 Now before we even start the chapter, I am going to introduce something which we have already learned in the bridge program, Scala product, okay, because in this chapter you will be dealing with Scala products many, many times, write down Scala product. So I will assume that you will have basic knowledge of the vectors, so I will try to move little faster here in case you are not able to get it, immediately message me, you can message me personally or you can message me on the WhatsApp group, whatever you may feel comfortable and I will not take your name, don't worry, okay. So suppose you have two vectors, let's say vector A is there and another vector B is there, fine. So these are the two vectors, Scala product is written as this, A dot with B, this is how you represent Scala product. So two vectors can be multiplied in different ways and this is one of the ways you multiply the vectors, fine. So when you multiply the two vectors like this, you will get a scalar out of it, fine. So you have two vectors which you have multiplied and you are getting a scalar out of these two vectors, fine. The magnitude of this Scala product or dot product is magnitude of A vector multiplied with magnitude of B vector into cos theta, okay, plain and simple, right. And if A vector is AX i cap where AX is the component of vector along x axis plus AY j cap plus AZ k cap, okay and B is BX i cap plus BY j cap plus BZ k cap, okay. Then what is A dot B? When you do A dot B, you will be getting AX BX plus AY BY plus AZ BZ, okay. All of you remember this, okay. Shushant is saying I didn't understand why we multiply by cos theta. There is no particular reason why you multiply cos theta when you take dot product, okay. Just like, you know, in physics you have many physical scenarios where, you know, a multiplication of cos theta was required. So we define a product between two vectors in such a way that Scala product is defined like this, okay, because it is useful in many physical scenarios. That's why I can as well define that A dot B is A into B into tan theta, okay. So fine. Now, this is A dot B and what is magnitude of A? If you guys remember when we did this chapter, magnitude of A was root over AX square plus AY square plus AZ square, fine. Similarly, magnitude of B, okay, don't call me while I'm taking class. I don't want to take the name, okay. This is root over BX square plus BY square plus BZ square, okay. So this is mod A, this is mod B, this is A dot B, right. This is the magnitude of A and B and the dot product is given by this expression, okay. So I hope you have a good enough understanding of all this, okay. I think in maths, you have vectors already started, right. Yes or no? Akhil has started vector chapter for you guys, okay. But then we have done this in our bridge program, right. I have covered vectors in greater detail. You have any doubts till now? Anyone of you, if you have any doubts, please ask or type in no, I don't have any doubts. You can type in no, if you don't have any doubts. Any doubts, guys? Okay, no doubts. I have heard that you have ray optics in your syllabus now. Is it correct? Is ray optics in your syllabus now? Okay, great. So your class 12 will be a lot lighter. Anyways, so let's solve a particular question on whatever we have learned till now. You have to find out angle between this vector, 3i plus 4j minus 5k. This is one vector and this is another vector, okay, which is 5i plus 4j plus 3k, okay. Find out the angle between these two vectors. What is the angle between these two vectors? 37, someone is saying cos inverse of 8 by 25, that's correct, cos inverse of 8 by 25. The f dot d, first let us do that. f dot d is what? 3 into 5, 15 plus 4 into 4, 16 minus 5 into 3, 15, okay. This will come out to be 16. Now, magnitude of the force is under root of 3 square, which is 9, 4 square 16 plus 5 square 25. This is magnitude of force and magnitude of displacement, the second vector, 5 square 25 plus 4 square 16 plus 3 square 9, okay. So these are the magnitude of the two vectors. So you can say that f is a and d is b, fine. So a dot b comes out to be 16, which is equal to magnitude of f, which is under root of, this is under root 50 and this is also under root 50. Both are under root 50, okay. So magnitude of a is under root 50, b is under root 50 into cos of theta, all right. So cos of theta will come out to be 16 by 50, which is 8 by 25. So theta comes out to be cos inverse of 8 by 25, all right. Okay, now why we have just brushed up dot product or scalar product, you will soon understand, okay. Now I am going to start the physics part of the chapter, work. This is our first topic, work, fine. Now as we have already understood that work should represent change, okay. It should represent a change and physics, the change in is represented by displacement, let us say s vector is a displacement vector, okay. And the amount of work which you are doing, nobody will talk about direction of work because it is a scalar quantity, right. So every time they will be saying amount of work, fine, because it is a scalar quantity. So amount of work should be proportional to what is the displacement, how