Guys, let's check this one out. We've got a 2 kilogram block that is launched up a 40 degree ramp. So I want to just draw this out for 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, v0, of 10 meters 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 3 meters vertically above the bottom of the ramp. So that means that basically vertically above the bottom, that means this height of this triangle here is Δy. I'm going to call it Δy and it equals 3. So what we want to do is 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 going to be my free-body diagram here. So basically, I'm going to start out with the weight force that's acting straight down. That's mg. Then I also have the normal force, which acts perpendicular to the surface like that. I don't have any applied forces, no tensions. I've pushed it, and it's already starting with an initial 10 meters per second, so there's no applied force or anything like that. But we do know 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 going to be fk. Alright. And the last thing we have to do is just break down this mg into its components. So I've got my mgy here, and then I also have a component mgx. So mgx acts down the ramp. Alright? So there are these two forces like this. 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 2. We know it's kinetic friction. So we want to figure out this coefficient of kinetic friction here. That's going to come from fk. So let's go ahead and write out our F=ma here. So I've got my F= <math><mrow><mi>m</mi><mi>a</mi></mrow></math>. And now I just want to choose a direction of positive. So if the block is going up the ramp with a velocity like this, then that's the direction I'll choose as positive, right? It's the direction it's moving. So what this means here is that when I expand my forces, I have these two downward forces, mgx and fk, and so they both pick up negative signs. So what I've got here is I've got mgx which is negative and then minus fk is equal to mass times acceleration. Alright, so I can expand both of these terms out. Remember, mgx is really just mg sin(θ), and then fk has an equation as well, it's <math><mrow><mi>mu</mi><mi>k</mi><mspace linebreak="newline" /><mrow><mi>m</mi><mi>g</mi><mi>cos</mi><mo>(</mo><mi>theta</mi><mo>)</mo></mrow></mrow></math>. So that's equal to ma. So what I can do here is say this is mg sin(θ) and then remember that this <math><mrow><mi>mu</mi><mi>k</mi><mo>times</mo><mi>the</mi><mi>normal</mi></mrow></math>, in inclined planes, the normal is just equal to mg cos(θ). So I've got minus muk mg cos(θ) is equal to ma. 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 sin(θ) is; those are all just numbers. I know what mg cos(θ) 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 0 for just a second. And I know that it's going to do this over some displacement, which I'm going to call Δx. So how do we solve this acceleration? Well, if we get stuck using forces, we're going to 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 kinetic friction. Then we're going to list out our 5 variables, and we're going to figure out 3 out of 5 and then pick an equation to solve for a. So we know the initial velocity is 10, we know the final velocity is 0, and then we don't know anything about a, t, or Δx. So unfortunately, this is 2 out of 5 variables, and I've gotten stuck here. I'm going to have to solve for one of them. So usually when we get stuck, we would use F=ma to figure out the acceleration, but that's actually where we just came from. We just came from F=ma. So we're really stuck between solving for the time or the displacement. We're 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 Δx is. So bascially, if we're trying to figure out this Δx, which is the hypotenuse of this triangle, and we know the Δy and the angle, we can always solve for that other sign just using our sine and cosine equations. Basically, what we can do is say that Δx times the sine of 40 degrees is equal to Δy. I want to use the sine because the sine of this angle is going to give me the opposite side, which is the one I know. Alright, so we use this expression here, and what we can do is we can solve #x# or Δx because this is just equal to Δy, which is 3. Right? It's 3 meters above the bottom of the ramp. So this is 3 divided by the sine of 40 and what you get is 4.67. So there it is, there's my 3rd out of 5 variables, 4.67. So we can finally set up an equation to solve for this acceleration here. So if we do this, we're going to use equation #2, which is the one that ignores my time. So we get that the velocity squared is equal to initial velocity squared plus 2 a times Δx. We want to solve now for this acceleration. So the final velocity is 0, the initial velocity is 10, and then we have this 2 times the acceleration times 4.67. So when you move this to the other side to solve for a, you're going to get negative 9.34 a equals 100. And actually, I'm missing an a here. So this is negative 9.4 or a. So when you solve for a, you're going to get negative 10.7. So this is our answer of, sorry. This is our acceleration. And remember, we came here because we wanted this last variable so we can plug it back into our F=ma. And now we can go ahead and solve for muk. So what I'm going to do is I'm just going to plug in all the values that I know. This is just 2 times 9.8 times the sine of 40, and this is going to be minus muk times 2 times 9.8 times the cosine of 40, and this is going to equal the mass, which is 2 times negative 10.7. So all of these really just become numbers here, right? So let's go ahead 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, muk. And then this 2 times negative 10.7 is going to be negative 21.4. So if we notice here all of these numbers are actually negatives. So what we can do is basically just turn them all into positives. It's kind of as if you like multiply the equation by negative one. It doesn't change anything. So what we've got here is we've got 15 muk plus 12.6 equals 21.4. And then if you subtract this 12.6 over, what you end up getting is you end up getting 15 muk equals 8.8. And so when you solve for this, you're just going to get 0.59. So you go to your choices, and that is going to be answer choice c. That's your coefficient. So, hopefully, this made sense, guys. Let me know if you have any questions, and let's move on.
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7. Friction, Inclines, Systems
Inclined Planes with Friction
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