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Solving Systems Problems with Friction using Energy

Patrick Ford
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everybody. So let's get started with our problems. So we have a system of objects and we want to calculate the speed of the three kg block just before it hits the ground. So I'm gonna start things off by drawing are sort of diagram here. So we've got this sort of, my initial, I've got these two boxes, I'm gonna call this one A and this one be What happens is once you release the system this three kg Block is gonna fall like this and then eventually right before it hits the ground it's going to be traveling with some speed. I'm going to call this V be final. So that's really what we're interested in now. As a result, the A block is also gonna be sliding to the right because these things are connected as one falls, it pulls the other one to the right like this. So it's gonna be somewhere over here like this and it's gonna have its own speed which is V. A. Final. But remember that from systems of objects, these speeds are actually gonna be the same because they're connected. So the whole system here has the same final speed. So in other words VB final is equal to V A. Final. So really what we're interested in is just the v final of the whole entire system. Alright, so how do we do this? We're gonna write out our energy conservation equation. So I'm going to write this out, this is going to be K initial plus you initial plus work. Non conservative equals. Okay, final plus you. Final And a written out sort of spaced out like this because remember we have to consider all of the individual energies from each object. Alright, so we're gonna have lots of terms here. So what's the initial kinetic energy? Well, it's gonna be the kinetic energy from A. And B. So in other words it's going to be one half M A. V. A. Initial squared Plus 1/2 and be vb initial squared. And now with the potential energy, remember that includes things like potential spring energy and gravitational potential. There's no springs involved but that we do have some changing heights for the B. Block. So we're gonna have some gravitational potential but the initial potential energy for both objects is going to be just MG. Y. Right? So it's going to be a um So it's going to be M A. G. And then Y. A initial plus M B. G, Y B. Initial. Alright, and finally we have any work done by non conservative forces. We do have a non conservative force which is friction because we're told the coefficient of friction here. So what happens here is that friction is going to be acting not on both the blocks, just on one of them because one of them is gonna be sliding across the tabletop that's sort of rough, like this? So here's what happens if you have some force of friction that's going to be sort of pulling back against the block and that's going to do work. So this is really just gonna be friction. The work done by friction on block. A So in other words, this is just gonna be I'm gonna call this negative F. K. Um times and we're gonna call this deed that distance. All right. And then finally I have my my kinetic energy final, which is gonna be one half of M A V. A final squared Plus 1/2 MB VB final squared. And then the final potential energy which is going to be M A. G. Why a final plus MB G Y B final. Alright, lots of terms there, but we're gonna see that some of them actually are going to cancel out. All right. So, let's see what's going on here. Is there any kinetic energy? No, there's not. There's no kinetic energy. Is there any potential energy? Well, what happens is that really depends on how you sort of set your zero points. So, because we have some of the distances, the heights involved like this block here is let's say we know this is two m. This y here that we can do here that this is going to call this delta Y. We can actually just set the ground to B. R. Y. Equals zero. Right? We usually want to set the lower points like that. All right. So, what does that mean then for why a initial So by the way, um Yeah. What does that mean for this block? Because this isn't two m, this is much higher. Well, it turns out that it actually doesn't matter because for this block, this four kg block, the block A. You have the MG. Y. And the M. G. Y. On the other side. But the block is only moving horizontally. So what happens here is there actually is no change in the potential energy because these two terms are going to be the same. So you have the same two terms over there and you can basically just cancel it at whatever height that is, it doesn't change. Alright, so then um we do have some initial potential energy for the B. Block, right? Because it's some height above the ground or zero points, there is some friction and there is going to be some kinetic energy because after the system starts moving, both the blocks are gonna be moving like this. Alright, so that's definitely there. And then what about this final potential energy for be? What happens is B is gonna be at the ground at this point? So therefore there's gonna be no potential energy. All right, so you kind of simplify all these terms there to really just maybe three or four of them. Okay, so here's what happens. Right? So we've got MB G times Y. B initial um Plus and then this is going to be the friction times the distance. Okay, so this friction force. Remember the friction force on this guy is actually just going to be um mu K. Times M A G. Right? So in other words it's going to be negative mu K M A G. Now, what about this distance here that it travels? Remember this distance here is D. But if you think about this, remember because these systems are sort of connected like this, whatever distance the B block falls is the same distance the A block travels. In other words it goes like this because the string has to remain sort of the same length. So in other words, these two things are actually equal to each other and see what we can do here is we can actually replace this D. Term and actually just becomes delta Y. So that's sort of what happens here, that actually just equal to the same thing. And then finally what happens is um you've got these two terms, Right, So the two kinetic energies, now, what we said here was that the initial the final energies or final speeds for both of them are gonna be the same. So we can do is we can say well, if these two terms are equal to each other the same, Then what happens here is this whole entire term can just become 1/2 and then you have M A plus MB times v final squared, right? So basically they're the same exact v finals, you can sort of group those terms together and that's really what we're after. Okay, all right, so we're almost done here. Let's just go ahead and start plugging in some stuff. Um So you've got the massive B. Which is going to be three times 9.8 times yB initial. So in other words why be initial is actually just going to be uh this term over here? So this is gonna be my Y. B. Initial. So this is gonna be two plus and then we have negative and then this is going to be 0.5. That's the coefficient times M. A. Which is gonna be the four times G. Which is 9.8 and then times delta Y. So in other words, the delta Y. Is just the two again like this, right? Because it travels at distance, this equals one half. And then we have this is gonna be four plus three and this is gonna be the final squared. Alright. So if you plug some stuff into your calculators, I'm just going to sort of simplify these into into numbers which you get over here, Is you just end up with 58.8 minus And then this is gonna be this works out to 39 points to and this is going to work out too 3.5, 3.5. The final squared. Alright. And now what happens is um If you work this out, what you're gonna end up with, is that the final is equal to um to .37 m/s. And that's gonna be your answer when you finally work all that stuff out. All right, so that's it for this one, guys, let me know if you have any questions.