much you have changed the status, and it should also be proportional to how much effort is applied, getting it. And then you should also write it that it should be proportional to the angle. It is not directly proportional. So let me, I will just modify this a bit. It is proportional to force and it is proportional to displacement in the direction of the force, in the direction of force, okay. Because if the displacement is because of the force, then there has to be a displacement along the direction of force, fine. If there is no displacement along the direction of force, then the work done should be 0, right, okay. So that is what it is. The amount of work done by the force, let us say this is the force, okay. Let us say that this is the force. The force is in this direction. This is the force, okay. This is just one of the force, probably there are multiple forces acting on the object. But my interest here is to find the work done by this force alone. So I am focusing on this force only, okay. But when it comes to displacement of the object, that is unique. Displacement cannot be different for different forces, right. Displacement will be same for all the forces. Forces could be multiple, but displacement will be always one, okay. So let us say that displacement is in this direction, okay. This is the displacement. Let us say this is s, fine. Now, when I am trying to find out the magnitude of the displacement, then the magnitude is what? This is the magnitude of displacement, right. The length of the vector is a magnitude, this is the magnitude of displacement. Now, if I have to find out how much is a displacement in the direction of force, then what should I do? I should take component of this displacement along the direction of force, right. So I should draw a perpendicular on it like this, okay. And this, this is the component of the displacement along the direction of force, fine. Which is equal to what? Magnitude of displacement cos theta, getting it. There is another component of displacement that is perpendicular to the force, fine. This is s sin theta, right. So in the direction of force, you have s cos theta as a displacement, right. So can I say that this displacement is because of this force? Yes or no? Please type in. Can I say that s cos theta displacement is because of this particular force? Yes or no? The answer is no. Why I am saying that? Because there can be multiple forces acting along the displacement. So this is just one of the force. Probably there is another force. Let us say f2 that is acting along the direction of displacement, fine. So this entire displacement may not be because of the single force f, okay. But then this is one of the force that is contributing towards this displacement, right. So that is the reason why in work done, the work done is not just proportional to amount of displacement in the direction of force. It also matters what is the force, fine. Because the displacement may not be because of that single force itself, there might be multiple forces acting along the same direction. So it is the collective effort that is creating s cos theta displacement. All of you understood this? Fine. Since it is a collective effort, the work done should always tell you or you should always understand that work done is because of a particular force, okay. So work done by a force f is equal to magnitude of force f into magnitude of displacement in the direction of force which is f s cos theta, fine. Suppose you have another force in the direction of s cos theta, then work done by that force will be what? Magnitude of that force f2 into s cos theta. All of you understood what is going on? Type in yes or no? So even though the displacement could be in the direction of force, you cannot say that the total displacement is because of that force alone. Because there might be other forces in the same direction, fine. So that is the reason why you have to specify which force you are talking about when you are finding the work done, fine. Just saying that work done is this much has no meaning. You have to specify or you have to say that work done by 5 Newton force or work done by the force which is acting horizontally of this magnitude. You have to specify, all right. Okay, great. So I will scribble here. So this is the magnitude of the work done, all right. But ultimately see we are dealing with vectors, fine. This particular expression for the work done by this force came because the way you have defined the work done, all right. Or the way we understood philosophically what should be the work done and in physics change means displacement and effort means force. So that is how it has come up and it finally came in this shape, okay. But then ultimately what is the force? Force is a vector and displacement is also vector. So that is the reason why you represent work done by the force as a dot product between the two vectors. Dot product magnitude is also equal to this only, okay. Now the assumption here is, the assumption here is that the value of force is a constant, not just value but the direction also. You are dealing with a constant magnitude and direction of direction force. So