 see when you deal with the when you deal with the variable force okay one very common variable forces the spring force fine so it will be very common to find the work done due to the spring force later on fine so since we will be dealing with the spring force or the work done by the spring force multiple times so let us just derive the expression for work done by the spring force and then we can directly use the expression for work done by the spring force okay so we are trying to derive the expression for a very common type of variable force which is the spring force right down work done by spring force okay or work done by spring all right so let us say that there is a spring with a spring constant k spring constant is k fine initially the spring was extended from its natural position it was extended by a value of x1 and finally it got extended to x2 fine so the spring is getting extended from x1 to x2 you need to find the work done by the spring when this process happens okay try to do it spend some time it's a variable force right and you know that the force because of the spring will be k times x okay and this force direction is opposite to the direction of change in x sorry the force itself depends on change in x so force is k delta x you can put a minus sign like this so force is equal to minus of k times x fine now work done if you suppose displace this spring by dx okay you further extended by dx so the displacement will be opposite to the direction of the spring force right so you can write the work done as minus of kx dx fine where x is the change or x is the so to say from the natural length how much it has deviated that is x okay so if the deviation increases by dx further that the work done will be kx dx and we are talking about a negative work done by the spring here because the force of spring when your displacement is this direction will be in opposite direction because the spring is already extended fine so this is the small amount of work done total work done will be an integral of this and k is a constant fine if k is not a constant it will not come out of integral so we can make a numerical in which k also depends on x fine but right now we will take a generic case k is a constant let's say its extension goes from x1 to x2 right so this will be equal to minus of k by 2 x2 square minus x1 alright so the work done okay Sukirt is asking why it is negative okay so we are taking a case in which the spring is extended already fine and you are further extending it right so when spring is extended this side the spring will apply force in that side and you're further extending that side so force will continue to act in backside so there is 180 degree between force and the displacement so that is why it is negative okay so this will come out to be half of kx1 square minus half of kx2 square fine so this is the work done by the spring alright now you're going to see something which is very unique to the spring suppose initially initially spring was compressed fine it was compressed by x1 right and then you are extending it now it is extended by x2 right so compared to natural length the spring initially was in compressed state and compared to natural length spring now has an extension of x2 okay so in this process when you go from 0.1 to 0.2 now what is the work done by the spring can you try it your own is the compression equal to extension see compression is right now x1 and extension is later on x2 okay bar that is not correct Sukirt no no Sukirt hmm others let's try it all on yes bar it will come out to be same the work done by the spring is same okay let us see how we can do this particular case see right now the spring is compressed it is compressed by x1 let us say that it goes to natural length this is step number one to step number two from compression it goes to the natural length and then it get extended fine this is this dotted line is natural length so this is x1 and that is x2 right okay now from one to two the work done is positive because the spring will try to push in this direction and the displacement is also in that direction so work done will be positive from one to two so let us say work done is w here and w two there so total work done will be w1 plus w2 fine let us try to find out work done w1 and w2 separately so w1 work done will be integral of kx dx the limits of x will be from x1 to 0 okay it will minus will come outside it net net it should come out to be a positive thing alright because the force is always on the right hand side and displacement is also on the right hand side okay okay fine so this is equal to half of kx1 square alright now w2 is what w2 is minus of kx dx now x is going from 0 to x2 you will get here as minus of half kx2 square fine so total work done will be sum of this plus that so w will be equal to w1 plus w2 this will be equal to half kx1 square minus half of kx2 square fine so it does not matter whether this compression or extension fine so if initial write down if initially impression or any of it if it is x1 and finally compression or extension it doesn't matter if it is x2 then the work done then the work done by the spring will be half kx1 square minus half kx2 square fine it doesn't matter whether x1 is compression or extension or x2 is compression or extension fine all of you understood this work done by the spring any doubt anything please type in yes or no no it did not freeze niranjan no doubts okay so now i think only 17 people are there so we'll focus on problem solving okay so let us take up questions now but that doesn't mean that problem solving is lesser importance this is actually more important than the concepts assume there is a fourth see you see that at times the problems from this chapter can be solved by applying concepts from the neuter okay so you can solve the question in two different ways but the thumb rule is you should first try to solve using concepts from this chapter because you are dealing with scalars and second whenever they talk about finding velocity okay so try to use work energy theorem okay do not find acceleration and then from acceleration you find the final velocity