Work done from kinetic energy - solved example

- [Instructor] 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 2,000 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 2,000 kilograms. So we know the mass of this truck, that's given to be, let's just write that down over here, 2,000 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 we were to plug in for F equals MA and somehow 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 have talked a lot about this equation in a previous video called Work Energy Theorem. It would 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 get ready to pause the video and see if you can try this yourself. And to recall, kinetic energy is half MV squared. All right, lets do this. So what's the final kinetic energy of our car? That's going to be half M into the final velocity squared. And we 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 900. So we get 1000 kilograms times 900 meters square per second square. And that gives us 900,000. Five zeroes, right? Kilogram meters 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 will 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, lets go down a little bit to make some space. Okay, so if we simplify we get again, goes one times, so 1000 kilograms into 20 square is 400. This time I will 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, 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 accurately use work energy theorem to calculate the work done. All right right, lets try one more. A pretty similar problem so great idea to pause and see if you can try yourself first. All right lets see, a 500 kilogram tempo, traveling at 30 meters per second, brakes 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 that's it's initial velocity. The unit brakes and comes to a stop. So that means this time it decelerates, so after a while it comes to a stop that means it's final velocity is zero. We need to calculate the work done on the tempo, again we are given the mass of the 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 energy 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 225,000 Joules. You can check that. You 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 does 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, 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.