Anderson Video - Work

Professor Anderson
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>> Let's talk a little bit about this idea of work. When you guys hear that word "work," what do you think of? When you hear the word "work," what do you guys think of? Anything that comes to mind. What's that? [ Laughter ] I missed it. I missed the joke. When you think of work, what do you think of? Do you think, oh, I got to get up? I got to go to work, right? You think, oh, end of the classes today. Now I have to go to work. So when you go to work, what do you do, right? You drive somewhere. You go to an office building. You push papers around all day, right. You make photocopies. You are doing stuff at your work, right, and this is why they're paying you because you need to do stuff. And that stuff that you have to do is really about moving things around, right. Moving things from here to there. In fact, if you go look at the construction guys that are building the engineering and interdisciplinary sciences complex, they are working really hard, right. They are taking these big, heavy things and they are moving them all the time. So work is intimately involved with this idea of moving something from here to there. And to move something from here to there, you have to apply a force. So let's think about how this works. Let's say we're going to accelerate a block, which we know how to do. If I have a table here, and I put a block on that table, and then I pull on it, I can accelerate it. So let's pull on it to the right with some force F. Let's say that it maybe initially was moving. It had some Vi. And now when it gets way over here, it's going to be moving with some Vf, okay. And so we will apply that force over this distance delta x. Now, you sort of know what's going to happen here, right. The box is going to speed up in this case, right. If it's moving along with Vi and we pull on it to the right, it's going to be moving faster than it was before. But how does all this stuff relate? One thing that we know is the good, old kinematic equations, those work. So let's go back to the kinematic equations and let's look at the one that has velocities and distances. And the one that I'm thinking of is this right here -- Vf squared equals Vi squared plus 2a delta x, okay. Everybody remembers that one. But we don't know exactly what a is. But if the box is mass m and we're pulling on the box with a force F, then, in fact, we do know what a is, right. It's just F over m. So this whole thing becomes Vf squared equals Vi squared plus (2 times (F over m)) times delta x, okay. Good, old kinematic equations. Now we applied Newton's Second Law. We put that in there as well. Okay, but we can rewrite this stuff because this looks a little ugly here, right, the 2(F over m). So let's rewrite that. Let's divide everything by 2. One-half Vf squared equals 1/2 Vi squared plus (F over m) times delta x. But now let's get rid of the m by multiplying everything by m. And if I do that, what does this become? One-half m Vf squared equals 1/2 m Vi squared plus F delta x. And that's starting to look a little familiar, right, because what is 1/2 m V squared? That's our good, old kinetic energy. And so let's rewrite this slightly. I'm going to move the 1/2 m Vi squared over to the other side. And then apparently, it's equal to this F delta x. But this is kinetic energy. This is final kinetic energy. This is initial kinetic energy. This is just a change in the kinetic energy of the object. What about this? Can we call that something special? And the answer is yeah. Let's call that something special. Let's call that thing the work W, okay. What is work? Work W is equal to F delta x. Force over a distance, okay. Of course, the Work Kinetic Energy Theorem, and what it says is that W goes to changing the kinetic energy, okay. So in that simple system that we just looked at, the work that you put into it went into speeding up the box, all right. Now, this is a very sort of select subset of the overall Law of Conservation of Energy, but it's an important one to understand. Okay, so if work W is equal to F delta x, then when we speed up the block by pulling on it to the right, we can increase its kinetic energy. But what if we're pulling up at some angle F and the block is still moving to the right delta x? How do we calculate the work in this case? Well, if that force is up at an angle theta, and it has components to it, horizontal and vertical, and it turns out that only the horizontal component is critical to calculating the work. And so the work W is, in fact, F cosine theta times delta x, okay. This was sort of a special case that we had over here because we were pulling it perfectly horizontal. But now if we're pulling up at an angle, you have to use cosine of theta in there. So what are these things? F is, of course, the force applied, right. How hard are you pulling on this thing? Delta x is the displacement, okay. How far have you gone and in what direction have you gone? And therefore, theta is the angle between those two things, F and delta x. F is a vector. Delta x is a vector. There must be an angle between them, and that's what you put in for theta. Okay, so let's go back to the case where we just had the force pulling directly to the right. So if the force is pulling directly to the right-- And the displacement is also to the right, now we can calculate the work. Work W is F cosine theta delta x, but what is theta in this case? Theta, remember, is the angle between the force and the displacement. So if they're both in the same direction, what do I want to put here for theta? Zero, right? They're both in the same direction, so the angle between those two vectors is zero degrees. What's cosine of zero degrees? One. And so we just get back to what we