WEBVTT physics/high-school-physics/selhorst-jones
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Hi, welcome back to educator.com, today we are going to be talking about work.
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This is going to be our first introduction to energy.
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Introduction to energy, but we are talking about work, you thought that this is going to be an introduction to energy, so what is going on here!
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Do not worry, both these things are actually really deeply connected.
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To be able to talk about energy, we are going to wind up talking about work, and to talk about work, we are going to have to talk about energy.
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But, first we can tackle the idea of work on its own, it is going to really help us understand energy.
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Bear with me, and let us just learn about work as its own idea, and then we will move on to using it as part of energy.
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Before we can rigorously define what work really means, let us look at a couple of scenarios before we try to figure out what we want, how we want to use work, what we want to find it as.
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Let us say we have got some 10 kg block sitting on a frictionless table.
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We are going to have a whole bunch of different scenarios, we have four different scenarios in the next page, and each one of those, we are going to slide the block some distance with some force, but these will vary in the scenarios, and we will talk about each one.
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In our first one, we have got 'f' and 'd'.
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We push it from here, to here, and we do that with that much force.
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So, that is one thing, but what if we did it for this much distance?
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Same force, but a much larger distance.
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If that is the case, by the end of it, because it is frictionless, it is going to pick up speed.
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Remember, the longer the force is going, the longer the acceleration, so it is going to have more speed in it.
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So, it makes more sense to think that, the longer the distance, the more energy, the more work that we put into it.
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You push for a longer period of time, it makes more sense that you are putting more work.
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If you had to do something farther, that is more work than having to do it for less distance, if you are putting in the same force every moment.
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What if instead, if we had a really big force and a small distance?
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Clearly, between this box up here, and this box here, we are going to have a much bigger amount of work put in to the yellow box, the bottom box, the big force and the small distance, than the small distance and the small force.
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It is not going to go as far, it is not going to move as far, at the end of that distance we will put way more work into it, because we are pushing so much harder.
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If we push lightly on an object for a little distance, versus we put our entire body against it and push as hard as we can for a little distance, at the end of it, the thing is going to be moving a lot faster, makes sense to think of it as putting in more work.
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What if we put both of them together, if we had a really big force, and a really big distance!
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Clearly, that is going to be the one that has the most work.
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We push really hard, for a really long time, that is going to be the most work.
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So, it is about force and distance together.
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More force means more work, more distance of pushing that force means more work.
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Put both of them together, and that is even more work.
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What if we consider two different blocks with masses that are different?
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Same force in both cases.
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So we have got the same force, and the same distance, but in this one, we have got a really big mass, whereas in this one, we have got a really small mass.
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Notice that, because F = ma, the amount of acceleration is relative to the masses, so one vision: we have got a low mass object traveling really quickly at the end of its distance.
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The other one, we got a high mass object traveling really slowly at the end of its distance.
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But, we had to push the same force for the same distance.
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So from the point of the view of the pusher, it is the same effort that we had to put into it.
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They changed the system of mass, having to do with the speed that it is going at.
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The two things are connected.
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But, from the point of view of the pusher, it is the same work, it is the same amount of push for the same distance.
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These are the same thing, so even though the result is different and how it comes out, it is a different mass with a different speed, we are going to have the same pushing, the same work in it.
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So, we are going to think of work as just force and distance connected together.
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The mass is not going to have direct effect.
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These are different results, but it is the same force that is put into the system.
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With this idea in mind, we have got a notion that work is force involved multiplied by the distance.
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Way more force and way more distance stacks up hugely.
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More force means more work, more distance means more work, these things that make good intuitive sense.
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If we had to push a car for 1 foot versus push a car for 1 mile, or push a car for a metre versus pushing the car for a km, clearly the really bug distance is going to be the one that is going to take more work, if you are pushing it with the same force the whole time.
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But there is one last thing that we have to consider before we really define what work means.
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Consider the idea that there is a giant 20 tons semi-truck, which I will illustrate with an incredible box as my figure, and you are standing in front of it, and you are pushing as hard as you can, you push so hard on that, but it is this huge 20 tons semi-truck, it does not move at all, it does not budge even a mm, so did you do any work?
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In one way, you definitely strained, and you put a lot of effort into it, you tried really hard, so the idea of looking from the point of view of the pusher is one thing, but what we really want to define is, we want to look at work as the way you change the world around you.
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Even in the case of boxes with different masses, we were changing the world, we were putting a velocity into it, which was not there previously.
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We created acceleration, we created change in velocity, by putting that work in.
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In this case, you put a lot of effort in, you tried really hard, you push and strain, but nothing happens.
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So we want to define work as 'change in the world'.
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You have expended a lot of effort, but you did not change anything.
