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.

2

example

Pulling Crate Up An Inclined Plane

Video duration:

4m

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Hey guys, when this problem we're pulling a crate of an inclined planes. I'm just gonna go ahead and draw that real quick about my incline plane like this. I've got my box and I know that the mass is equal to 19 kg. So M equals 19. I'm gonna go ahead and label all the forces that are acting on this crate. I have an applied force that is parallel to the ramp and it's 130 noon. So this FAA here is 130. Now I also have some gravity. Right? So I have this MG here, this is going to be my MG. And I have a normal force like this. Remember that are inclined planes we're gonna separate RMG and we're gonna have an MG. X. That points down the incline like this. And we actually are told that nothing about friction. So we're not gonna do this any friction in this problem. So in the first part here, I want to calculate the work that some of gravity. So the work that is done by MG. Remember that on an inclined plane? The work that's done by MG is the same exact thing. That's the work is done by MG X. This just allows us to simplify our equation and really this is just gonna be MG times the sine of the Theta X. In which data X. Is the angle of the incline here. What's hold? This is 36 degrees. So that's the end. We're gonna plug into this data X. Here. So it's MG. X. Sine theta. Sorry, MG sine Theta X times D. And we've already taken care of that co sign stated term. It's just this right here. Now what we have to do is we have to determine the direction of positive because that's going to influence whether our works are positive or negative now because we're pulling this 19 creates up the ramp. So that means that the displacement over here is actually gonna be this way. So, this distance here we're told is 15 m that I'm gonna choose this direction to be positive. And that means that this work done by gravity is actually gonna be negative. All right. So this is just gonna be negative And I'm just gonna plug it on my numbers. So, I got the mass which is 19 times the 9. times the sine of theta. Which is the sign of 36 times the distance that I'm pulling it over which is 15. Now again, there's no need to put this cosign Theta term here. This coastline between this MG. X. And D. Because the negative sign already takes care of that. I remember that MG. X. Is going to be parallel to displacement. So you actually don't need to do this. All right. So you just plug in all these numbers here. What you're gonna get is negative 1640 jewels. So that's the answer to part a. Alright, makes sense. It's negative because as you're going up the ramp, gravity is going to be doing negative work on you. All right. So, let's go ahead and we want to part B. And part You want to figure the final kinetic energy of the great. We just calculated to work. How do we relate that to kinetic energy? Remember, we just use the work energy theorem. The network on an object which is just the sum of all the works is just equal to the change in the kinetic energy. Now, in this case remember we have three forces we have are applied force, RMG X and the normal force. So when we go to some all of our works, the network is going to be the work done by the applied force. The work done by friction which we just calculated. I'm sorry that that the work of a friction work done by gravity which we just calculated plus the work done by the normal force. Remember the normal force is sort of a perpendicular. Right. It's gonna be pointing sort of in the new y axis, perfectly good to the incline and it's going to be perpendicular to the direction of motion. So there is no work done by the normal force. So it's just these two that contribute works and that's equal to change in the kinetic energy. That's kinetic energy final minus kinetic energy. Initial. We know that we're starting from rest. RV not equal zeros. There is no initial kinetic energy. So really the network, all these works here are gonna be really just do are going to equal the kinetic energy final. So I've already just calculated the W. M. G. In part A. Now all I have to do here is it just have to calculate the work done with the applied force and then I can figure out the kinetic energy final. Whether the applied force is just gonna be F. A. Times D. Co signed data. So that means the work done by the applied force here Is gonna be my apply force of 130 times the distance is 15. And then what about the angle? Well the FAA and the distance are going to be pointing in the same direction. They will be parallel. So this is just going to turn to the coastline of zero which is just one. And so if you plug this in, what you're gonna get is positive 1950 jewels. So now this is 1950 jules. It's positive. It makes sense because it points in direction of motion. Now we're just gonna plug that back into our F. A. So really the work nets, which is really just the W. F. A. Which is my 1950 plus. My work done by gravity, which is negative 16 40 is going to equal the final kinetic energy. So basically what happens is that the kinetic energy final is going to be equal to 310 jewels. And that's your answer. All right. So let's take this one guys, let me know if you have any questions.