this force which you are writing here, you are assuming this force do not change by the time the particle is moving from point number one to point number two, fine. So till now we have only learned work done by a constant force. So this is work done by a constant force, fine. So even though it's a very specific scenario, you know that we are talking about work done by a constant force, but it is found in physics many a times like gravitation force, MG is a constant force, right. And in a particular scenario, if a normal reaction, tension, everything could be just a constant force. So constant force can feature many a times. So that is the reason why we are talking about specific case before talking the generic case, all right. So f dot s is the work done by the force f, okay. Now if you look at the work done which is magnitude of force into magnitude of displacement into cos theta, you know physically you can understand it like this. This is first way you can understand you can take component of displacement along the direction of force and then you multiply force with displacement along the direction of force you get the work done, fine. Or another way you can visualize this, you know, you can do one thing, you can draw the same force, you know, you draw it like this, this is the force and you draw a displacement this direction. So this is displacement and that is the value of force. Now rather than taking the component of displacement along the direction of force, now what I am doing I am taking component of force along the direction of displacement, okay. So displacement vector could be small, but then I can always extend it, fine because I am talking about direction only. So this is the direction of displacement. So if this is the theta angle then f cos theta will be the component of force along the direction of displacement, right. If I draw right angle triangle like this, this thing the value of force along the direction of displacement this will be equal to magnitude of force into cos theta, right. So we have got, you know, component of force along the direction of displacement. Now you can see that the value of work will be same, you are multiplying displacement with component of force along the direction of displacement. Ultimately it comes out to be same, okay. So remember this, you can take component of force along displacement or you can take component of displacement along the force and then find the work done. In both cases you will get the same thing, getting it. Now when you solve the question you will be able to appreciate what I am talking about, okay. Because many a times it's simple to take component of displacement along the direction of force and many a times it is, you know, easy to analyze when you take component of force along the direction of displacement, okay. So that we can understand or appreciate only when we start solving the questions. But right now understand what I am telling you here. You can take component of force along displacement or displacement along force and multiply these two, okay. All right. So let us try to understand this work done in more granular level. Work done is force, magnitude of force into magnitude of displacement into cos of theta. What is theta? Theta is angle between two vectors, force vector and displacement vector, fine. So it is the angle between these two vectors and when you talk about angle between the vectors always remember you find out angle between the two vectors by connecting two vectors tail to tail. This is the angle between the two vectors. You can say this is F and this is S vector, fine. You cannot say that if you write like this, F and S, this is not the angle between the two vectors. This is the angle between two vectors, fine. You need to move this vector parallel to itself like this, connect it tail to tail and then say that this is the angle, fine. So make sure you follow this when you are solving any particular question, okay. All right. So now we will be taking different kinds of cases and we will discuss it one by one. Write down case number one. Case number one is when the angle between, when the angle between force and displacement is zero degree, head to head is angle correct. Yeah, you can connect head to head also. Like in this particular case, you can move the displacement vector and connect it like this. Even this you can say angle, they both are same. Okay, but usually people connect tail to tail. You know, that's more intuitive. Anyways, so case number one is theta is equal to zero. So the direction of force is this and the direction of displacement is also the same, fine. So force and displacement both are making zero angle, all right. And this is a very common scenario to happen, fine. So example could be when you are pushing this block forward and this block moves by a distance x. This displacement x is parallel to force, right. So that is the example. And when you drop a ball, it is going down, right. So downward direction, you have mg force, right. And the ball is also moving down. So even displacement is downward direction, fine. So in this case, also when the ball is dropped, the angle between force