 By the way, nobody's going to Clarkson for transfer, the ball, and $20,000 we just got to throw away. Oh well, not my $20,000. What? I can't, I don't think so, no, you have to be accepted, you have to be half full-time, full-time, so. That's Jim Claes's. Full-time. Major in criminal justice for all I care about. I don't know. All right, no questions, no concerns over the last problem yet. So, besides what was it? Well, you're ready for a new one then. Here's the setup. Blinding along, well, I guess it's a smooth surface. It will make it even easier on you for that. But then there's a cable that goes over there too, a weight, and you have a spring connected to it there. All right, so that's the basic setup, fresh length, 12 inches, distance, X. But the system will go kind of a warm up, but a step beyond what we were trying before with the work energy equation. 18 inches is to roll. Spring is attached. What was that? X max, maximum distance that will travel. But the spring is straight up and down, is that a good question? No, it's a legitimate question. All right. Are we supposed to take that tension from A to B? Can we use that as an outside, like a word force? Well, good question for once. Yes, you can. You could take A as a system itself, and then the tension would be an outside applied force. However, is that the easiest way to do it? Let's put up the work energy equation. We have actually here a system of two things. There's at least two things doing, moving, and undergoing all kinds of things. So we can use the work energy equation where there are more than one object undergoing any of the terms in here, which is real nice when you use this work energy equation. You just sum this stuff up. What was the next term? Delta T. If you've got more than one thing, no sweat. You just sum them up. Delta change in potential energy terms is the concern over there. All right. So are any of those terms zero? Well, that comes back to Jake's question. Jake's proposal is that we take block A, make it into a system of its own. All right. So you'd have the tension there. You'd have the spring coming back there, and then you treat it like any other problem that we've done before. And then you'd have to finish it. You'd have to work with it. And then you'd have the tension on B. It's weight, which obviously need to be unbalanced because otherwise the system wouldn't accelerate. It'd just stay at rest. So we'd have to find the connection between those two things. So that's one possibility. What about the possibility of the thing as a single system? Now what's the work term? Y zero. There are outside forces. We've got reactions here. We've got support there. However, it's not just that they're conservative, but actually they're not. Reaction forces aren't conservative. But what's more important to us, how much work do those forces do? None. Because those reaction forces, the force holding this and the force there in the wall, they're not moving anywhere. So they don't go anywhere. They don't move anywhere. They don't do any work. So if we do that, that work term becomes zero. But then how do we handle the tension in that line? How do we handle the tension in the line? B has some potential energy change. So that term's not zero. At least for B it's not. For A it is. But for B it's not. How do we handle this unknown tension in that cable connecting the two blocks? Maybe it's already handled from one of those terms. Which one? The tension wealth. This is, which term is this? Kinetic energy. Kinetic energy. So this only matters if a mass, any of the masses is changing speed. Which clearly A and B are. So we're just going to have to sum those two things together there. But the tension itself is not a mass that's moving. So there's no kinetic energy associated with the fact that the cable's moving. These remember we take these to be stretchless, massless cables. Since it's massless we don't care what the height of the cable is. Cable's stretchless. So we couldn't put it in there. We don't care what the tension is within the cable. If we make this entire, one entire system it's an internal force. We don't care about internal forces. They're equal in office. They cancel each other. We're not in the least bit concerned. So we need to do this for the kinetic energy term for both. The gravitational energy only applies to B. And then there is a spring. So we do need to include that one in there to take into account that we have a spring that changes length. So what we have here actually is a system where there's no outside forces that do any work. What we have is a system, an isolated system, and a system that undergoes conservation of energy. No energy is being introduced by the outside forces. All that can happen then is energy can transfer between these three groups. And that's indeed what we have, a decrease in the potential of gravitational potential energy is going to contribute to an increase in kinetic energy and an increase in the elastic energy because we will be stretching the strength. What do we do? Any trouble with this one? That you're forcing. One half MB squared for the two of them. Change in one half MB squared. What about this one? Pretty straightforward. Do you know how far it's falling? At least for this first calculation we know what's going forward. What about this one? You've got the spring. Let's see, the system starts with A here so the spring is straight up and down. Later the spring moves to here and in fact as it's always moving the spring is at a changing angle. I'm confused that's something we said to define a line at x equals four inches. Always. Starts here, goes four inches. So you need to find the speed at four inches after it's gone four inches. You okay