3

example

Finding Speed of a Crate Using Work-Energy Theorem

Video duration:

5m

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Hey guys, when this problem I have a crate that sliding down an inclined plane and it has some forces acting on it and we have to calculate some works. So let's go ahead and just draw a quick little sketch here. So I've got this incline plane like this. Going to draw a nice and big here. I've got my box that is seven kg and I've got some forces acting on it. So let's go ahead and label all the forces. So I've got MG. I know that's that's acting. I've got the normal force, we've got a component of MG that's acting down the ramp, this is MG. X. And this is going to be my MG. Why? And then I also have the force of friction as the block slides down the ramp, friction wants to oppose that and want to go to play up the ramp like this. So I got my friction force. So now what I want to do is I want to calculate the work that's done by gravity if the crate slides 2.5 m down the ramp. So basically that's this distance like this d equals 2.5 here in which it's going to slide down the ramp so it's going to end up over here somewhere. So let's go ahead and do that in part a and part. I want to calculate the work that is done by gravity. So it's W. M. G. Now remember that on an inclined plane W. M. G. Is the same thing as W. M. G. X. And so I can just say this is just gonna be MG times the sine of theTA X. This angle here is the 26 degrees. So this is day to exit the sequel 26 degrees. And um it's gonna be MG. Sine theta X times the distance. So what I have to do is first I have to pick a direction of positive. Now what happens is my MG. X. Is gonna actually point in the direction of my displacement. And so what I'm gonna do, I'm just going to have my positive direction be this way. So what this means here is that your W. M. G. Is just going to be, this is gonna be positive and so therefore I'm just gonna start plugging stuff in. So I've got seven times 9.8 Times the sine of 26°.. And then I've got the distance here, which is 2.5. If you go ahead and work this out what you're gonna get, you're gonna get 75.22 jewels. Makes sense that this is positive because the block is going to start going faster because of this work here. All right, so what about part B. And part B. Now? I want to calculate the work that's done by friction. So, remember that when we calculate the work done by friction, it's always going to be negative F. K. D. So, this is just gonna be once I expand this out negative mu kinetic times the normal times distance. So really the trouble is that I wanted to figure out this normal force on an inclined plane. Now, on an inclined plane, if you're only if you're only forces act horizontally, then your normal force is going to equal M. G. Y. Which is just gonna be MG time to the co sign of theta. X. So, what I can do here is I can see that the work that's done by friction is gonna be negative mu K. And then the normal force really just becomes MG co signed tha tha X. And then I have D here. So, I looked through all my variables. I have Mieux que I have the mass gravity, I have the angle and the distance. So I can calculate the work done by friction. So my work done by friction is gonna be negative Times 9.8 times. Oh, I'm sorry, the mass, the mass is seven massive seven. Now we have 9.8, so now we have the co sign of the angle which is 26 degrees and this times the distance, which is 2.5. So you go ahead and work this out. What you're gonna get is 55.5 except it's gonna be negative. So it's negative. 55.5 jules makes sense that it's negative because as the block goes down, grab so the first force of friction is going to want to remove energy from the block and slow it down so it's negative. All right. So now, in this last part here, we have to find the speed of the crate if it starts from rest. So the idea here is that the initial velocity is going to be zero here and then when it gets down to this point is going to have some final velocity. And I want to figure out what that is. So I just calculated a bunch of works and I can relate them to speeds by using the work energy theorem. Remember that the network is just equal to the change in the kinetic energy. So the work that we just calculated the work that we just calculated 75.2 and negative 55.5. We can combine them to find the network. So our network is really just gonna be our MG plus the work done by friction. Plus the work done by normal. Now, in this case the normal doesn't do any work right? We know that. So this is equal to the change in the kinetic energy. So there's really just K final minus K. Initial. However, because the initial speed is equal to zero, there is no initial kinetic energy and all of it just goes to the kinetic energy final. All right. So that's all we got to do. So really we just have this 75.2 that we just plug in plus negative 55.5. And this equals the kinetic energy. Final. This is gonna be one half. M. V final squared. So we want to find this V final here. We're just gonna have to solve everything else. So agenda getting When you subtract these two things is you get 19.7 this equals one half and then times seven times the final squared we rewrite that again. So this is seven here. So when I move everything over to the other side, what I'm gonna get is 19.7 times two times two divided by the mass of seven. And this equals DV final squared. So when you go ahead and plug this in and take the square roots, what you're gonna get is that your V final is equal to 2. m per second and that's your final answer. All right guys. So that's it for this one. Let me know if you have any questions.

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