and displacement is zero, right. So so many such scenario happens where the angle between force and displacement is zero. In this case, simply the work done is magnitude of f into magnitude of displacement because cos of zero is one, okay. Now magnitude cannot be less than one, fine. So work done will be a positive quantity. It will be a positive quantity, fine. Okay, so this is case number one, right. Let us go to case number two. Case number two is when theta is equal to 180 degrees, okay. What does it mean? It means that force is suppose in this direction, this is your force but the displacement is happening in opposite direction, okay. This is possible if there are multiple forces acting on the object. But this particular force is acting opposite to the direction of displacement. There might be some forces that are acting in the direction of displacement, okay. Or the object already have the velocity against the direction of force. So an example of deceleration is valid over here, okay. Can you think of any other example where theta is 180 degree? You can just type in whatever comes in your mind. Give me a few examples where the angle between force and displacement is 180 degrees. Ball thrown up perfectly, right. When you throw the ball upward direction, the motion is upward, right. But mg force is acting downward. So displacement which is up makes 90 degree with mg force which is down, okay. Bharat has launched a missile. When missile is launched, you have to push it forward Bharat. Okay friction on a normal surface. Very true. When you have a scenario like this, when this block is sliding forward, let's say it has slid a distance of 1 centimeter. The friction acting on it all the while is backwards, right. So friction is in opposite direction of displacement. So even that is correct. Body moving backwards and a person is stopping it. Body is moving backwards. Correct. Krishna, you have given a very twisted example but that is correct, okay. It's like this. When this mass is coming towards you and you are applying a force to stop it, fine. Then you should be applying force in this direction. But the displacement is happening in left hand side. So the force applied by you is making 180 degree with the displacement, okay. This is case number two and case number two is as common as case number one, all right. So it is very, very common that the force makes 180 degree with the displacement, all right. Now let us take case number three. So case number three is when theta is equal to 90 degree, okay. When theta is 90 degree, the work done will be what? Work done will be magnitude of force into magnitude of displacement into cos of 90 degree, right. And cos of 90 degree my dear friends is zero, right. So work done will come out to be zero, okay. Now can you think of any physical scenario here when this is happening? Apart from that gulli example which you have been taking since your class ninth, yes Sukirt that is correct. A satellite in orbit, that's correct in a way. So when satellite, it's not a day-to-day life example, but then we'll discuss it. A satellite moving in a circular orbit, which direction is filling the force towards the center. This is gravitation force, right. But the displacement is always 90 degree. It always moves tangentially, okay. In fact, not only this, every uniform circular motion, whenever the object is moving in a circular motion, a uniform circular motion, there will be an acceleration that is always towards the center, fine. So if there is an acceleration towards the center, there has to be a force towards the center to cause the acceleration, right. So in a uniform circular motion, there always be a force towards the center and the displacement is always tangential, fine. So the force is making 90 degree with the direction of motion, fine. But then why I was not taking this example because the direction of displacement is continuously changing. Every moment it is changing, okay. Even the direction of force is changing every moment. Any other example that you can think of, just a routine example, anything. An airplane flying at a constant altitude, if it is flying at a constant altitude, which force we are talking about? Are we talking about gravity? Then you are correct. If this is an airplane, not a very nice looking airplane, you do not want to sit in here. This is moving with velocity v, okay. So suppose it moves horizontally, let's say it moves by a distance x, the gravity is acting downwards, right. So work done with the gravity will be zero because gravity is making 90 degree with the displacement, fine. Another example which I was expecting that you should be taking is when you have a block and it moves forward, there is a normal reaction and acting, but then this block is always moving forward, fine. So this scenario is very common that normal reaction makes 90 degree with the displacement, fine. So work done by the normal reaction, many cases like 90% of the cases will be zero, right. But don't take it as a rule just like an observation that many a times work done by the normal reaction is zero because the angle normal reaction makes with the displacement vector is 