**Example:** Energy in Connected Objects

Hey guys so in this video I want to show you how to use the conservation of energy equation to solve problems where we're going to have multiple objects moving together as a system let's check it out, so something like this here I have multiple objects moving together and it says here when we're dealing with multiple objects we'll still we use only one equation if you remember when we did this with forces F=MA if you had two objects you would write F=MA twice, here we're going to use just one equation the big energy equation but you have to consider the energy of each object individual independently so let's check it out.

Here I have a system that is released from rest so the initial velocity is 0, the 5-kilogram object is initially 3 meters above the floor so this gap here it's going to be released it's going to fall so I'm going to do this, 3 meters let's call this object A let's call this object B, OK? So, if the 5 kilograms is initially 3 meters above the floor calculate its speed just before it hits the ground so I want to know what is the final speed of object B (Vb final)? But remember this is a system and in systems when you have a connected system like this all the objects move with the same speed and they also have the same acceleration so when I ask for the final velocity of the 5 kilogram I could have just as well have asked for the final velocity of the 4 kilogram or the final speed of the 4 kilogram it would have been the same and I could have also asked for the final speed of this system so it's all the same I'm just going to think of this as V, they're all going to have the same V final, OK? Now once the in 5 kilogram drops 3 meters this guy has to go up it's going to go up the same amount if this drops 3 this goes up 3 so I'm going to draw this over here as well, 3 meters so the block B starts here so the initial height of block B is 3 and the final height of block B is 0 and for block 4 block A it's the opposite the initial height starts here so the initial height is 0 and the final height is 3 so they basically swap places, cool. Let's write the energy equation, Kinetic initial + potential initial + work non-conservative = kinetic final + potential final, the reason why we're going to use the energy equation is because I'm being asked for a final velocity or final speed and the energy equation deals with that, it deals with changes in speeds with changes in heights, right? So it's a good use of the energy equation here but I have to consider every object individually I'm going to first write the long version of this but don't scared you don't have to do it every time but I just want to show you all the different elements you have, there are 2 objects so when I talk about kinetic initial, it's kinetic initial of A plus kinetic initial of B, Potential energy here is going to be just the gravitational potential energy MGH but I have two objects so it's Ua U initial A and U initial B and then there's the work done by non-conservative forces either one of the two objects which here will be simply 0 because there's no friction it says here that we can disregard friction and there is no external force, right? So I'm going to have kinetic and potential so it's the same thing K final A + K final B + U final A + U final B, these are all the possible energies there are 8 of these things but again as you get better with this you don't have to write all of them because you're immediately going to recognize some of them are 0 for example the system starts from rest which means nothing is moving so there is no initial kinetic energy you could have very well have just crossed this out, OK? Now the potential energy in the beginning this guy starts at 3 and this guy starts at 0 so there's no potential is for block A only for Block B at the end you get the opposite of the inverse of that they flip places so at the end the A has potential energy but the B has no potential energy so you can U final B is 0 so I have initial B and final A, OK? Initial B and final A, alright? So now I have kinetic energy of both of these guys because they both have speeds after this thing moves, right? So, I can start writing this out here this is going to be MbGH initial this is the initial height of B so if you want to put all the little letters equals 1/2MaVa final 1/2MbVb final and they're all squares plus this guy here MaGHa final those are all the letters. one thing to realize is that this V here is the same so instead of calling it I'm going to say same V instead of calling it Va final or Vb final I'm just going to call it V final which means that if these two are the same I'm going to be able to combine them, OK? So, let's start plugging in some numbers here, the mass at B is right there 5 gravity 'm going to use 10, the initial height is 3, OK? This is going to be half the mass of A is 4V final squared this is what I'm looking for + 1/2 mass of B which is 5 V final squared this is what I'm looking for, + mass of A 4(10) final height is 3 as well notice that the initial heights of one is the same as the final height of the other because they swap places, now all we have to do is solve for V final in this equation so I have here this is going to be 150 this whole thing here is going to be 120 so it's going to come here as a negative and then I can combine these guys I get a 2V final from the first plus 2.5V final from the second this adds up to very simple 4.5 V final squared so V final will be this is a 30 over here divided by 4.5 and then you got to take the square root of both sides when you plug this into your calculator you get approximately 2.58 meters per second and that is your V final for both objects again it didn't matter that I asked you for the final speed for one it's the same for both of them, OK? Now one last point I want to make here is I want to talk about the kinetic final right here IÕm going to do a different color I want to talk about kinetic final right here which becomes this which then eventually becomes this, OK? Kinetic final you can think of it as 1/2MaV final + 1/2MbV final, right? You can think of it that way or notice what happens the V's are the same so really what's going on is you end up with something like this 1/2Vfinal squared (Ma + Mb). Notice that the masses combine into a single velocity so you can think of it this way the kinetic final in the system is just 1/2 big M V final squared where big M is just the sum of all the masses M1 + M2 + all the masses etc. OK? So, form now to make life a little bit simpler we can just write kinetic final as treating all objects or the entire system as just one big mass and you can do that because all the masses move together instead of writing this and then taking the bit longer to combine everything, cool? So here in this case big M would have been 4+5=9, alright? So I have another example here let's jump into that