then? Do we have the height that B drops? Is it going to be that 18 inches? Well, if for this first part are these two the same problem? Well, how do we know where x max is? That's where the system starts from rest, speeds up for a bit and then finally the string gets stretched out enough and brings the system back to a start. Now stop. So if these two are the same then the A is zero. Because the velocity has got to be zero at x max. If it's not zero then it's still going. So these are two separate problems. So for this one if A moves four inches B drops four inches. That makes that term a pretty straightforward calculation. What do you do about the spring that's pointing in one direction here in a completely different direction here? Delta vk times minus del one squared. Where del is what? That del is the length at any instance. So del one would be l one, the length of the spring at point one, which is 18 inches. So we can do that. Minus its rest length. 18 inches is just how long the spring is at point one. It's rest length is 12. So del one is simply 6 inches. What about del two? Well that's how long the spring is when the block has moved to four inches. What about the angle of the spring? Doesn't matter. All we want to know is how much energy is in the spring. We don't care what angle the spring is. We just care the fact that it took some energy to stretch it. It doesn't matter what angle it was stretched at. It just needed some energy to stretch it. So find those three terms. Then I'll leave you with x max because then I have another problem I want you to look at instead. Do these three terms separately. Remember too, you've got some units you have to watch in this problem and you've got some minus signs you've got to watch if anything goes down something else has to go up because no energy enters or leaves this system. Only outside forces can bring or take energy out. I know it was a stretch length and that spring was 18 inches. Because I gave you that. This dimension is 18 inches and the problem starts x equals 0 and b equals 0. It starts from rest there. It was released from rest. That's what's great about the work energy system is just really an accounting problem. You can divide it up into small parts as you want. Just add them all together. As long as you've got the minus signs right, the units right, and when you add things together you're adding the correct things to each other. You can't just mix them and match them. But if you have more than one thing moving figure them out separately, add them together. Actually for the kinetic energy term when you write that out they're both moving the same velocity. You will know they're different masses. But b, in this case, will have the same velocity. So that will reduce to since the velocity at 0.1 is 0. This will reduce to the velocity 0.2 times the sum of masses. That's the mass or the weight. These are going to need the mass. It's the mass or the weight I've given you here. I told you that any time we see the pound it's going to be... What about the units? Leave them that way. That's my suggestion. You can convert them to slugs or pound, mass or whatever it is you want to. But if you just leave them like this then when you put it in the equations the units will just come right back out and you'll be fine. In fact on the gravitational energy term g delta h, mg is the weight. So just put the 8 pounds in there because that's block b is the only one going through a change in height. I want to check each of the three terms with somebody before you go any farther. No sense adding them together because that's all the problem. If you've got any one of those three terms wrong then that means minus signs and units. A little bit of mix of inches and feet. The kinetic energy term, anything with inches unless you're doing the velocity of inches per second if you're doing the velocity of feet per second you don't have to do it anyway. Can I get energy terms in terms of unit conversion? The only time you need to do unit conversion is this since it moves four inches and we have gravity usually in feet per second squared rather than inches per second squared. And the spring one we've got its lengths and inches but its strength is in feet. So for the kinetic energy term you don't have any unit conversion for inches and feet. Well it's up to you. However, the velocity, let's say feet per second have any of the three terms you want to check? Do you want to check with somebody else first? No. Plus don't forget all of these you can tell whether they're getting whether they're positive terms or negative terms so always check that too. Don't forget the square stuff. I'm actually still starting with del one and del two. I'm still starting on that first term. How to get this? Yeah, I'm not sure about del one. This is the length of the spring at any time of the problem. This is the rest length of the spring. How long it was when you bought it from the store and took it out of the box. Which I gave 12 inches. So at 0.1 where the box is up here to start with the spring is 18 inches long. That's L1. So L1 is 18 inches minus the rest length 12 inches gives you a stretch. That's what this is, the stretch in the spring the over stretch if you will is 6 inches. 0.2 is when it's moved 4 inches so it's an aggregate term. It's an aggregate term. We've got 4 inches by 18 inches you can figure out how long the spring is and then that's L2 you subtract from it L0. Watch your units because the spring strength is in per foot, ounce per foot. What do you have? Remember this work energy term the energy quantities are all changes in energy don't leave the delta out or you've got a different equation. What is it? What do you tell me? Delta VE will tend to what? Inches or feet? Doesn't matter if you want me to look. If you give me any one of those 3 terms I can check it. Which one? The kinetic term. Would you just put in 0.37 VG squared? What's 0.37? 