so suppose there is a force of 100 newton applied here this is a block of 1 kg mass okay initially its velocity was one meter per second you need to find out the final velocity when this block got displaced by a distance of two meters all of you clear about it if you if you have any doubts please type in or start doing it okay there is a block it has initial velocity of one meter per second you are applying 100 newton horizontally fine coefficient of friction is 0.1 between the block and the ground all right when this block moves by two meters what is the final velocity of this block use work energy theorem okay don't use newton's law of motion concept to solve this because you're just learning how to apply work energy theorem root of 395 that's what Niranjan also got okay so see here basically you need to find total work done on this one kg block work done by 100 newton force work done by friction work done by gravity and work done by normal force all these work done when you add up will be equal to change in kind of energy that is half m into v square minus half m into u square I hope you all have got this particular thing work done by the 100 newton force is what 100 into 2 okay work done by friction is what first of all we have to find friction force which is mu times normal reaction normal reaction is what mg right so which is 1 into 10 so 10 newton's is mg so mu into 10 is 1 newton okay so friction force is 1 newton and good thing is we know the direction which is opposite to the displacement minus 1 into 2 is a work done by friction work done by gravity is 0 and work done by normal reaction is 0 because both of them make 90 degree with the displacement okay this is equal to half m is 1 so v square minus half m is 1 u is 1 so this is just half okay so v square by 2 minus half is equal to 200 minus 2 200 minus 2 198 okay so v square is equal to 198 into 2 plus 1 fine so this is 397 v square so v is equal to under root of 397 fine okay we'll take up one more question before you know closing this session any doubts on this message any doubts on whatever we have done just now we have simply applied work in ag theorem any doubts okay guys so let us move to next question I'll take up a j don'ts level question see we have learned everything that will take you I mean that will take to answer any question from this chapter okay although the definition of potential energy is something which we haven't done yet but then if you apply work energy you can solve almost every question okay so I'll just find out a question from previous here this thing I think online session like this is lot more efficient than the physical class otherwise you guys talk so much I waste my energy in just keeping you quiet you can type in your uh mess your your feedback on whatever I just said so what I will do here is that I'll just discuss with you the scenario initially and then you have to solve it yourself this is m and that is capital M by the way we have derived work energy theorem for a particular shushant I doubt anyways so we have discussed work energy theorem but that work energy theorem is applicable for a single mass right now okay so work done on mass one one change in kind energy of mass one right so this is mass m ones okay great so this is for mass one k one minus k two okay I am talking about mass one this is m one m one okay and suppose there are two masses in a system like in this particular scenario two masses are there then work done on the second mass will be equal to change in kinetic energy for the second mass I'm writing it in the wrong way let me correct it change is always k two minus k one okay work done work done on the second mass is focus on the question k two for mass two minus right now I'm discussing the concept okay I'll describe the question little bit later k one m two fine so these are the two work done on two different masses if you add up these two work done if you add basically two equations you'll get here k two for mass one plus k two for mass two minus k one for mass one plus k one for mass two okay now basically there will be some internal force between mass one and mass two right now the internal forces tension between capital M and small m fine so when you add up the work done separately like this work done by the internal force cancels out so this will be only this is work done by external forces so external forces what mg is the external force over here all right so total work done on the system okay it is change in kinetic energy of the total system the system may have multiple masses so change in the final kinetic energy of mass m one plus final kinetic energy of mass m two and so on minus you know initial kinetic energy of mass m one initial kinetic energy of mass m two so we just modified work energy theorem so that we can apply it for a system which has multiple masses okay now let us see what this question is about Bharat is in a hurry focus on the concept okay anyways a string with with one end fixed on the rigid wall so this end is fixed on the rigid wall passes over the fixed frictionless pulley so this is a fixed pulley through which this string is passing and this distance is two meters okay capital M has a mass of two kgs small m has a mass of 0.5 kgs all right now the string is horizontal that capital M this string is horizontal fine this one from here till here the string is horizontal okay and it is vertical from here till here so this string is vertical that string is horizontal okay you need to find out what will be the speed what will be the speed with which capital M will hit the wall when you release the small m what will happen this is small m will start moving upwards okay when this small m starts moving upward since the capital M is heavier the heavier mass tries to come down okay and when it comes down it will try to move in a circle like this and the entire string as I you guys try please first then we'll discuss the solution try it your own does it have friction otherwise where you see friction can be and there is no contact surface pulley is frictionless which is your usual case there is no surface contact so there's no