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Current time:0:00Total duration:12:31

Video transcript

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 going to push this hamburger to the right with a force of 4 Newtons and you're going to do this you're going to push this a distance of 5 meters to the right so a question we could ask 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 could just plug straight into the work formula but I'm not going to do that I'm going to show you an alternate way to think about this because what we learn from this alternate approach is going to 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 5 meters that we push 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 we started at zero it moved 5 meters to the right and we exerted a constant force of 4 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 going to 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 I'm going to say that the force here was four Newtons so we can calculate for Newton's 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 for Newton's times five meters that's just the area of this rectangle so notice that this line it's a straight line and 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 that was just the width of this rectangle and if you multiply height times width you know you get you get the area of the rectangle and so what we found was when when the force is constant one way to find the work done is just by using the work formula but another way to find the work done is just 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 going to equal the work done now you might be like all right 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 to 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 going to 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 for Newton's 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 Newton's 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 is still made it to five meters how do we find the work done by our force now well I'm still going to make the claim that the work done is going to 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 going to give me the work done well physicists and mathematicians are clever what they did is they said this all right they're 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 4 Newtons but I pushed for like only 10 centimeters with 4 Newtons 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 of the work done for this rectangular case is just going to equal the area of all of these rectangles added up right because they're each constant forces so no 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 I'll 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 push with 3.9 8 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 going to just equal the work done but that's the same area and process it's just the area underneath this triangle now you might not buy that you might be like wait a minute look at 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 got a little hole in here that's under estimating the area you got these all over the place is that really going to 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 going to get infinitesimally small and if the error is getting infinitesimally small that means that the area under all these infinitesimal rectangles is going to 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 going to equal the work done by that force and now we have a powerful tool to determine the work done for cases where the forces 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 derive 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 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 I know how to find the area of a triangle so the area under this triangles one-half base times height well so I'll have one half the base here is going to 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 Newton's times five meters times a half is going to give me positive ten joules of work 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 going equal to work so there's one more thing you got to 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 let's 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 going to 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 all right I'm gonna go underneath we'll keep going this way how much area is down here but that's just crazy that that goes on infinitely I did not do infinite work on this cheeseburger no the area I'm talking about is going to be this area right here so I got to 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 going to be 1/2 base times height so it's going to 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 at this down here it's negative to Newtons that means I'm going to have a negative area is that okay yep that just means I did negative work on this cheeseburger during that time so I have negative one Joule meter of me exerting force on that hamburger I did negative one Joule of work that for this portion of the trip I'll find the area underneath it's also a triangle so one-half base times height I'll have one half 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 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 3 joules so recapping if your force is constant you can just use this formula to find the work done it's just F D 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