V0, cos2, GL, cosθ-1. You won't get mass. Brother, you made a drive twice.

It was a little bit of a printing mistake. Did you make a drive twice or not? Tell me.

Say sorry, Shaktimaan. You made a drive twice. Next goal.

Listen to me. P is equal to dw by dt. What can I write dw? P into dt.

Now listen carefully, what will be the slope of the work time graph? The slope of the work time graph will be power. The area of the work time graph will be nothing. The slope of the power time graph will be nothing.

The area of the power time graph will be work done. Is this right? See, it is written here. See, it is written here.

What? Area of power time graph is work. Area of power time graph is work. Slope of work time graph is slope of work. time graph is power slope of work time graph is power can you understand from here differentiation means slope integration means area you can write dw as f dot ds if constant force is out then ds by dt will be velocity so power is force into velocity dot word we can write it as fb cos theta if theta is positive sorry power is scalar but power may be positive negative and zero suppose theta is is 0 degree or is less than 90 degree then how will your power come?

It will come negative. And negative power means that energy is coming out of it. If your angle is greater than 90, whose angle is it?

Angle between force and velocity. Instantaneous power is every cost hitter. Second thing, remember that if you ask the power of any engine, remember that engine's power is directly proportional to the power of the pump is directly proportional to V cube.

And the force that the pump is applying to flow the liquid, that force is proportional to V square. For example, let's ask you this question. Let's assume that there is a power P of an engine, from which X liter of water is coming out. If I want to take out 2X liter of water per second, then how much power will I have to do?

I will have to do power 8 times. Because the amount of water will come out as soon as the velocity. Is this thing in your mind?

Have you thought about this? Okay. Next, this is a question from NEET 2016. A question, pay attention.

Body of mass 1 kg begins to move under the action of a time dependent force F, where I and J are the unit vector along X and Y axis. What power will be developed by the force at the time t? Now see, he is asking you whether it is force or power.

No problem, listen carefully. This is force. How much? 2Ti cap plus 3T square j cap. If I get velocity, then power is the dot product of force and velocity.

Power is the dot product between force and velocity. Does it happen? Yes sir, it happens. Now I know the force here.

Can I calculate the velocity? Yes, because the mass is 1 kg. If the mass is force divided by acceleration, then the acceleration will be this. 2Ti cap plus 3T square j cap.

And if you integrate the acceleration, then your velocity will be this. Don't do anything, integrate it. Can you write the result of direct integration without writing?

Yes. It is 2T, so its integration will be 2T square by 2. This is 3T square, so its integration will be 3 times. The integration of T square is 3T cube divided by 3. 2 from 2 will be divided by 3 from 3. This is your T square I cap plus T cube J cap.

This is your velocity. Sir, very good. Let's multiply this velocity with this. So 2T to T square. So this will be 2T cube plus 3T square to T cube.

So it will be 3T 5. 2T cube plus 3T 5. 2T cube plus 3T 5. This will be your answer. Hi, everyone. This will be your answer.

This will be your answer. Tell me yes or no. Yes. Tell me yes or no. Is it clear?

Next. Next question is from Mains 2012. What? A car of mass M starts from wrist. acceleration so that instantaneous power delivered to the car has a constant magnitude P naught. Then instantaneous velocity of the car is proportional to.

So initial velocity, sorry, instantaneous velocity of the car. Such questions are asked a lot. Listen carefully.

Listen carefully. Is power constant? We can write power as force into velocity dot.

Now, power is given as constant. It's okay. We write constant p naught. We can write force as dv by dt into m.

Mass into acceleration. m into dv by dt into v take this dt here p0 into dt equal to mv into dv integrate both sides p0 out, dt's integration is t equal to m out, v into dv's integration is v square by 2 ignore the constant can I say v square is proportional to t so v is proportional to v is proportional to root t second answer See how easy it is. Power half. Finished.

Next, see. A body is projected vertically from the earth, reaches the height equal to the earth radius. Before returning to the earth, the power exerted by the gravitational force is greatest. See, when you were throwing it up from here, its speed was maximum.

Gradually, the speed was decreasing. Here, it became zero. Then, when it came down, it was running here.

Gravitational force was falling down here. So, what will happen here? MG. V into cos 180 degree because force and velocity have the same angle here the power will be zero because the velocity is zero when it will collide here, then the velocity below and the force below here the power of gravitational force sorry I am putting the vector on the power, mg into V0, so where will be the power maximum?

just instant the body is projected, no it remains constant all the time, the velocity is changing at the instant just before the body hits the earth now you, assume When the body was hitting the earth, then the power was there. And when it was thrown, then the power was there. Are the magnitudes same in both places? Yes, the magnitudes are same in both places. But one is plus and one is minus.

So if you have to choose between the two options, then your second option is the best. By the way, while throwing and when hitting, the magnitude of the power is the same in both cases. Next, look, son. Ask the question.

Engine pumps water through the horse pipe. Water passes through the pipe and leave it with velocity 2 meters per second. When Mars per unit length of the water of the pipe is 100 kg per meter.

What is the power of the engine? See, in this kind of question, if he asks you the power of the engine, then remember the power of the engine will be mass per unit length into VQ. Mass per unit length is 100 and velocity is 2Q.

So, it will be 100 multiplied by 2Q which will be 8. So, this will be 800 watt. up your answer will be fourth. Okay? But sir, A is also sometimes given, wrongly given. Yes, if this is given, rate at which, rate at which, at which, kinetic energy, supplied, to the, to the water, then this will be, half mu V cube.

This will be 400 watts. Meaning, if we ask the power of the engine, So that will be µVQ. µ is mass per unit length.

If you are asked to tell the rate of kinetic energy, the rate at which we are providing kinetic energy to the water, tell that rate. So that will be µVQ. Okay, sometimes you are asked to tell the average power. If you are asked to tell the average power, what are you asking?

If you are asking the average power, then also you will write If you are asking the instantaneous power, then you will write µVQ. Clear? If you are asking the power of the pump, then you will write µVQ.

Clear? Clear? So, we have done some questions about power from here. Now, this portion ends here.

Transcript for:Physics Work, Energy, Power Overview

Transcript for:
Physics Work, Energy, Power Overview