1.5 times the masses divided by G 0.37 12 that's the 2 masses together divided by 32.2 0.186 times B2 squared since D1 is 0 terms I can check on. Did you get that for delta T finally? What do I have? Check the entire problem but if I check any one of the 3 pieces independently because it's unlikely you've got all 3 of them wrong so we can find out just which one of them is in the small problem to correct. For the mass term would that be 8 over 32.2 or would that be 8 plus 4 over 32.2? Well, you tell me. What's changing speed? One thing or two things? Two things. And they're directly connected so they always have the same speed since they always have the same speed and if D1 is 0 then you just add the 2 you can just add the 2 masses together even for a more sane way of changing the potential energy term the MG of H Oh, here? Well, how much does A change? None. None. It's only the height change in B so this would be 8 pounds only B I was confused since they were still directly connected Yeah, but for the potential energy term gravitational potential energy term A has no change What do you got? G-negative? What? 2.67 Yeah. B per second? Yeah. I had 2.64 so that might be a little bit of a reservoir Delta V what? V-E What do you have? Foot pounds? Earth to feet from inches still wouldn't necessarily have that V-E I had 1.36 so that might be round off since V-G that's just what I see you can multiply It is. It is. You don't want me to lie to you, do you? I do. You probably do. That's right. 2.36 1.36 is okay? Yeah. You're multiplying fine. Look at the physics. No, no, no, no, no, no. It's not the height is 4 It is a change in height. This hurts. How are we getting past negative 4? Well, we each have to. It doesn't matter what is positive or negative direction. This is a loss in height in a gravitational field that's got to be then a negative term. Even if you choose down the positive then that makes G Delta V-G 2-V-G 1 and there's no way that can come out with anything but negative no matter whether you choose up as positive or down as positive. That's going to be a negative term. Feet per second. There's 2. There's 2 potential energy terms. Gravitational and elastic. Gravitational. You got what? Even so, if you were listening to Jake's problem a second ago it's got to be negative. I don't know why. There was a negative time. Work is still zero. Work is still zero. Delta T is zero as well. Starts at rest, finishes at rest. So you only have those 2 terms left over. So all of the gravitational energy it loses goes into stretching the spring until they mount. Yeah, you know what I'm talking about. Yeah, you know what I'm talking about. Is it what? Negative energy. You can have it negative. You can find and grab. The downward direction is negative. You're not watching it? No, because it's not how it sounds to be. How many inches it might be about that? The downward direction is positive. The upward direction is negative. Are you swearing? Are you swearing? Where did you find the downward direction is positive? Remember, you square before you subtract the 2 terms. Yeah. Downward direction is negative. Because it didn't be reversed. G itself never has a negative sign on it. Only your geometry has a negative sign on it. So when you use this this is never negative. We're talking about a downward direction of negative. No. There's no negative in here. Don't introduce one. I thought it depended on which direction you need to find this negative. No. It's always mg. Then it's going to be h2 minus h1. It doesn't matter whether you have up or down as negative. It's always going to finish low. Then it starts. Sometimes you have a decrease in height and gravitational field. You want potential energy. There's another way to look at it. By 4 inches. It'll be 18 inches by x max squared as an unknown. And height will also be x max as an unknown. So you have a quadratic in x max squared per feet. One of these terms are either foot pounds or inch pounds. Times probably makes more sense since I said at the start we'll look for a v in feet per second. One of these terms has got to be in foot pounds or inch pounds. No squared is in there. A gravitational term and elastic term. It moves back to here. Which it could I guess move back. No, actually it wouldn't. It would stop there because we all have concentration of energy. 0.6, 0.62. 0.65 could be around about my inch. And it's 8 and x separating the thing. The quadratic. You have to graph it. For later. The x max in it could also have another term so it won't square root plainly. Square root of half. You're going to have this term. This term has a one half with the x max under it with another term that won't square root plainly. And then you have this one here. And then that's what gets squared. Leave that one further. We're checking you at home to solve it if you want. But I got 0.62 feet. There's a number of different ways but I don't think you'll be able to solve it directly. There's less let over that complicates itself. Okay, Jay? Ready for a harder one? There's no tricks in solving this one. But again, I recommend you do each of the three terms or each of the four terms that will exist separately. Different pieces moving will have to sum the motor on the inside that worked this time. So how you're going to do the spring not essentially anything different than that problem. The mass is moving at different speeds not essentially anything different. There's a little bit of friction in this problem. What about the motor? Well, what if you put it