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Work as area under curve

David shows how the area under a force vs. position graph equals the work done by the force and solves some sample problems. Created by David SantoPietro.

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  • starky ultimate style avatar for user Alex the Chrononaut
    So if the line was a curve, than the work done would be the integral of the formula of the line, right?
    (29 votes)
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  • starky ultimate style avatar for user Gilleon Khonglah Ropmay
    But What if the graph was a quarter of a circle, how do you find the work done in that case?
    (14 votes)
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    • blobby green style avatar for user jamie_chu78
      Ah, true, the question is not primarily focusing the shape of the graph, just the area.

      However, Howeveeeeeer, there is a reason why the shape of the graph is chosen in the form of rectangles/triangles, because they're easy to work with, the equations which describe them are simple, relatively.

      So is the equation for a circle (thus any fraction of said circle), but where this breaks down for us, is where the equation for the graph is in a shape that we cannot easily calculate via conventional means.

      There is a whole system in mathematics dedicated to just this, just this one feature of graphs, it's so important, an entire system has been based around it, which you will need to learn at some point if you delve far enough into physics. This system is called calculus, which is split into two parts, differential calculus, and integral calculus. Sal Khan has done his thing, and has made video series on calculus. You need not delve into calculus right now, though I would recommend watching at least the first section of differential calculus, as it explains why it was invented. It's origins has to do with the motion of objects, and was invented by Newton to explain said motion.

      By the way, you'll need to know differential calculus to understand integral calculus. So start there :)
      (28 votes)
  • old spice man green style avatar for user Mark  Hill
    @ he takes away the negative work of pushing the burger to the left then to the right from the total force. However in reality work is work, right or wrong? it still takes extra energy (fuel) to push something back in the same direction you came from, you don't get a credit upon return. Or am I missing the point or taking the term "work" too literally?
    (11 votes)
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    • blobby green style avatar for user Bogdan Petrica
      If I understood it right, work = energy put into the system/object.

      For the first meter, we apply a negative force in the oposite direction of motion and that means slowing the burger down and this equals removing of some of the kinetic energy from the burger, thus negative work for the first meter.

      After that 1 meter we start applying a positive force in direction of motion and that means accelerating the object, increasing its kinetic energy and thus having positive work done on that object.

      So negative work means we steal some kinetic energy from the moving burger.
      (5 votes)
  • mr pink red style avatar for user lilymontpetit
    If you push something to the left, does that make it negative work?
    (3 votes)
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    • blobby green style avatar for user Teacher Mackenzie (UK)
      If I may add a contribution:

      The thing about work is, its a scalar. not a vector. So, the direction of work does not make it posisitve or negative

      The positive and negative thing about work is more about WHO does the work. For example. If You do work on something, then the work you have done on that thing is positive. If the thing does work on you, then the work that YOU have done on that thing is negative.

      Hope that makes sense

      IM
      (8 votes)
  • leaf green style avatar for user colin
    What if the graph was a changing curve with different slopes at different times?
    (4 votes)
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  • blobby green style avatar for user IzayahF
    how eat hamburgur
    (3 votes)
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  • leaf yellow style avatar for user Jose Molina
    To find the area under the graph wouldn't you still need to know calculus depending on the degree of the change in the force?
    (4 votes)
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  • male robot hal style avatar for user Aditya Vikram Roychowdhury
    How would you determine work done by a varying force by using calculus??
    (5 votes)
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  • leafers ultimate style avatar for user RoelRobo
    At , Why is the work being done to the burger negative? If the force acting on the burger is going in the same direction as the displacement of the burger, shouldn't the work be positive?
    (6 votes)
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    • blobby green style avatar for user Teal Harbour
      The X-axis is the position of the hamburger. The graph starts out below the X-axis with a negative value meaning it was initially moving in the opposite direction. therefor, causing negative displacement and ultimately negative work. To find the net work you must add the two together.
      (3 votes)
  • piceratops ultimate style avatar for user H.J.
    I don't understand. Why are there less work in the triangular graph than the rectangular one?
    (4 votes)
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    • eggleston blue style avatar for user IsotonicFlaccidCell21
      area shows displacement/distance, depending on whether it is a speed or a velocity time graph. Work done is directly proportional to distance, hence as rectangles have a larger area, given that the time (length) and magnitude of speed/velocity (height) is the same, more work is done in the rectangular graph.
      (4 votes)