90 degrees. Work done by the gravity on the car, correct. Assuming car is moving on a level ground, not on, you know, on our Bangalore roads which are filled with potholes, then your car will be moving up and down and the gravity will do the work. And when you run, running is more exhausting than walking because you go up and down also. Anyways, so we will talk about case number four. Now, case number four is the angle is acute. So angle is greater than zero degrees and less than 90 degree, okay. Now, this can happen, right. So if these three cases can happen, even this can happen, you can apply a force on this block, you know, you can apply it like this, this angle theta is acute and this block moves forward. This is direction of displacement, suppose, fine. So then also the work will be done and the work done will be greater than zero. It will be a positive work, fine. So this is case number four, a positive work done when the angle is acute. Let me try this yellow color. No, that will not work this one. Okay. All right. So case four, sorry, case five. Case five is when theta is more than 90 degree, but less than 180 degree, I am talking about obtuse angle between force and displacement, then you will get work done as magnitude of F into magnitude of displacement into cos of theta. Now F is positive, displacement magnitude is positive, but cos of that obtuse angle is negative, right. So work done will be less than zero. All right. So work done being less than zero is a very, very common scenario. In case number two also, I'll write here, it is minus F into S because cos of 180 is minus one, all right. So work done can be positive, work done can be negative and work done can be zero. Work done can be anything by a particular force, fine. So do not create any notion in your head. So it's a very generic thing. You just have to go by the definition of the work, all right. So I hope all the cases are very clear to you and also in your notebook, please write down whatever cases we have discussed. We have discussed for constant force. Okay. We have discussed for constant force scenarios. Okay. Any doubts you guys have, you can type in yes or no. Example of work done less than zero. See case number two where angle is 180 degree is an example of work done when it is less than zero, right. Work done by friction. This example was a negative work done example. Okay. What if we have multiple angles? What does it mean Ananya? See when I'm talking about work done, you know, when I'm talking about work done, I am right now trying to find only for a particular force which is constant. Okay. So direction of force is a constant and displacement is also unique. It's a straight line always. Displacement is a straight line. So you have two lines, force, line of force and line of displacement. Fine. So if you talk about work then by a particular force, it will be always a unique angle, a single angle. Fine. But then if you're talking about Ananya, I think you mean what if there are multiple forces and multiple angles are involved. Fine. Then what you do is you just find out work done individually by all the forces and add it up. You will be able to find the total work done. Fine. But angle between one force and the displacement. If force is constant, it will always be unique angle. Multiple angles will not exist. So I hope I have answered it. Any other doubt is negative magnitude possible for any other scalars? You tell me. Anyone want to answer it is negative magnitude is possible for any other scalars? Okay. Omkar, I'll answer you after this. Is negative magnitude possible for any other scalars? Is it possible? Have you ever heard about a scalar magnitude being negative? What else other than power? See, generally scalars represents amount. Okay. It represents amount. Usually it's like amount of substance, number of substance, the amount of matter which is mass. Okay. So usually it will be positive scalar. But it doesn't mean that it will always be positive. All the scalars. There is negative scalars that are there. There are negative integers. Right. So there is a use of those negative integers. For example, potential gravitational potential that we will be learning later on and electrostatic potential that also we are going to learn later on. And suppose you don't take SI unit, you measure the temperature in degree Celsius. Okay. Then even temperature, if you measure in degree Celsius can be negative. Okay. So it's a very common scenario, not as common as what we are talking about here. All right. But then you'll encounter it soon. In class 12, there are many, many scalars that are negative. Any other doubt? Okay. Omkar has asked, is it an exception that at work is a scalar quantity yet it can be negative? No Omkar, just what we have discussed. Right. I hope you have got it. There are many scalars that are negative. In fact, there are theories which says that, you know, anti-matter exist, which have negative mass. But there was no observation, you know, to prove its existence. But theoretically, mathematically, it came out that there has to be an anti-matter that has negative mass. But then we'll not get into all that.