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Lecture Comments (5)

1 answer

Last reply by: Professor Selhorst-Jones
Wed Mar 19, 2014 9:11 AM

Post by Nathan Lipinski on March 18, 2014

How come for example five we don't have to put the little h a negative? The big H is a positive?
Thanks

0 answers

Post by javier chichil on October 8, 2013

Good explanation. Thanks.

1 answer

Last reply by: Professor Selhorst-Jones
Sun Jul 28, 2013 9:18 PM

Post by KyungYeop Kim on July 27, 2013

Why is work FxDxCos(x) as opposed to just F times D? it seems cosnine is redundant.. cosnine= Adjacent/Hypotaneous.= it ends up being (adjacent)^2 which is also (distance)^2 ??

Work

  • The idea of work is deeply connected to the idea of energy, as we will see in coming lessons.
  • Qualitatively, you've done more work on an object if you push with more force than less force. Similarly, you do more work if you push for a longer distance than less distance.
  • We want to define work as how much change you put into the world. Even if an object moves a distance, your force has to have some effect on the object's motion. No effect, no work.
  • This means that if your force is perpendicular to the movement, it contributes nothing-no work.
  • Work is the distance traveled multiplied by the force parallel to the motion. If the angle between them is θ, we can use trigonometry to get
    W = |

    F
     
    | ·|

    d
     
    | ·cosθ.
  • The unit for work is the newton·meter, which we call a joule (J).

Work

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Equivocation 0:05
    • Equivocation
  • Introduction to Work 0:32
    • Scenarios: 10kg Block on a Frictionless Table
    • Scenario: 2 Block of Different Masses
  • Work 4:12
    • Work and Force
    • Paralleled vs. Perpendicular
    • Work: A Formal Definition
  • An Alternate Formula 9:00
    • An Alternate Formula
  • Units 10:40
    • Unit for Work: Joule (J)
  • Example 1: Calculating Work of Force 11:32
  • Example 2: Work and the Force of Gravity 12:48
  • Example 3: A Moving Box & Force Pushing in the Opposite Direction 15:11
  • Example 4: Work and Forces with Directions 18:06
  • Example 5: Work and the Force of Gravity 23:16

Transcription: Work

Hi, welcome back to educator.com, today we are going to be talking about work.0000

This is going to be our first introduction to energy.0004

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!0008

Do not worry, both these things are actually really deeply connected.0015

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.0018

But, first we can tackle the idea of work on its own, it is going to really help us understand energy.0022

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.0026

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.0034

Let us say we have got some 10 kg block sitting on a frictionless table.0045

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.0051

In our first one, we have got 'f' and 'd'.0062

We push it from here, to here, and we do that with that much force.0068

So, that is one thing, but what if we did it for this much distance?0073

Same force, but a much larger distance.0077

If that is the case, by the end of it, because it is frictionless, it is going to pick up speed.0081

Remember, the longer the force is going, the longer the acceleration, so it is going to have more speed in it.0085

So, it makes more sense to think that, the longer the distance, the more energy, the more work that we put into it.0090

You push for a longer period of time, it makes more sense that you are putting more work.0095

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.0100

What if instead, if we had a really big force and a small distance?0107

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.0110

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.0131

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.0139

What if we put both of them together, if we had a really big force, and a really big distance!0150

Clearly, that is going to be the one that has the most work.0155

We push really hard, for a really long time, that is going to be the most work.0158

So, it is about force and distance together.0162

More force means more work, more distance of pushing that force means more work.0164

Put both of them together, and that is even more work.0169

What if we consider two different blocks with masses that are different?0173

Same force in both cases.0177

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.0179

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.0192

The other one, we got a high mass object traveling really slowly at the end of its distance.0202

But, we had to push the same force for the same distance.0207

So from the point of the view of the pusher, it is the same effort that we had to put into it.0212

They changed the system of mass, having to do with the speed that it is going at.0216

The two things are connected.0221

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.0222

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.0231

So, we are going to think of work as just force and distance connected together.0242

The mass is not going to have direct effect.0246

These are different results, but it is the same force that is put into the system.0249

With this idea in mind, we have got a notion that work is force involved multiplied by the distance.0253

Way more force and way more distance stacks up hugely.0257

More force means more work, more distance means more work, these things that make good intuitive sense.0261

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.0266

But there is one last thing that we have to consider before we really define what work means.0282

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?0287

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.0311

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.0325

We created acceleration, we created change in velocity, by putting that work in.0330

In this case, you put a lot of effort in, you tried really hard, you push and strain, but nothing happens.0335

So we want to define work as 'change in the world'.0342

You have expended a lot of effort, but you did not change anything.0346

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.0349

It is force × distance, zero distance means no work, even if it is a giant force.0357

With this idea in mind, there is one thing to consider.0363

What about this scenario!0367

We have got a block, and this lock moves in this direction.0368

But the entire time, we have got a force moving this way, perfectly perpendicular to the motion of the block.0373

Does the force do any work on the block?0382

We talked about the fact that, if you did not change anything, then you did not put in any work.0385

If there is no change from the force, then the force does no work.0393

Force has to be connected to the distance.0397

So in this case, no work is done by the force, because it is perpendicular.0400

That motion to the side is going to happen whether or not the perpendicular force does anything or not.0404

It is not able to change its distance, because it is perpendicular.0410

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.0414

So, this one does not happen.0427

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.0433

Forces perpendicular to displacement, contribute no work.0443

Force has to be at least partially in the direction, the amount that is perpendicular will contribute no work.0447