**Example:** Energy in Connected Objects

Alright so here I have a 3 kilogram block right here 2 meters above the floor so this gap between the initial position and the final position is 2 meters I like to draw this little arrow and I like to indicate that the initial height is 2 and that the final height is 0 so that when I plug this stuff into my energy equation I already have some of these numbers figured out, now if this block moves 2 this way it means that this block on top has to move 2 this way, right? So, this guy is going to move 2 meters like this, OK? And the problem says that there's a coefficient of friction here, coefficient of friction here is 0.5 I only give you on coefficient so that means it's both kinetic and static friction we don't have to worry about it so what happens here? Well remember the way we think about these problems is what would happen without friction? Without friction this would certainly go this way because there is no force opposing it therefore friction is going to be going this way, OK? And we're going to use the since I'm being asked for the final speed or the speed just before I hit the floor I have changes in height I have changes in speed and I get a friction I'm going to use the conservation of energy equation which means I can solve this much faster, an alternative way to solve this by the way would have been to write F=M.A for both blocks combine those two equations like we did back in the day and find the acceleration of system and then I would be able to use one of the motion equations to figure out how fast this thing is so I could have done one written F=MA twice to find my A and then two I could have used motion equation to find the final velocity but what's better is I can just do this in one step using the big energy equation and that's much better that's what we're going to do here, alright? I just want to point that out you could do this in other ways. Kinetic initial + potential initial + work non-conservative = kinetic final + potential final, right away here the system starts from rest, right? It's released from rest which means the initial kinetic energy for both objects, the potential energy there are two objects let's treat them separately potential energy we're going to be a bit careful I'm going to call this guy A and I'm going to call this guy B so it's Ua initial + Ub initial, the work done by non-conservative forces is the work done by friction because here we have friction but we don't have any external forces, right? And remember non-conservative is friction plus the external so the only one here is the work done by friction, the kinetic energy we talked about this in the previous video kinetic energy I'm going to treat it as an entire system of one 1/2 of the total mass times the final velocity of the system and this is what we're looking for, this total M here is going to be Ma+MB which is 3+4=7, OK? Plus So we're going to treat the kinetic energy as if you have just one body because they move together so they have the same velocity they're able to do that it's just a mass simplification there and the potential have to treat them separately because they have different heights, Ua final I'm sorry Ub final let's go through it and see if anything else is 0, so I'll go 4 kilogram certainly has a height above the floor but it has no height relative to the table and even more importantly its height doesn't change so because the floor doesn't change heights it doesn't even matter what its height is if it doesn't change heights it doesn't change potential energy so we're going to just cancel out the potential energies, so look at Ua here whatever number this is would be the same number here so we can just cancel the two of them out because they have the same height so I want to do a little thing here to sort of so you have it your notes the change in height for A is 0 therefore I can eliminate both potential energy it doesn't matter what the potential energy is it's going to be the same in both places so we just get rid of it make our lives simpler. Is there a potential energy for Block B? well in the beginning it does it has a height but not at the end, OK? So, this simplifies a lot and I have just these types of energies continuing here this is MbGHb initial so that's just MGH with all the subscript don't get lost in all of the letters, the work done by friction this is kinetic friction because the block is sliding so it can be -FD, OK? And I can calculate this friction is mu normal the little side thing here and normal in this case is MG because I got this block here it's being pulled this way by tension it's being pulled this way by kinetic friction it's been pulled down my MG there are no other forces in the Y axis except for normal and MG therefore they have to exactly cancel and normal = MG so I can write that friction is mu MG and it's the mass of the block A so it's a 4-kilogram I can calculate this real quick friction, friction mu is 0.5, mass is a 4 gravity we're going to around that as a 10 so friction is simply 20 newtons and I'm ready to plug this in here for now let me just continue here FD 1/2 Big M V final squared, OK? I have all these numbers so let's start plugging them in, mass of B is right there 3 make sure you be careful be very careful with the letter so you don't pick out the wrong numbers, gravity is 10 height of B is a 2, friction is 20 what is the distance? So again, if you move 2 down this block over here is moving 2 sideways so this distance here happens to be the same as your drop because they're connected so they have to move the same distance so this D Here is a 2 1/2 the combined mass is 4+3=7Vfinal is what we're looking for, alright? This number is a 60 this number is a 40, 60-40 equals this is 3.5 V final squared so Vfinal=20/3.5 and we take the square root of that 20 divided by 3.5 we get 2.4 meters per second and that's the answer, now one last point I want to make here is in both of these problems we calculate the same thing we asked for the final speed obviously you can get a different question you can have a problem where I give you the final speed but then I ask you for the mass, right? So, there's all kinds of combinations of things you can do even though I'm asking in both of these for the final speed I could have given you that is taken away and asked you for a different number the process would have been almost the same thing, OK? That's it for this one let me know if you guys have any questions.

**Problem:** The 4-kg block is 1 m above the floor, and the surface-block coefficient of friction for the 3-kg block is 0.4. If the system is released from rest, find its speed just before the 4-kg block hits the floor. (The string and pulley are massless)

**Problem:** The system below is released from rest. Calculate the speed of the system after the hanging block has moved 1 meter. (The string and pulley are massless, and you may disregard any effects due to friction)