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So, we are to going to define work as change in the world, so no distance, if you did not make any distance, even if you had a huge force, no distance means no work.
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It is force × distance, zero distance means no work, even if it is a giant force.
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With this idea in mind, there is one thing to consider.
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What about this scenario!
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We have got a block, and this lock moves in this direction.
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But the entire time, we have got a force moving this way, perfectly perpendicular to the motion of the block.
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Does the force do any work on the block?
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We talked about the fact that, if you did not change anything, then you did not put in any work.
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If there is no change from the force, then the force does no work.
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Force has to be connected to the distance.
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So in this case, no work is done by the force, because it is perpendicular.
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That motion to the side is going to happen whether or not the perpendicular force does anything or not.
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It is not able to change its distance, because it is perpendicular.
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The only way it would be able to change its motion, is if the motion was going somewhere like this, but the entire time it slides along, the effect of the force has no acceleration, because its motion is this way only the entire time.
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So, this one does not happen.
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Force and distance are perpendicular, so from that we see that the force does no work, it does not change the motion of the object, no change means no work.
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Forces perpendicular to displacement, contribute no work.
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Force has to be at least partially in the direction, the amount that is perpendicular will contribute no work.
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With all this thinking, we got things down pretty well.
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It is the length of the displacement times the amount of force parallel to the displacement, the amount that is perpendicular has no effect.
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Now we can finally create a formula.
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If we have got some object, it does not matter what the mass is, remember the mass has an effect on the outcome of what happens in the world, but the work that is put in, is going to be the same whether it is a tiny mass or a really large mass.
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The work = the size of the force × the size of the displacement × cos θ , θ is the angle between the two.
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Why is that?
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Remember, basic trick, since this is the hypotenuse, and this is the side adjacent, that side = force × cos θ .
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So, the amount that is parallel, is going to be F × cos θ , so the amount that is the sin θ , the side opposite has no effect, so we can completely get rid of it.
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So the only one we have to care about is the force × cos θ and that appears here, and here, and then we take the amount f the displacement in here, so the work = force × the distance × the cosine of the angle between the two, and that tells us what the work is.
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That is the formal definition of work, it allows us to look at all the ideas that we have talked about so far and make sense of them.
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One alternate formula you can use, in addition to force × distance × cos θ , we can also formulate it as a dot product from math.
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The work = (force).(distance), (as vectors).
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So the dot product is, if you take, **a**.**b** = (the x components multiplied with one another) + (the second two components multiplied with one another)
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It might seem surprising at first, but it turns out actually having the exact same effect.
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If you look at a formulation where, if one of them is lying on the x axis, then you can actually quickly see why if this is here, then the amount that this is out is the x axis amount here, x-y, well, it is going to wind up being F, if this is F again, then Fcos θ = its x component, because that is how much it is, because we can see that the way the vector breaks down.
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We can break the vector into its constituent perpendicular and parallel pieces.
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It will be a little bit more complicated to prove this in a different angle, but you can trust me on this, work = (force vector).(distance vector)
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We can also use just force × distance × cos θ if we know the magnitudes and the angle.
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Sometimes one is going to be more useful than the other, it depends on the specific conditions, and what you need to do.
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As always you got to pay attention to what you are trying to solve fro in Physics, and figure out what is the best thing for you to use right there.
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Finally, what units does work use?
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From our formula, work = force × distance × cos θ .
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cos θ , θ comes in angles, angles do not really have a unit, they are radians, but they are unitless, cos θ is just a scalar.
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The only things that come with units are our force and our distance.
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Force's unit is newtons, and distance's units is metres, if we are working in the S.I. system.
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That means, work = N m , for ease we call this a joule, which we shorten as J.
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J on its own, that is what we do for work, and later we will find out it is also what we do for energy when we talk about how those two are connected.
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We call it a joule in honor of James Joule, who did pioneering work in heat and energy, in the 1800's.
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We are ready for our examples.
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Real simple, real easy one to start off with.
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We got a bus, and we push it a distance of 10 m with a parallel force of 20 N.
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If it is parallel, what is the angle?, the angle = 0, so cos θ = 1.
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So, work = force × distance × cos θ = 20 N × 10 m × 1 = 200 J, and that is our answer.
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If we wanted to, we could have also done that in dot product form, because we know that they are parallel, so the force would just have a vector of (20 N,0 N) , and the displacement vector would be (10 m,0 m).
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So, we wind up getting 200, because 20 × 10 = 200, and 0 × 0 = 0, so we get 200, the exact same answer.
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Second example: Ball of mass 0.25 kg is dropped at a height of 12 m.
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When it hits the ground, how much work has the force of gravity done on the ball, what if the mass was 'm,, and the height was 'h'.
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First off, we have got some ball, and we drop it 12 m, so the ball is up here, and mass = 0.25 kg, what is the force pulling on that ball?