had a second ago, okay. Work is F delta x if the force is perfectly horizontal. All right. Let's try a slightly different problem. Let's say that we're going to lift the object up by applying a force, F lift, but we're still going to move it to the right, okay. So here's the box. You lift up the box, and then you move it nice and slow to the right. How much work have we done in that case? Well, we know that it is. Work is F cosine theta delta x. The force that we apply is F lift. What's the angle between those two? Ninety degrees. What's the cosine of 90 degrees? Zero. So in this case, what we're saying is we haven't done any work at all. The work is equal to zero. Does that seem right? Omar, what do you think? [ Inaudible Speaker ] So I'm lifting it up, and then I'm just going to move it real slowly to the right, okay. And we just said that the work is equal to zero. You like that answer? [ Inaudible Speaker ] Okay, mathematically, he likes it. But let's think about it from the point of view of myself lifting up this box, right. I have this box right here, and now I'm going to move it slowly to the right. Why do I mean slowly? Because the Work Kinetic Energy Theorem says that if you don't have any work, you can't have any change in kinetic energy. We know if I push it really fast, then I am applying a force in that direction, and that will change the kinetic energy. And therefore, you have done work. But if I move it slow like this, then my force is basically up. The displacement is to the right. And so I've done zero work on it. But I can't do this all day, right, even with something as lightweight as my cell phone. Although with this wood case, it suddenly got a lot more massive. I can't do this all day, right? Why can't I do this all day? Because eventually my arms are going to get tired. It gets heavy. So what's the difference? How come it feels like I'm doing work and yet we just said that there's no work being done? How do we reconcile that? Are you guys getting hypnotized? You're getting sleepy. [ Inaudible Speaker ] Okay. It's not doing work on me because gravity is straight down, and so the gravitational work, if it's moving horizontally, is also zero, right. The difference is rather subtle. It's this. When I hold up an object as a human, I have to exert some energy to do that because it takes me physiological work to lift it up. Why? Because I have a heart that's pumping blood, going through my system. There's friction. It's generating a lot of heat. My muscles are, in fact, twitching ever so slightly as you go. And all of that burns energy, and that takes physiological work, okay. So physiological work is not the same as physical work, how we define work in physics, but it is there as a human being, right. As a human being, it just takes a lot of energy to be alive, right. You're 98.6 degrees. You're radiating all this energy all the time, okay. But physically, we say it takes zero work. All right.
>> Let's talk a little bit about this idea of work. When you guys hear that word "work," what do you think of? When you hear the word "work," what do you guys think of? Anything that comes to mind. What's that? [ Laughter ] I missed it. I missed the joke. When you think of work, what do you think of? Do you think, oh, I got to get up? I got to go to work, right? You think, oh, end of the classes today. Now I have to go to work. So when you go to work, what do you do, right? You drive somewhere. You go to an office building. You push papers around all day, right. You make photocopies. You are doing stuff at your work, right, and this is why they're paying you because you need to do stuff. And that stuff that you have to do is really about moving things around, right. Moving things from here to there. In fact, if you go look at the construction guys that are building the engineering and interdisciplinary sciences complex, they are working really hard, right. They are taking these big, heavy things and they are moving them all the time. So work is intimately involved with this idea of moving something from here to there. And to move something from here to there, you have to apply a force. So let's think about how this works. Let's say we're going to accelerate a block, which we know how to do. If I have a table here, and I put a block on that table, and then I pull on it, I can accelerate it. So let's pull on it to the right with some force F. Let's say that it maybe initially was moving. It had some Vi. And now when it gets way over here, it's going to be moving with some Vf, okay. And so we will apply that force over this distance delta x. Now, you sort of know what's going to happen here, right. The box is going to speed up in this case, right. If it's moving along with Vi and we pull on it to the right, it's going to be moving faster than it was before. But how does all this stuff relate? One thing that we know is the good, old kinematic equations, those work. So let's go back to the kinematic equations and let's look at the one that has velocities and distances. And the one that I'm thinking of is this right here -- Vf squared equals Vi squared plus 2a delta x, okay. Everybody remembers that one. But we don't know exactly what a is. But if the box is mass m and we're pulling on the box with a force F, then, in fact, we do know what a is, right. It's just F over m. So this whole thing becomes Vf squared equals Vi squared plus (2 times (F over m)) times delta x, okay. Good, old kinematic equations. Now we applied Newton's Second Law. We put that in there as well. Okay, but we can rewrite this stuff because this looks a little ugly here, right, the 2(F over m). So let's rewrite that. Let's divide everything by 2. One-half Vf squared equals 1/2 Vi squared plus (F over m) times delta x. But now let's get rid of the m