by Patrick Ford

Hey guys, so that's probably gonna be calculating the works done by a bunch of these forces, kinetic friction, weight and normal. So, let's go ahead and check out this problem here, we have the box that weighs 50 newtons, which means that your MG, which goes down, is already equal to 50 newtons. Don't assume that to mean 50 kg and then multiplied by nine point, exhibit at the wrong answer. The MG is already 50 which means that the normal force, right? If these are the only two forces that are acting in the vertical direction, have to cancel, that's our normal force. We actually have one last force to consider which is the force of friction. So we're pulling this block to the rights and friction is going to oppose that by acting to the left here. So this is going to be RFK, that's the friction force. Now, what we want to do this first part here is we want to calculate the work that is done by this kinetic friction. So, that's the work that's going to be done by F K. Well, remember if we're trying to figure the work done by F. K. That's going to be F. K times D. Co signed photo. So that's gonna be the equation for that. We have this data here. That's the angle between F. K. And D. All right. So, how do we figure out this force of friction will remember the force of friction is always going to be um U. K. Times the normal force. So, in UK times normal times D. Co sine of theta. All right. So, we have this coefficient of friction here. This mu K is 0.7. Now, we just have to figure out the normal force. Well, actually, we already know that it's equal to 50. So already start plugging in the work that's done by the friction force is going to be, this is going to be 0.7 times 50. That's the normal force. Yeah 50 or here. Now we got the distance right? So the box actually travels a horizontal distance of eight m. So basically we've got this distance over here this delta X. Or D. Is going to be eight. So that means that we're just gonna plug in eight. And to here and now we're just gonna look at the co sine of the angle. Remember the co sine of the angle of the angles between the force of friction and your displacement. Your friction force acts to the left but your display your displacement acts to the right. So here what happens is your force of friction of your F. K. Is this way. But your displacement is this way this is your delta X. Or your D. And so what that means here is that the angle is equal to degrees. So your co sign is going to be the coastline. 180. Remember what happens is that when you plug in the coastline of 180 you're gonna get a negative one always. So what this means is that you're gonna get, the friction force is equal to negative 280 jewels. Now what I want to point out here is that the friction force is always going to be negative. Your friction force is always going to oppose your direction of motion. We were going to the right in this example in a friction force are friction force pointing to the left. If it was reversed, if we were going to the left are friction force would actually point to the rights. So because friction always opposes, your motion is always in the opposite direction. Is V. It always does negative work. So, what you're always going to see when you calculate the work done by friction because you're gonna see F. D. Co sign and the number of him here is always going to be 100 and 80 degrees. They're always gonna be opposite of each other. So one way we can simplify this from now on is all right. That the work done by friction is just negative F. K. Times D. All right. That's a pretty simple way to figure that out. So, you're plucking those numbers and you'll get negative 280 jewels. All right, So, let's move on to part B. Now. Now we're going to calculate the work done by weight force. So, the work done by weight is really going to be the w work uh Sorry, the work done by gravity work done by MG. So here we're just gonna do MG. That's the weight force times D. Co sign of data. But we can quickly tell here that your MG is gonna point downwards And your displacement is going to point to the right. So this angle here is 90°.. So what happens here is you're gonna have MG. Which is 50 Times the displacement which is eight. But it doesn't matter because the coastline 90 is gonna wipe everything out and the whole entire thing is gonna become zero. So what happens is the work that's done by MG is going to be zero jewels. That makes sense. Your weight force always points down if you're going to the right that way force doesn't actually do any work on you. All right now I've got one last one last part here which is the way to work done by the normal force. So this is the work done by this end. So this is going to be our Normal force times D. Co sign of theta. And we're going to see is that this basically is going to be exactly the same as part B. Your normal force points up. Your displacement is to the right. So this angle here. So this angle here is 90 degrees. So you have your weight or start your normal 50. Your displacement is eight. But it doesn't matter because you're gonna have the coastline of 90 again. And so this whole thing is gonna go away. You're gonna have to work the sun is zero jewels. All right? So that's it for this one. Guys. Let me know if you have any questions.

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