Video transcript

- [Instructor] There's a hamburger sitting right in front of you, but you don't want it. Maybe it's because you're a vegetarian, maybe you're already full, maybe you found someone that needs it more than you do, so you're gonna push this hamburger to the right, with a force of four newtons and you're gonna do this and you're gonna push this a distance of five meters to the right. So a question we get asked would be how much work did we do in passing this hamburger to the person to our right? Now since this is a pretty simple problem we can just plug straight into the work formula, but I'm not gonna do that. I'm gonna show you an alternate way to think about this, because what we learn from this alternate approach is gonna help us in more challenging, complicated work examples. So I ask you to imagine this. Instead of just plugging straight in, consider the fact that we exerted a force of four newtons for the entire five meters that we pushed this hamburger to the right, so if we were to plot what the force was on our hamburger as a function of its position, it would look something like this. So if it started at zero it moved five meters to the right and we exerted a constant force of four newtons, That's why this is a horizontal line. It's horizontal, that means it was a constant amount of force, and the reason I'm doing this is because there's gonna be a geometrical significance to this work. Check it out. We can just use the work formula. We know work is the force in the direction of motion times the displacement or you could say the entire force times displacement times cosine theta. If you don't like this cosine theta, this just makes sure that you're singling out the force in the direction of motion and only that component of force that's directed in the direction of motion but for our hamburger example, the force already was in the direction of motion so we don't really need this. In other words the angle here would be zero. Whenever you take cosine of zero, you just get one and one times anything is just that thing so whenever your force is already in the direction of the motion of the object you don't really need that cosine theta and we don't need it here. So I'm gonna say that the force here was four newtons so we can calculate four newtons was the force for the entire displacement of five meters and we get the work that we did on the hamburger was positive 20 joules of work. Now if you're clever you might be like wait, four newtons times five meters, look it, that's just the area of this rectangle. So notice that this line is a straight line in this force graph just forms a rectangle in here and all we did, we took four newtons, that was just the height of this rectangle and we multiplied by five meters and that was just the width of this rectangle and if you multiply a height times width, you know what you get, you get the area of a rectangle, so what we found was when the force is constant one way to find the work done is by using the work formula but another way to find the work done is just to find the area enclosed by the graph. So from this line that determines the force down to this x-axis, if you find the area, that's gonna equal the work done. Now you might be like, alright, well, fat lot of good that does us, we can already find it with this formula. Why would I ever want to know that the work is equal to the area under a force graph? I'll show you why. Because in this case, yes, it was easy, we had a formula, we could just plug the force in here, we could just plug the displacement in there, we get our value, but think about this. What if our force was not constant? What if we had a varying force? In that case, what force would we plug in here? It'd be changing. One way to handle that scenario is using calculus, but if you don't know calculus that doesn't do you any good. Fortunately, there's another way to determine the work done by a varying force and that's to take this idea seriously. That's why I showed you this. It turns out the area underneath any force versus position graph is gonna equal the work, not just ones where the force is constant, even where the force is varying, if you can find the area underneath that graph, that's a quick and easy way to get the work done by that force. So what I mean is this. Instead of just pushing with a constant four newtons for the entire duration of this trip, let's say you started pushing with four newtons but you were getting tired and your force was diminishing and you were pushing with a weaker and weaker force until your force became zero newtons. This would be zero newtons of force when it hit this axis. Let's say the hamburger still made it five meters. You might be confused. How could it make it five meters if we're pushing with less force? Well it may have taken more time to get there, but let's say it still made it five meters. How do we find the work done by our force now? Well I'm still gonna make the claim that the work done is gonna be equal to the area underneath this force versus position graph, but you might be skeptical, you might be like wait a minute, all we really showed in that previous example was that the work done for a constant force was equal to the area underneath. How do I know for this varying force that the area underneath this graph is still gonna give me the work done? Well physicists and mathmeticians are clever. What they did is they said this, alright, they were like, the only thing we know is that for a constant force the area underneath is equal to the work, so let's just say this, instead of just considering this case where I diminish my force continuously, let's say I pushed with a whole bunch of constant forces but for a small displacement, so I started with four newtons but I pushed for like only 10 centimeters with four newtons, and then I dropped that down to maybe 3.9 newtons and I pushed for another 10 centimeters, and I keep doing this over and over, reducing my force but keeping it constant, and then dropping it again and here's why that's useful. Because the work done for this rectangular case is just gonna equal the area of all of these rectangles added up, right, because they're each constant forces so I know the work done is just equal to the area underneath, so if I add up the area of all these rectangles I get the work done during that rectangular process and here's the really clever idea. If we made these rectangles infinitesimally small they would add up to the total area just underneath this triangle, it'd be the exact same area. Now the area under this triangle would be exactly the same as the area under all of these rectangles and it'd be the exact same process we had before, because think about it, you'd be pushing with four newtons for