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Force pulling on that ball is the force of gravity, we assume no air resistance for ease, actually in this problem we can have air resistance, but we are paying attention to just the work done by the force of gravity, if we have to look at the energy later on, we would have to take the air resistance into account, but the force of gravity is going to do the same work no matter what, as far as it does move those 12 m.
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In any case, force of gravity = mg, so in this case, if it travels 12m, and what is cos θ?, θ = 0 (since parallel).
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So, work = force × distance × cos θ = mg × 12 × 1 = 0.25 × 9.8 × 12 × 1 = 29.4 J.
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What if we wanted to solve this in the general case, what if we want to talk about, what if we were dealing with an arbitrary mass 'm', what if the height was just an arbitrary 'h'?
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If that is the case, the work once again, the fall is parallel to the force of gravity (same direction), so work = F × d = mg × h = mgh, is the work done by the force of gravity for an object dropping.
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Third example: A box moves a distance 5 m to the right, a force of 10 N pushes on it in the opposite direction.
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How much work does the force do on the box?
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The first thing to think about is, what is the angle.
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Here is the 5 m, what is the angle between those two vectors?
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5 m is a vector, 10 N is a vector, so what is the angle between them.
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They are parallel, but they are pointing in opposite directions, we got to pay attention to the fact that they are going opposite.
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So, 180 degrees.
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So, if θ = 180 degrees, what is cos(180)?
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cos(180), remember in your unit circle, it is pointing in the opposite direction, so it is going to be -1.
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We drop this all in our formula for work, we got, work = force × distance × cos θ = 10 N × 5 m × (-1) = -50 J.
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This is a totally new idea we have not encountered before.
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We got the idea that we can actually take work out of a system.
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If we had it going with it, that means we would be making it go faster, you put in work into it, because you would be making it going with it.
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Bu this time, we are actually resisting the motion that it has.
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It is going to move forward 5 m, but this time we are pushing in the opposite direction.
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If we push in the opposite direction, this means that we are actually resisting it.
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We are using our work to take the total work in the system out, the total energy in the system out.
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We will talk more about the connection between energy and work, but right now, before, work was contributing to the distance it was moving, it was contributing to motion.
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In this case, our force was going against the motion, so it is actually taking away from the motion, so it is a negative work.
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If we want to do this with the dot product, work = F.d, if we make this the positive direction, then what is our distance vector?
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It is going to be equal to (+5m,0m), what is the forces?
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It is going to be going in the opposite direction, so (-10 m,0 m).
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We put these two together with our dot product, and we got, 5× (-10) + 0 × 0 = -50 J.
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Two different ways of doing it, both equally valid, gives you the same answer, the idea is the fact that, one you work against the motion of it, you have negative work, you are taking work out.
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Fourth example: We have got a box traveling a distance of 50 m, 30 degrees South of East.
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So it is moving South of East by 30 degrees, and it travels 50 m.
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The box is acted upon during that motion by a force of 50 N, in a direction of 30 degrees North of East.
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Even though there is a force moving on it, it does not change the displacement.
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We know the displacement vector beforehand, it is given to us, we can be sure of it.
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For some reason, there is something keeping it on that track, we are just worried about what the work that force does is.
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We do not have to worry about the displacement changing, displacement is given to us in the beginning.
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The work is going at an above direction by another 30 degrees, so it is 30 degrees North of this.
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How much work does the force do?
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We are looking at this from above, so it is flat, so we do not have to worry about the gravity, that is what the North and East and South all tell us.
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In this case, what is θ?
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The angle between the two is not just 30, it is the total between the two, so it is 60 degrees.
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So, θ = 60 degrees.
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Use our formula for work, work = force × distance × cos θ = 60 × 50 × cos(60) = 1500 J, is the answer.
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Now, in this case, we could also do this in vector mode.
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For this one, we were given the angles and magnitudes, so it is less useful, but I want to show you how to use the dot product, because sometimes, you are going to get things in vectors, and it is way more useful not have to convert into angles and magnitudes and see if you can do the problem, it is useful to go, "We have got vectors, let us use the dot product!"
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We will convert this first into vectors.
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For the top one, if it is 60 N on the hypotenuse, and 30 degrees angle here, then over here, it is going to be 30 N vertical, and what is its horizontal going to be, it is going to be 51.96, which is what we get when we take cos(30) × 60.
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What about for the triangle representing the displacement?
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It has got 50 m on its hypotenuse, so what is its vertical, its vertical = 25 (remember, we got 30 degrees here, and in 30-60 triangle, the side opposite to 30 is 1/2), and up here, cos(30) = sqrt(3)/2, is going to be 43.30.
