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Work By Gravity On Inclined Planes

Patrick Ford
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Hey guys, So in this video 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 you make sure to make sure that you actually don't make that mistake. So we're actually gonna start with the example and we'll come back to this point in just a second here. So in this problem we have 100 kg block that's on an incline and it's gonna slide down a distance of 12. 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 work is done by MG X M G Y N MG. So basically the works done by each one of these uh 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? So I an 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 gonna be my M. G. Y. So, I'm trying to figure out the works done by the components of MG and then MG itself. All right, So how do I figure out works? I'm always just going to use F. D. Co signed data. So we're just gonna use MG. X. Times D. Co signed data and we're gonna use em G. Y. D. Cosign theta and then M. G. D. Cosign Theta. All right so what I want you I want to remind you is that the theta term that we use in our F. D. Cosign Theta always refers to the angle that is between the force and the displacement or the distance. Basically the the angle between the force and the motion. That's why I always draw an arrow between this data 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. And then the coastline between those two vectors here. So let's get started. How do I figure out MG. X. In an inclined plane? Remember MG. X. Is going to be MG. Times the sine of the incline angle. This 37 degrees here is actually tha X. It's the angle with respect to the horizontal. That's my 37° there. So really what happens is I can replace this MG. X. With MG times the sine of theta X. Here. So we're doing MG sine theta X times D times the co sign of data here. So here's what I want to point out is that these problems for with inclined planes be careful not to plug in the incline angle this data X. Here, don't plug in this data X. Here for cosign data which is really in your F. D. Cosign data. Right? So don't get these two angles confused. This is the angle of the incline your data X. This is the angle that is between these two forces. They're not the same thing. All right, so let's go ahead and do this. Right. We can actually go ahead and plug in all of our numbers. We have em is 100. We have not 9.8 for G. And then we have the sign of Fedex, that's my 37 degrees here. Now I multiply it by the displacement and the or the distance and that's 12 m down the ramp. Now the co sign is gonna be the coastline of the angle between your MG. X. And your displacement. If you take a look at your diagram here your 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 zero not 37°.. And remember that the coastline of zero is just evaluates to one. So once you plug all the stuff into your calculator you just get jewels. So that's the work done by MG. X. Now let's take a look at M. G. Y. Basically we're gonna do the exact same thing here except now we're gonna do MG. Times the co sign of data. X. Right? Because that's M. G. Y. Times design deco signed data. We plug this we're just gonna plug in all the numbers. This is going to be 100 times 9.8 times the co sign of 37. Now we have times 12. And now we just figure out the co sign of the angle between these two forces right between my MG. Cosign Theta or my M. G. Y. And the distance. What is that angle? Well 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 coastline term not 37. Again. However, the coastline and 90 actually always evaluates to zero. So even though you have a bunch of numbers out in front of here, you're gonna multiply it by zero. And the work that's done by M. G. Y. Is always just gonna be equal to zero jewels. 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. All right. So the last thing we do is M. G. D. Cosign theta here. So we have to do is now we actually have to look at this force which is RMG and we have to figure out the angle between this MG. And this displacement here. And this is where it gets really complicated because this angle here and are inclined plane problems is actually fatal. Why it's the angle with respect to the Y axis which were 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, there actually is another way that we can solve for it. We know about gravity or the work done by gravity or so what we know about works in general is that 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 M G 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 scholars, they're not vectors. And what this allows us to do is our work done by M. G. Y. Remember is always just equal to zero so we can just cross it out. So really what you 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 gonna I'm gonna write W. M. G. 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 7000 and jewels just like your MG was. And that's the answer. So you have 7080 jewels. And so your work done is equal to MGX. By the way, this is also always going to be the case. All right. So that's it for this one guys. Hopefully that made sense. Um, and let me know if you have any questions.