 fine let us move to the next concept fine now graphically what does it imply graphically how we can see this work done suppose on y axis you have force fine so this is the value of the force and this is the displacement along the direction of force fine this is displacement along the direction of force and the force is changing its magnitude like this okay it goes from here to there so you have started from point number one this is point number one and you went to point number two okay now at any moment let's say you're talking about this particular point at any moment let's take a very small displacement if I take a very small displacement I'll get a strip whose width is dr fine so this is a strip I'm talking about whose width is dr so dr is the displacement okay and let us say at this moment this is the value of force let's say f fine so this particular length this length is f because on y coordinate it is f and the width is dr because on x axis there is r fine so f dr is nothing but area of strip isn't it it's an area of the strip and if you integrate this if you integrate this what does it mean you're going everywhere and drawing a strip you're slowly and slowly moving and you're drawing strip and summing it up fine so when you sum the strip you'll get what this will be the total area fine so the area under force and displacement along the direction of force will give you the work done okay in thermodynamics we have already seen such scenario the area under pv graph for gas is work done fine similarly here area under force and the displacement along the direction of force will give you the work done by the force okay all of you clear about it type in yes or no you have any doubts is this applicable when force and area are not force and area see here there is we are not taking that okay force and displacement uh no then this area will not represent we are assuming that uh the displacement is in the direction of force here okay otherwise what will happen there will be a term of cos theta also comes in f dr cos theta will come now this integral is not the area of the graph between f and r okay now even though force is variable or force is a constant it doesn't matter area under force and displacement along the direction of force will always give you the work done okay and at times force can be negative also it can be like this then what you have to do you have to find this area and subtract that area from it fine just like what we did for pv graph this is negative force right from here till here this force is negative it means the forces against the direction of displacement okay let us take a question on whatever we have learned till now fine so here goes the question all of you please listen a woman pushes a truck on a railway platform okay there is a woman that pushes a truck on a railway platform which has a rough surface okay she applies a force the woman applies a force of 100 Newton over a distance of see the woman applies a force of 100 Newton over a distance of 10 meters okay thereafter she gets progressively tired after she applies a constant force of 10 sorry 100 Newton for 10 meters she got tired and her applied force reduces linearly with distance so the force after 10 meters reduces linearly with distance okay and it reduces to 50 Newton linearly okay the total distance through which the truck has moved is 20 meter so total displacement is 20 meters okay 10 meter is with the constant force and rest 10 meter is with this variable force that reduces linearly with the distance to 50 Newton okay you need to plot a graph between force versus plot for yeah plot a graph between force versus displacement let's say displacement is on the x axis okay find out Sukirt that is wrong answer find out the plot the graph between force and displacement Sukirt you don't need to hurry up every time okay focus on the learning plot the graph first done see till the displacement is 20 meters sorry 10 meters let's say this is 10 meters okay 50 meters okay so till the displacement is 10 a constant force is applied right let us say this is the force force this coordinate is 100 Newton's okay then this will be 50 Newton half of the distance so if you draw this line this represents 50 Newton's right and we know that linearly the force has reduced from 100 Newton till 50 Newton's right so this point represents 50 Newton and since it is linear reduction the graph will be like this okay so I hope all of you got this graph okay now find out the work done work done by the lady will be nothing but area of versus displacement graph right so if you are able to find this area that is the answer okay you can divide this entire area into two parts a trapezium and a rectangle fine so you can get area as 100 into 10 which is still 10 meters and then half into some of parallel sides which is 100 plus 50 100 plus 50 into distance between them which is 10 okay this will come out to be 1750 Newton's sorry again and again I'm taking it as Newton the unit says joules laws of motion chapter still in our heads okay this is what you'll get any doubts on whatever we have done till now please ask any doubts yes or no all right guys so we have discussed about the work in greater detail okay now let's talk about energy now energy is the capacity to do the work all right now you will be knowing many energies okay there will be multiple types of energy there will be sound energy there will be light energy there will be kinetic energy potential energy there will be you know many many other spring potential energy gravitation potential energy okay there will be multiple types of energies that are there okay our focus when we are in this chapter is okay so we are dealing with mechanical energy fine and there are only two kinds of mechanical energy one is kinetic energy and the other is potential energy okay potential energy is usually represented by a letter capital U and kinetic energy is represented by a letter of capital K straight forward right now potential energy is the outcome of the force only so we can leave this potential energy for