 Let's solve a couple of problems on calculating work when velocities are given to us. Here's the first example A truck accelerates from 20 meters per second to 30 meters per second. Find the work done on the truck Given its mass is 2000 kilograms Okay, let's go ahead and draw a situation for this. We are given that the truck accelerates from 20 to 30 So let's say here is our truck Before accelerating so before it accelerates. It's going at 20 meters per second. Let's write that down and Then after accelerating it goes a little faster. So now It is at 30 meters per second We are asked to find the work done on the truck. How much is the work done on this truck? Given its mass is 2000 kilograms. So we know the mass of this truck. That's given to be Let's just write that down over here 2000 kilogram. We need to calculate the work done So, how do we calculate work? Well, we have seen now for quite a while work done is Calculated as force acting on a body multiplied by the displacement of that body Okay, now do I know what the force is? No, that's not given in the problem. Do I know what the displacement is? That's also not given. We are only given the initial and the final velocities and the mass. What do we do? Well, maybe we can use this data and Calculate what force and displacement is and then substitute and calculate, right? But guess what? We've already done this before If you if you were to plug in for f equals ma and somehow, you know substitute all of this We will eventually arrive at an equation called the work energy theorem Which says that work done equals change in kinetic energy. So work done equals final kinetic energy minus the initial kinetic energy and This equation can be derived from here and The equation is basically saying that the work done on a body basically tells how much kinetic energy is added to the body So in this example, if we find out that 10,000 joules of kinetic energy was added to this body Then it means that the work done was 10,000 joules So I just have to calculate how much kinetic energy was added when the car accelerated and that itself represents the work done And if you need more clarity on this equation, we've talked a lot about this equation in a previous video called work energy theorem It'll be a great idea to go back and watch that video and then come back over here Anyways now by calculating how much the final kinetic energy is and how much the initial kinetic energy is We can now calculate what the work done is so great idea to pause the video and see if you can try this yourself and To recall kinetic energy is half MV squared All right, let's do this. So what's the final kinetic energy of our car? That's going to be half M Into the final velocity squared. Let me call that final velocity as V So we can now plug in the values and if we simplify we will get Two goes one times. So we get a thousand here and 30 square is nine hundred. So we'll get thousand kilograms times 900 meter square per second square That gives us 900,000 Five zeros right kilogram meter square per second square Now this is the SI unit of energy. Isn't it? We also call this as Joules. So we'll just call it as 900,000 Joules Similarly the initial kinetic energy Would be half Into M Into the initial velocity squared. We'll call that as U call this as U and Again, if we substitute and simplify, let's go down a little bit to make some space Okay, so if we simplify we get again goes one times. So thousand kilograms Into 20 square is 400. This time I'll get 400,000 Joules So this is the initial kinetic energy. This is the final kinetic energy now work done just represents Final minus initial meaning how much kinetic energy got added? So we can do that now So work done on this truck should equal final which is 900,000 900,000 Joules minus 400,000 Joules How much is that equal to? That is 500,000 Joules That's our answer So let's put everything in one frame now So this equation is basically saying that the car gained the sorry the truck the truck gained 500,000 Joules of kinetic energy and so that must be the work done on the truck And so this is how even velocities are given. We can directly use work energy theorem to calculate the work done All right, let's try one more A pretty similar problem. So great idea to pause and see if you can try yourself first All right, let's see a 500 kilogram tempo traveling at 30 meters per second Breaks and comes to a stop Calculate the work done on the tempo So we have a tempo which is initially traveling Initially traveling at 30 meters per second. So let's say that's its initial velocity It's given that breaks and comes to a stop. So that means this time it decelerates So after a while it comes to a stop that means its final velocity is zero We need to calculate the work done on the tempo again We are given the mass of that tempo the mass is given to be 500 kilograms So we can do the same thing what we did before We can say work done is how much kinetic energy got added to this tempo So from work in the theorem we say work done equals the final kinetic energy minus the initial kinetic energy This time, what's the final kinetic energy? Well, the tempo is at rest velocity is zero. So kinetic energy becomes zero Okay, what's the initial kinetic energy? Again, we can say it is half times mass times the initial velocity squared and Again, if we substitute and if we simplify we will get this to be 225 1,000 joules You can check that we can pause the video and check the calculation and so the work done becomes Final kinetic energy, which is zero minus the initial kinetic energy, which is 225,000 joules and that gives us minus 225,000 joules that minus sign is important, okay So this time the work done on this tempo is a negative number So what is this negative number saying? Well, remember work done represents how much kinetic energy gets added, isn't it? But this time the kinetic energy was removed Since the kinetic energy is removed We are basically saying the negative now the negative sign is saying that the kinetic energy is removed Anyways, this much kinetic energy is removed. So by definition. This is the work done on the tempo So when velocities are given work done can be calculated directly by using the work energy theorem