like a millimeter but if it's infinitesimal I mean it's even smaller than that, but for the sake of just conceptually thinking about it, let's say you push with four newtons for a millimeter and then for the next millimeter you pushed with 3.99 newtons and then for the next millimeter you pushed with 3.98 newtons. That's basically just you diminishing your force continuously and going through this exact process that we described earlier. But now we know the area under all these rectangles is gonna just equal the work done, but that's the same area and process that's just the area underneath this triangle. Now, you might not buy that. You might be like, wait a minute, look it, these edges are throwing everything off, look at this little edge, this shoots out a little too far, that's a little overestimating the area, and then you gotta little hole in here, that's underestimating the area, you got these all over the place. Is that really gonna be equal to the area? And it will be if you let these widths of the rectangles become infinitesimally small, the error you're incurring from all these little overshoots and undershoots are also gonna get infinitesimally small, and if the error is getting infinitesimally small that means that the area under all these infinitesimal rectangles is gonna exactly equal the area under this triangle which is great news for us, because that means that the area under any force versus position graph, it didn't have to be a triangle, we could use infinitesimally small rectangles to represent the area under any graph. That means the area under any force versus position graph is gonna equal the work done by that force and now we have a powerful tool to determine the work done for cases where the force is not constant. This formula here where you have work equals Fd cosine theta, this only gives you true answers if you use a constant force. If the force is constant you can use this equation, but if the force is varying you couldn't just use this, but now we derived this. This is great. This showed me that if I can find the area under my force graph, my force versus position graph, I can find the work done. So, in other words, for this case right here, if I push with four newtons on my hamburger and then I reduce that force to zero and it went five meters, how much work was done? Well now I know how to find it. I just need to get the area under this force versus position graph and it's a triangle and I know how to find the area of a triangle. So the area under this triangle is 1/2 base times height, well so I'll have 1/2, the base here is gonna be five because this is five meters, so I have five meters and the height of this triangle is four newtons and so I've got four newtons times five meters times a 1/2 is gonna give me positive ten joules of work done during this process where my force diminished continuously, and that's why this idea is so powerful, because if your force versus position graph happens to take a shape that you know how to find the area of, like a rectangle or a triangle or some combination of rectangles and triangles, then you can quickly get the work done without having to know anything about calculus because you just get the area underneath. Let me just clear up any confusion. You don't actually have to draw any rectangles. That was just to prove this relationship. Once you know this relationship you don't need to draw or think about those little infinitesimal rectangles. That was just how we were proving to ourselves that even for cases where the force was varying the area underneath, it's still gonna equal the work. So there's one more thing you gotta be careful about. When we say the area underneath the graph is equal to the work done, we mean the area from the line that represents the force to that x-axis, and you might be like, well of course that's what you mean, what else could you mean? Well sometimes it gets a little unclear when you have a case like this. So let's say your force started down here. So we'll say we first started pushing on our hamburger to the left and then we're like oh there's no one there, awkward, we'll start pushing to the right, so our force starts off negative and then it becomes positive. Now if we wanted to find the total work done during this entire process, how would we do it? We can still use this idea that the work done is gonna equal the area underneath the curve, but for this first portion what do I do? How do I find the area underneath this portion? Some people are like underneath, alright, I'm gonna go underneath, we'll keep going this way, how much area's down here, but that's just crazy. That goes on infinitely. I did not do infinite work on this cheeseburger. No. The area I'm talking about is gonna be this area right here, so I gotta take all this area right here. That area is finite. It's from this line that represents the force to the x-axis. That's the area we're talking about, just like over here the area we'd be talking about is from this line that represents the force to the x-axis. It's all of this area. So how would I find these areas? They're both triangles. I can do this, so this one's gonna be 1/2 base times height, so it's gonna be 1/2 the base, this base right here is one meter, so it'd be one meter times the height. I've got negative height. Look it, it's down here. It's negative two newtons. That means I'm gonna have a negative area. Is that okay? Yep, that just means I did negative work on this cheeseburger during that time so I'm negative one joule. During the first meter of me exerting force on that hamburger, I did negative one joule of work, and then for this portion of the trip I'll find the area underneath. It's also a triangle, so 1/2 base times height. I'll have 1/2, the base this time is not three, the base of this triangle is only two, because it goes from one to three, so that's two meters times the height, well that is four newtons, so I'll multiply by four newtons and then I get for that portion of the trip, I did four joules of work. So, positive four joules of work for this portion, negative one joule of work for that portion. The total work done for the entire three meters would be four joules minus one joule and that would be positive three joules. So, recapping, if your force is constant you can just use this formula to find the work done. It's just Fd cosine theta, but you can also find the work done by determining the area underneath a force versus position graph and this is useful because this works even when the force is varying which would render this equation somewhat unusable but if the shape of the graph is one that you know how to find the area of, you can still find the work done by determining the area underneath the force versus position graph.