Guys, in this video, I want to show you a specific example of calculating the work that's done by gravity, but in a specific case where you have an object on an inclined plane. The reason I'm doing this is because these problems can be kind of confusing and tricky. They sometimes try to trick you into plugging the wrong number into your equations. So I want to go over this so to make sure that you actually don't make that mistake. So we're actually going to start with the example and we'll come back to this point in just a second here. In this problem, we have a 100 kilogram block that's on an incline, and it's going to slide down a distance of 12 meters. We also know that the angle of the incline is 37 degrees. What we want to do in this problem is we want to calculate the works done by mg_{x}, mg_{y}, and mg. So, basically, the works done by each one of these, the components of mg and then also mg itself.

So how do I do that? Well, I want to go ahead and draw those forces in my diagram. Right? In the inclined plane, my mg points straight down. My normal points perpendicular to the surface. And because we tilt our coordinate system, then we just have to break up this mg into its components. So this is going to be my mg_{x} and this is going to be my mg_{y}. So I'm trying to figure out the works done by the components of mg and then mg itself. Alright? So how do I figure out works? I'm always just going to use fdcosine theta. So we're just going to use mg_{x} × d × cosine(θ), and we're going to use mg_{y} × d × cosine(θ), and then mg × d × cosine(θ).

What I want you to remember is that the θ term that we use in our fdcosin(θ) always refers to the angle that is between the force and the displacement or the distance. Basically, it's the angle between the force and the motion. That's why I always draw an arrow between this θ and my f and my d just so I don't forget that. Okay. So what I have to do is I want to figure out mg_{x} × d × cosine between these two vectors here. So let's get started. How do I figure out mg_{x} in an inclined plane? Well, remember, mg_{x} is going to be mg × the sine of the incline angle. This 37 degrees here is actually θ_{x}. It's the angle with respect to the horizontal. That's my 37 degrees there. So really what happens is I can replace this mg_{x} with mg × sin(θ_{x}) here. So we're doing mg × sin(θ_{x}) × d × the cosine of θ here. So here's what I want to point out, is that these problems, with inclined planes, be careful not to plug in the incline angle, this θ_{x} here. Don't plug in this θ_{x} here for cosine(θ), which is really in your fdcosine(θ). Right? So don't get these two angles confused. This is the angle of the incline, your θ_{x}. This is the angle that is between these two forces. They're not the same thing.

Alright? So let's go ahead and do this. Right? We can actually go ahead and plug in all of our numbers. We have m as 100, we have 9.8 for g. And then we have the sine of θ_{x}. That's my 37 degrees here. Now I multiply it by the displacement or the distance, and that's 12 meters down the ramp. Now the cosine is going to be the cosine of the angle between your mg_{x} and your displacement. If you take a look at your diagram here, mg_{x} points down the ramp, and your displacement also points down the ramp, which means that the angle that we plug in here is actually 0, not 37 degrees. And remember that the cosine of 0 always evaluates to 1. So, once you plug all this stuff into your calculator, you get 7,080 joules. So that's the work done by mg_{x}. Now let's take a look at mg_{y}. Basically, we're going to do the exact same thing here except now we're going to do mg × the cosine of θ_{x}, right because that's mg_{y} × d × cosine(θ). And we plug this we're just going to plug in all the numbers. This is going to be 100 times 9.8 times the cosine of 37. Now we have times 12. And now we just figure out the cosine of the angle between these two forces, right between my mg × cosine(θ) or my mg_{y} and the distance. What is that angle? Well, if you take a look at your diagram, your mg_{y} points into the incline and your displacement points down the incline. Those are perpendicular to each other. Right? This is actually a right angle like this. So what happens here is we actually plug 90 into our cosine term, not 37 again. And remember that the cosine of 90 always evaluates to 0. So even though you have a bunch of numbers out in front of here, you're going to multiply it all by 0. And the work that's done by mg_{y} is always just going to be equal to 0 joules. So this is actually always going to be the case. And you can think about it like this. Right? Your mg_{y} always points into the incline, but you're always moving down the incline so it can never do any work like that.

The last thing we do is mg × d × cosine(θ) here. So what we have to do now is actually look at this force, which is our mg, and we have to figure out the angle between this mg and this displacement here. And this is where it gets pretty complicated because this angle here in our inclined plane problems is actually θ_{y}. It's the angle with respect to the y-axis which we're almost never given. So this actually ends up being really complicated here and we're not going to use this approach to solving the work done by gravity. Fortunately, we can add them up just like regular numbers. So if I want to calculate the work done by gravity, it's really just the addition of the work done by mg_{x} and the work done by mg_{y}. I can add these two works together even though they come from forces that point in the different directions like x and y because works are actually scalars, they're not vectors. And what this allows us to do is our work done by mg_{y}, remember, is always just equal to 0, so we can just cross that out. So, really, what happens here is that the work done by gravity is always just the work done by mg_{x}. And this should make some sense. So this is going to write W_{mg} is always equal to the work done by mg_{x}. And the reason for that is if you think about this, the component of gravity that pulls the block down the ramp is mg_{x}. That's the only thing that can do work on the box. So what happens here is that our work done by mg is just 7,080 joules, just like your mg was. And that's the answer. So you have 7,080 joules and so your work done is equal to mg_{x}. By the way, this is also always going to be the case.

Alright. So that's it for this one, guys. Hopefully, that made sense. And let me know if you have any questions.