question of friction knows okay the string is not cut see that this small m is very light so it automatically this capital M will go down and when capital M goes down this will move up this capital M is hanging in air right now are you guys clear about the question type in yes or no now try solving it this came in you know very very long back when you guys were not even born 1985 I was two year old that's a very valid question this distance the distance of capital M from the wall is one start solving the question for you forget what what is asked huh we have to find with what velocity capital M hits the wall okay with what velocity capital M hits the wall just write questions don't need to solve it and just message that you have written the equation then I'll write my equations you can compare message done when you're done with writing the equations oh somebody got the answer velocity root of 2.1 Niranjan I've read it hmm answer is 3.29 meter per second is that what you you guys are getting 3.29 5 root 5 is like can say 10 so root 10 is what yeah yeah so get close to that that's correct okay let us analyze the scenario I'll just analyze the scenario and see what it is when this block moves how it will move it will move in a circle so it will complete a quarter circle and hit here getting it so when it hits here the string becomes like this isn't it so now see what is happening this block comes down okay this block comes down and this block moves up suppose this is x the distance it moves up I need to find out the distance why because I need to find out the work done by gravity and for that I need to know what is the displacement right so this is clear how much capital M moves along the direction of gravity this is one meter only because the distance of capital M from this is one meter right I need to find out what is x all right so for that I need to first know what is this length this length is root over 2 square plus 1 square which is under root 5 fine so initially that same length was one meter and now it has stretched this distance one meter got stretched to root 5 right so the extra distance that is root 5 minus 1 this must have been coming from this end fine so this is the value of x all right now gravity is doing work on capital M as well as on small m all right so work done by the gravity on capital M will be mg into what into one all right because that is a displacement of capital M along the direction of gravity fine work done by the gravity on small m this is what now minus sign will come because gravity is acting down but this is small m is going up by what distance by x right so this will be small m g into root 5 minus 1 all right there is no other external force fine or you can say that this tension might you may feel it is external force okay which is correct also but then since we are saying that it is moving in a circle tension always remains perpendicular to the displacement so work done by this tension will be zero okay so total work done in the system will be work done by the gravity on capital M plus work done by the gravity on small m right this total work done will be equal to change in kinetic energy so if initial velocity was zero and final velocity is v1 and v2 you can say that kinetic energy is half small m into v1 square plus half capital M into v2 square where v1 is the velocity of small m and v2 is the velocity of capital M all right now we need to find the relation between velocity of capital M and velocity of small m all right now in order to find the relation between the velocities I need to use constraint relation here that the string is fixed length fine so string is of fixed length hence along the string relative velocity should be zero if along the string if there is any relative velocity it means that the length of the string is changing fine so if the relative velocity along the string is zero then what will be the scenario so for small m we are saying v1 is the velocity okay and for capital M we are saying velocity is horizontal because it is moving in a circle fine so after quarter circle it will try to hit the wall horizontally fine so if this is the angle theta fine and this is the velocity let us say this is v2 is theta I can say that v2 cos theta is the velocity component along the direction of string okay this should be equal to v1 all right and what is cos theta you have a nice right angle triangle here for which cos theta is 2 divided by root 5 fine is this how you guys have done it Krishna Sukheer those who got the correct answer is this how you have done this yes or no is this how you guys have done it okay anyways hope you have learned a few concepts from this particular question so in j advanced questions like these you can expect all right so today was just a basic session on introduction to work and energy all right so next class probably you know we will our focus is more our focus will be more on problem solving okay we'll start with the basic level of problems and then we'll build on it and we'll also introduce something which is called potential energy okay so we'll introduce potential energy and we'll see that how we can use potential energy in problem solving also okay at times you know in question itself it will be written that potential energy is this much so you cannot avoid potential energy at times all right otherwise if there is an open-ended question like this where there's no mention of using potential energy you can solve this particular question without using gravitational potential energy formula okay because you have work energy theorem which is more generic all right so thanks for joining in and that's it for today's class and I hope you have enjoyed today's online session the homework for for this class is you have to finish to finish objective one from hcverma at times you'll see some questions which for which we haven't done theory yet but that's okay you you try to do it still and then you have to solve at least 30 questions from exercise one okay so that is a homework for next class and come with all of your doubts and we'll discuss it first and then we'll continue with the session okay thank you