some time okay we let us first only talk about kinetic energy right and I mean there is no surprises here kinetic energy is equal to half m v square okay this is how we write kinetic energy okay now this kinetic energy we have written this is the kinetic energy of a single mass moving with a speed v okay speed is equal to the magnitude of instantaneous velocity at that moment okay fine a don't message unnecessary comments if you have any doubts then please message fine so kinetic energy is half m v square where m is in kgs v is in meter per second okay so this is the kinetic energy that we are dealing with this will be in joules right now energy is capacity to do the work okay now if I get some relation between how much energy is required to do what amount of work then I can plan accordingly right I will be exactly knowing what is the energy that is required to do this much amount of work right so we have learned about the work done okay with the formula for work done we have taken different and we have solved problems to find the work okay anything which you can create right now energy is something which you can convert also one form of energy can be converted into another form of energy so energy also we have introduced here this is the kinetic energy now we are going to learn what is the relation between the kinetic energy and the work done by the forces okay so this relation is very important almost every numerical from this chapter will be solved from this particular relation that we are going to derive right now okay so we'll take up different cases case number one is constant force okay so for a constant force acceleration is constant okay so I will be able to write v square equals to u square plus 2 a dot s I will be able to write down this okay now if I multiply this entire thing with m by 2 what will I get I will get half m v square is equal to half m u square plus m a dot with displacement s okay now see left hand side is what this is kinetic energy finally let's call it as k2 okay and this is what kinetic energy initially let's call it k1 okay and what is m into a m into a according to Newton's second law should be equal to sum of all the forces right so if you rearrange the terms you'll be able to write down this as sum of all forces dot with displacement okay this is equal to k2 minus k1 which is what change in kinetic energy all right now sum of all forces this you can write down it as f1 plus f2 plus f3 you know like this you can keep on writing depending on how many forces are acting on it dot with s this is equal to change in kinetic energy fine now let's open the brackets if you open the brackets you'll get f1 dot with s plus f2 dot with s plus f3 dot with s like this if you keep on doing you'll be adding it up and then that is equal to change in kinetic energy all right now tell me what is f1 dot s what it is what it is correct f1 dot s is a work done by f1 force fine f2 dot s is work done by f2 force and so on if you add up all the forces sorry all the work done that will be equal to change in kinetic energy okay so basically work done by all the forces this is total work done total work done on a block of mass m will be equal to change in kinetic energy k2 minus k1 fine write down this you just individually take all the forces one by one you take all the forces find out work done by all the forces okay and add it up you'll get total work done that total work done should be equal to change in kinetic energy okay this is also called as work energy theorem work energy theorem fine any doubt on this this is is the heart of the chapter any doubt you can notice here one thing that potential energy is not there in the expression okay fine so we will later on introduce the potential energy right now when you talk about gravity force okay you don't take gravitation potential energy what you'll do you'll find out the work done by gravity also once you define gravitation potential energy then you will not take work done by gravity okay so let's keep that aside all right i'm not assuming that you know anything with which is potential energy okay so potential energy is something which we will talk later right now the theorem is plain and simple work done by all the forces will be equal to change in kinetic energy okay and this is the case we have discussed for a constant force okay now if force is variable then what will happen let's take case number two case two is variable force okay now kinetic energy is given as half m into v square fine this is the kinetic energy this is how kinetic energy is defined all right now if mass is constant i'm assuming mass is constant when you're applying variable force okay so if mass is constant and you differentiate kinetic energy what you'll get dk by dt is equal to half m times 2v dv by dt okay so this two is also gone fine so you will get m into v and what is dv by dt dv by dt is acceleration okay so if you write it down you'll get dk by dt there will be dot product also coming in over here once you take it as vectors this will be equal to m a dot with velocity okay now the velocity you can write down in terms of displacement it will be equal to rate of change of the position vector fine now you can cancel out dt from both sides so you will get dk is equal to m into a what is m into a m into a is the force now this force is a variable okay force is not constant it continuously changes so you just take it as f and this is dot dr now if you integrate you'll see that this side is what work done right and that side you'll go from k1 to k2 so that side will become delta k fine so again you're getting the same thing work done is equal to change in kinetic energy the work energy theorem does not depend on whether you are having constant force or a variable force if you are finding total work done that total work done will be equal to change in kinetic energy be it constant force or a variable force fine any doubts till now typing yes or no okay great good to see you guys are clear about the concept i hope you'll be as clear when you solve the numericals no doubts