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In this case, this means that we know our force vector, what is the horizontal component?, it is 51.96, and let us say this (right) is positive, and up is positive, and vertical is +30 N .
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We look at the displacement vector, and that is going to be equal to, (43.30 m, -25m).
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This one is pointing down.
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We take the dot product of these two, force dotted with distance, **f**.**d**, that is going to wind up equaling (51.96 × 43.30) + (30 × -25) = (2249.9) - (750) = 1499.9 J .
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The only reason this wound up being any different from this answer which they are approximately equal, is because of rounding errors.
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Rounding errors when we wound up figuring out what these two horizontal components were, we got slight answers off, because when we are using our calculator we wound up having to round it, because we did not use the entire thing, which we should, because remember we want to take some care about how significant digits work when you use an entire 10 digit long expansion.
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Because that is the kind of extreme amount of accuracy to have, so we wound up having a slight rounding error, but when we consider the fact that this is 5 digits long, and we are only off by the very last digit, that is really close to 1500 J, that is as precise as we will be able to measure anything, in any lab we do.
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And probably any lab you would wind up doing very long time, unless you are working in seriously experimental Physics.
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Example 5: This is our last example.
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Similar to example 2, remember in example 2, we talked about dropping a ball from a height 'h' or 12 m, we have got the same mass and the same height as we did in example 2.
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We drop a ball 12 m, it is going to be the mass of the ball times gravity because that is the force of gravity times the height that it falls and cos θ just winds up going away turning into 1, because cos(0), because they are parallel is just 1, so we can just pay attention to the force × distance, so mgh.
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What about this case, a ball of mass 0.25 kg is tossed out of a window at a height of 12 m.
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Now, it travels up, and then travels down.
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So, we do not know how high it gets, and I did not tell you precisely what it was.
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It could be that it winds up getting really tall or it could be practically flat and then falling immediately.
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It could be either one of these, I did not tell you how it is going to look.
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So how could we figure out what it is going to be.
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Remember, we know about how negative work works.
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Notice that for the amount up here, for this portion right here above the height of the window, no matter what happens, the ball is going to wind up going positive.
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Let us call it 'H', so there is the h that it falls to the ground, it should be 'h', that is the amount that is guaranteed, and then there is H which is the variable amount that it winds up going depending on how we throw it.
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It has to go up by H, but then to be able to make it down, it has to also go down by H.
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We have got a positive H and a negative H.
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What happens if we look at the work done over the positive H and the negative H section?
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It is traveling to the side, gravity is only going to be caring about the component, about up and down because everything else is going to be perpendicular, so we can just toss it out.
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We only have to care about the up and down components.
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Positive H is the amount that it travels up, that is the parallel amount, the perpendicular amount, the motion sideways we can just get rid off because it does not matter, that is the amount that is perpendicular and we can throw it away because we only care about the parallel amount.
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In this case, we only have to care about the up by H and down by H, and then h.
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We already know that the amount of work done in the h is going to be mgh.
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What about the amount of work done by H?
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This one, g is negative number.
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So, +H, if we go this way, they are actually going to have an angle, θ = 180 degrees.
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This is going to give us, -mgH.
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What about the direction where it goes down?
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That is going to be θ = 0, because now they are going in the same direction.
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So this is going to be, +mgH.
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We have got the idea that you go up by some amount of height, but you are fighting against gravity, so gravity is taking work out of the system, that means a negative work.
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-mgH, the amount that you travel up, but then we wind up having travel the exact same amount down if we are going to make it to the ground, so that means, that amount that we just lost in work is going to be regained in work.
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The H, mgH is going to wind up cancelling the -mgH, the two are just going to hit each other, and they are going to disappear, and we are going to wind up getting, these two things, just cancel each other out, and in the end, the only thing that we are left having to care about, is the h that we got right here.
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That is the important part, mgh.
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No matter how crazy a throw we have, if we go way up, or it is really flat, it does not really matter because the amount of work done by the extra arc from where it, the amount that is not just where it starts falling to 12 m we are guaranteed, is going to be canceled out.
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The two works go opposite to one another, so it just gets canceled out.
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In the end, it is mgh, the end of our answer is going to be the exact same answer we got for example 2.
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The work = mgh, which is going to be, 0.25 × 9.8 × 12 = 29.4 J, is what we got.
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The reason why is because the amount of arc that goes above where we started gets canceled out, because it winds up having to travel up, but then it travels that same amount back down, before it can do the real fall of h.
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So H is, wind up cancelling one another out, because they do the same thing, they are doing with the same gravity , but they go in opposite directions, so they just cancel one another.
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All we have to worry about is where we started, if we want to figure out the amount of work that gravity is going to put into it when it hits the ground.
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Hope you enjoyed this, hope you got a good understanding of work.
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This is going to be really useful in the next section when we talk about energy.