7. Friction, Inclines, Systems

Inclined Planes with Friction

# Block Launched Up Ramp

Patrick Ford

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Hey, guys, let's check this one out. So we've got this 2 kg block that is launched up a 40 degree ramp. So I want to just draw this out for just a second Here. We've got this incline like this. I've got a box that's at the bottom. It's given an initial launch speed this V not of 10 m per second. So I know that the angle of this is 40 degrees, and basically what happens is that it's going to come to a stop at some point that is m vertically above the bottom of the ramp. So that means that basically vertically above the bottom, Uh, that means this height of this triangle here is Delta y. I'm gonna call it Delta Y and equals three. So we want to do is we want to calculate the coefficient of kinetic friction. So what I want to do first is I want to draw a free body diagram, so this is gonna be my free body diagram here, So basically, I'm gonna start up with the weight force that's acting straight down. That's mg that I also have the normal force which acts perpendicular to the surface like that. I don't have any applied forces, no tensions. Basically, I've pushed it and it's already starting with an initial 10 m per second. So there's no applied for us or anything like that. But we do know that there is a coefficient of kinetic friction. So what happens is if the block is sliding up the ramp, that means that the friction has to oppose this and it's gonna be f k. All right. And the last thing we have to do is just break up this mg into its components. So I got my M g y here and then I also have a components mg X, so m g X acts down the ramp. All right, so these two forces like this. So we already actually figured out what kind of friction that we're dealing with here, so we actually don't need to even do Step two. We know it's kinetic friction, so we want to figure out this coefficient of kinetic friction here that's going to come from F. K. So let's go ahead and write on our ethical dilemma here. So I've got my F equals m a. And now I just want to choose the direction of positive. So if the block is going up the ramp with a velocity like this, then that's the directional. Choose as positive rights, the direction it's moving. So what this means here is that when I expand my forces, I have these two downwards forces mg X and F K. And so they both pick up negative science. So what I've got here is it got mg exit is negative and then minus f k is equal to mass times acceleration. All right, so I can expand both of these terms out. Remember, MG X is really just mg times the sine of theta. And then f k has an equation as well. It's mu k times the normal. So that's equal to M A. All right, so what I can do here is say this is mg sine theta. And then remember that this mu k times the normal in inclined planes, the normal is just equal to mg times the coastline of data. So I've got minus mu k mg co sign of data is equal to N A. So if I want to figure out what this coefficient of kinetic friction is, I need to know everything else about this equation. I know what mg sine theta is. Those are all just numbers. I know what mg cosign theta is again. Those are all just numbers here and I know what the mass is. The one variable that I don't know is I don't know what the acceleration is. I know that it starts with some velocity. And when it gets to the top of this ramp here, the final velocity is going to equal zero for just a second. And I know that there is going to do this over some displacement, which I'm gonna call Delta X. So how do we solve this acceleration? Well, if we get stuck using forces, we're gonna try to solve it using motion equations. And that's exactly what we do here. If we want to figure out the acceleration so we can plug it back into this formula here and solve for the coefficient of static friction, then we're going to list out our five variables, and we're gonna figure out three out of five and then pick an equation to solve for a So we know this initial velocities. 10 Final velocity is zero and then we don't know anything about A T or Delta X. So, unfortunately, this is two out of five variables, and I've gotten stuck here. I'm gonna have to solve for one of them. So usually when we get stuck, we use f equals M A to figure out the acceleration. But that's actually where we just came from. We just came from F equals M A. So we're really stuck between solving for the time or the displacement were not given any information about the time. But we do know something about the distance vertically above the triangle or the ramp or something like that. So maybe we can figure out what this Delta X is. So basically, if we're trying to figure out this Delta X, which is the hypothesis of this triangle, and we know the Delta y and the angle we can always solve for that other side just using our sign and co sign equations. Basically, what we can do is say that Delta X Times the sine of 40 degrees, is equal to Delta Y. I want to use the sign because the sign of this angle is going to give me the opposite side, which is the one I know. All right, so use the this expression here, and what we do is we can solve for X or Delta X, because this is just equal to Delta Y, which is three rights 3 m above the bottom of the ramp. This is three divided by the sine of 40. And what you get is 4.67 so that there it is. There's my third out of five variables, 4.67. So we can finally set up an equation to solve for this acceleration here. So if we do this, we're gonna use equation number two, which is the one that ignores my time. So we get that the velocity squared is equal to initial velocity squared plus two, eight times Delta X and we want to solve now for this acceleration. So the final velocity is zero. The initial velocity is 10, and then we have this two times the acceleration times 67. So when you move this to the other side to sulfur A, you're gonna get negative. 9. 34 move there. So you're gonna get negative. 9.34 equals 100. And actually, I'm missing an a here, so this is negative. 9.34 A. So when you saw for a you're gonna get negative 10.7. So this is our answer. Uh, sorry. This is our acceleration. Remember, we came here because we wanted this last variable so we can plug it back into our f equals m a. And now we can go ahead and solve from the U. K. So I'm gonna do is I'm just gonna plug in all the values that I know. This is just two times 9.8 times the sine of and this is gonna be minus mu K times two times, 9.8 times the co sign of 40. And this is going to equal the mass, which is two times negative, 10.7. So all of these really just become numbers here, right? So let's go ahead and, um, and solve for each of them. So if you look at, let's see, the first term just becomes negative. 12.6, and then this term here is going to become negative. 15. Exactly, um UK and then this two times negative 10.7 is gonna be negative. 21.4. So if we notice here, all of these numbers are actually negatives. So we can do is basically just turn them all into positives. It's kind of if you like, multiply the equation by negative one. It doesn't really change anything. So we've got here is we've got 15 m. U K plus 12.6 equals 21.4. And then if you subtract this 12.6 over what you end up getting is you get a new getting 15 u k equals 8.8. And so when you solve for this, you're just gonna get 0.59. So if you go to your answer choices and that is going to be answered choice, See, that's your coefficient. So hopefully this makes sense. Guys, let me know if you have any questions and let's move on

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