inside the system like we did here then none of the force of the motor is from the outside? Then you have to count for it. It becomes a little bit trickier problem because what happens then is you're worried about energy coming into the system. That's what this term is. Is any energy coming in or leaving the system? Then the energy just bounces back and forth between these ones like the problem we just had. If you put the motor on the inside though there's energy coming to it. You've got to plug it in somewhere. So we have energy coming into the system by either the electrical connection on the motor or the fact that you've got fuel that's consuming you. It's that kind of energy. So you can't take the motor out of this work term. It can only be as a work term. But that's fine because you've got the force it exerts. And you know how long it exerts that force or the work term becomes pretty straightforward for the work loader. Were you worried about the friction when it was the work term? Yes, you were funny. You've got two work terms now. The motor is doing positive work and the friction is doing negative work. Is there a friction in this one? Yes, there's a friction here. That thing runs in the chute. I put that in there. See if I gave you that and we're going to know what the normal force is and what's the normal force when something is squeezed into the chute like that. So I'm not going to have this force, friction force directly. This one's still a sum of each of those individual parts. It's still going to be done that same way. Different things are zero this time. What Bobby said you had a question? Are any of them zero? In this case none of them are zero. We have work, always negative work. It doesn't matter if something's going up, down or sideways. Friction is always negative work because it always opposes the direction of motion from the dot problem. In the kinetic energy term we've got both A and B are moving or are having changes in speed. Only B has any change in height and delta B is the spring. Only to the right. Point two comes to a stop. Watching your minus signs. Watching your units. It'll go too far. It's not heavy enough. There's errors. As soon as you stop just guessing. I'm done. I hope you won't guess. What was your guess? It was D1 of B half. I didn't hear the whole thing and I started nodding. You just said you wanted a head nod so you got it. But I don't know what it was nodding to. It's like signing a paper ad in red. It can't be held responsible. I'm going to start thinking. Right before lunch. Lost a momentum. Only one more week before spring break. That's pretty cool. Southwest Airlines. Down there. Four a lot of hail. What's next Friday's videos before you leave? Yeah. Right. We bet that's going to happen. Let's see. I think I'm going to fix it too. Or can you just cancel Friday's class and leave early and just post a slideshow and things like that. So you just take that through the end of the term? Quite as well. Trouble is it took me about two hours more than just me driving in and doing the class and driving home. So I'm going to live one mile away. That's not that exactly. I'm going to do the route to the faculty parking lot. Oh, right. I don't know what Friday's see for route is going to be. I don't get that until that morning. It automatically shows up in our GPS to the school tables. 2.64 Yeah, 2.64. Was yours feet per second? Yeah. No, no, I said that. We agreed it was the inches per second. I mean feet per second. Your inches per second? Well, that's all right then. No, I'm not quite sure. The second problem is SI units. 2.64 feet per second is the first problem. We're a little too much for roundup. How would you get for the work done by the motor? All right. Would you get for work done by friction? No. More work than the motor is doing. For what? For 6. But that could be roundup on those delts. 4.493 4.493 4.78 I just want everybody to get double check. Could you do a 16.7 on these terms? Close. 7.73 more. What squared is that? Minus. Minus is the squared. Square the velocity and then it's the minus. Plus, you look at that term and you say, well, all of it has to be negative. The system was moving and then it finishes at a stop. So it's got to be negative potential energy. I mean, the kinetic energy change. Now my math is way round. So, did you what did you have for del 2? 4.78 Okay. So now the change of potential energy. 34.6. Alright, what did you have for delta VG? 1.9632 Okay. What about delta T? Maybe 10.08 minus 0.72. 0.72? I didn't have that. Okay. Alright, so minus 1.5 times the mass of V. 1.5 times the mass of V. That's what? 0.32 minus nothing. He isn't moving. I mean, V is moving. He isn't moving 1.2 square. So there's actually a factor of 4 on that one. It's 1.5 square. 7 minutes. 17.1 Let's check. Where are we? Minus 12. Minus 10.1. Okay. 35. You're missing the term. 70.6 Beautiful. That's why you remain my second doctor student. That's great. Time's up. You still got 7 minutes left? We're going anyway. VK? Colin? What? Checking in? For the final mass. Could be. I had 17.1. Could be round off. The spring stretch terms have different precision depending on what you think. So what? You go look at a garage door when you're there. That's what they've got on. Both of them are minus. Are you doing work? Yeah. You like doing that one. Okay. You know what you're challenging me to do is come up with a problem. Even harder than this. This is a problem I had all four terms and multiple parts to some of the terms. And now Colin is challenging me that the difference between yours and mine is round off. But we can't check. Because I can't check your individual terms which is the easiest way to to debug this problem. No. 17.1.