by multiplying everything by m. And if I do that, what does this become? One-half m Vf squared equals 1/2 m Vi squared plus F delta x. And that's starting to look a little familiar, right, because what is 1/2 m V squared? That's our good, old kinetic energy. And so let's rewrite this slightly. I'm going to move the 1/2 m Vi squared over to the other side. And then apparently, it's equal to this F delta x. But this is kinetic energy. This is final kinetic energy. This is initial kinetic energy. This is just a change in the kinetic energy of the object. What about this? Can we call that something special? And the answer is yeah. Let's call that something special. Let's call that thing the work W, okay. What is work? Work W is equal to F delta x. Force over a distance, okay. Of course, the Work Kinetic Energy Theorem, and what it says is that W goes to changing the kinetic energy, okay. So in that simple system that we just looked at, the work that you put into it went into speeding up the box, all right. Now, this is a very sort of select subset of the overall Law of Conservation of Energy, but it's an important one to understand. Okay, so if work W is equal to F delta x, then when we speed up the block by pulling on it to the right, we can increase its kinetic energy. But what if we're pulling up at some angle F and the block is still moving to the right delta x? How do we calculate the work in this case? Well, if that force is up at an angle theta, and it has components to it, horizontal and vertical, and it turns out that only the horizontal component is critical to calculating the work. And so the work W is, in fact, F cosine theta times delta x, okay. This was sort of a special case that we had over here because we were pulling it perfectly horizontal. But now if we're pulling up at an angle, you have to use cosine of theta in there. So what are these things? F is, of course, the force applied, right. How hard are you pulling on this thing? Delta x is the displacement, okay. How far have you gone and in what direction have you gone? And therefore, theta is the angle between those two things, F and delta x. F is a vector. Delta x is a vector. There must be an angle between them, and that's what you put in for theta. Okay, so let's go back to the case where we just had the force pulling directly to the right. So if the force is pulling directly to the right-- And the displacement is also to the right, now we can calculate the work. Work W is F cosine theta delta x, but what is theta in this case? Theta, remember, is the angle between the force and the displacement. So if they're both in the same direction, what do I want to put here for theta? Zero, right? They're both in the same direction, so the angle between those two vectors is zero degrees. What's cosine of zero degrees? One. And so we just get back to what we had a second ago, okay. Work is F delta x if the force is perfectly horizontal. All right. Let's try a slightly different problem. Let's say that we're going to lift the object up by applying a force, F lift, but we're still going to move it to the right, okay. So here's the box. You lift up the box, and then you move it nice and slow to the right. How much work have we done in that case? Well, we know that it is. Work is F cosine theta delta x. The force that we apply is F lift. What's the angle between those two? Ninety degrees. What's the cosine of 90 degrees? Zero. So in this case, what we're saying is we haven't done any work at all. The work is equal to zero. Does that seem right? Omar, what do you think? [ Inaudible Speaker ] So I'm lifting it up, and then I'm just going to move it real slowly to the right, okay. And we just said that the work is equal to zero. You like that answer? [ Inaudible Speaker ] Okay, mathematically, he likes it. But let's think about it from the point of view of myself lifting up this box, right. I have this box right here, and now I'm going to move it slowly to the right. Why do I mean slowly? Because the Work Kinetic Energy Theorem says that if you don't have any work, you can't have any change in kinetic energy. We know if I push it really fast, then I am applying a force in that direction, and that will change the kinetic energy. And therefore, you have done work. But if I move it slow like this, then my force is basically up. The displacement is to the right. And so I've done zero work on it. But I can't do this all day, right, even with something as lightweight as my cell phone. Although with this wood case, it suddenly got a lot more massive. I can't do this all day, right? Why can't I do this all day? Because eventually my arms are going to get tired. It gets heavy. So what's the difference? How come it feels like I'm doing work and yet we just said that there's no work being done? How do we reconcile that? Are you guys getting hypnotized? You're getting sleepy. [ Inaudible Speaker ] Okay. It's not doing work on me because gravity is straight down, and so the gravitational work, if it's moving horizontally, is also zero, right. The difference is rather subtle. It's this. When I hold up an object as a human, I have to exert some energy to do that because it takes me physiological work to lift it up. Why? Because I have a heart that's pumping blood, going through my system. There's friction. It's generating a lot of heat. My muscles are, in fact, twitching ever so slightly as you go. And all of that burns energy, and that takes physiological work, okay. So physiological work is not the same as physical work, how we define work in physics, but it is there as a human being, right. As a human being, it just takes a lot of energy to be alive, right. You're 98.6 degrees. You're radiating all this energy all the time, okay. But physically, we say it takes zero work. All right.