With all this thinking, we got things down pretty well.0454

It is the length of the displacement times the amount of force parallel to the displacement, the amount that is perpendicular has no effect.0457

Now we can finally create a formula.0465

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.0468

The work = the size of the force × the size of the displacement × cos θ , θ is the angle between the two.0480

Why is that?0490

Remember, basic trick, since this is the hypotenuse, and this is the side adjacent, that side = force × cos θ .0491

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.0505

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.0515

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.0534

One alternate formula you can use, in addition to force × distance × cos θ , we can also formulate it as a dot product from math.0541

The work = (force).(distance), (as vectors).0549

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)0554

It might seem surprising at first, but it turns out actually having the exact same effect.0579

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.0582

We can break the vector into its constituent perpendicular and parallel pieces.0608

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)0614

We can also use just force × distance × cos θ if we know the magnitudes and the angle.0623

Sometimes one is going to be more useful than the other, it depends on the specific conditions, and what you need to do.0628

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.0634

Finally, what units does work use?0642

From our formula, work = force × distance × cos θ .0644

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.0648

The only things that come with units are our force and our distance.0660

Force's unit is newtons, and distance's units is metres, if we are working in the S.I. system.0664

That means, work = N m , for ease we call this a joule, which we shorten as J.0672

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.0679

We call it a joule in honor of James Joule, who did pioneering work in heat and energy, in the 1800's.0685

We are ready for our examples.0694

Real simple, real easy one to start off with.0695

We got a bus, and we push it a distance of 10 m with a parallel force of 20 N.0697

If it is parallel, what is the angle?, the angle = 0, so cos θ = 1.0707

So, work = force × distance × cos θ = 20 N × 10 m × 1 = 200 J, and that is our answer.0713

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).0738

So, we wind up getting 200, because 20 × 10 = 200, and 0 × 0 = 0, so we get 200, the exact same answer.0757

Second example: Ball of mass 0.25 kg is dropped at a height of 12 m.0769

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'.0775

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?0781

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.0796

In any case, force of gravity = mg, so in this case, if it travels 12m, and what is cos θ?, θ = 0 (since parallel).0822

So, work = force × distance × cos θ = mg × 12 × 1 = 0.25 × 9.8 × 12 × 1 = 29.4 J.0841

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'?0872

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.0883

Third example: A box moves a distance 5 m to the right, a force of 10 N pushes on it in the opposite direction.0912

How much work does the force do on the box?0922

The first thing to think about is, what is the angle.0924

Here is the 5 m, what is the angle between those two vectors?0929

5 m is a vector, 10 N is a vector, so what is the angle between them.0935

They are parallel, but they are pointing in opposite directions, we got to pay attention to the fact that they are going opposite.0939

So, 180 degrees.0947

So, if θ = 180 degrees, what is cos(180)?0951

cos(180), remember in your unit circle, it is pointing in the opposite direction, so it is going to be -1.0955

We drop this all in our formula for work, we got, work = force × distance × cos θ = 10 N × 5 m × (-1) = -50 J.0961

This is a totally new idea we have not encountered before.0994

We got the idea that we can actually take work out of a system.0997

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.1002

Bu this time, we are actually resisting the motion that it has.1009

It is going to move forward 5 m, but this time we are pushing in the opposite direction.1012

If we push in the opposite direction, this means that we are actually resisting it.1016

We are using our work to take the total work in the system out, the total energy in the system out.1022

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.1027

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.1035

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?1042

It is going to be equal to (+5m,0m), what is the forces?1052

It is going to be going in the opposite direction, so (-10 m,0 m).1059

We put these two together with our dot product, and we got, 5× (-10) + 0 × 0 = -50 J.1064

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.1076

Fourth example: We have got a box traveling a distance of 50 m, 30 degrees South of East.1088

So it is moving South of East by 30 degrees, and it travels 50 m.1094

The box is acted upon during that motion by a force of 50 N, in a direction of 30 degrees North of East.1101

Even though there is a force moving on it, it does not change the displacement.1107

We know the displacement vector beforehand, it is given to us, we can be sure of it.1113

For some reason, there is something keeping it on that track, we are just worried about what the work that force does is.1116

We do not have to worry about the displacement changing, displacement is given to us in the beginning.1123

The work is going at an above direction by another 30 degrees, so it is 30 degrees North of this.1129

How much work does the force do?1134

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.1136

In this case, what is θ?1142

The angle between the two is not just 30, it is the total between the two, so it is 60 degrees.1144

So, θ = 60 degrees.1151

Use our formula for work, work = force × distance × cos θ = 60 × 50 × cos(60) = 1500 J, is the answer.1154

Now, in this case, we could also do this in vector mode.1183

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!"1186

We will convert this first into vectors.1205

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) × 60.1209

What about for the triangle representing the displacement?1232

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 43.30.1236

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 .1265

We look at the displacement vector, and that is going to be equal to, (43.30 m, -25m).1293

This one is pointing down.1306

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 .1309

The only reason this wound up being any different from this answer which they are approximately equal, is because of rounding errors.1347

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.1356

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.1374

And probably any lab you would wind up doing very long time, unless you are working in seriously experimental Physics.1390

Example 5: This is our last example.1397

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 2.1400

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.1412

What about this case, a ball of mass 0.25 kg is tossed out of a window at a height of 12 m.1432

Now, it travels up, and then travels down.1439

So, we do not know how high it gets, and I did not tell you precisely what it was.1444

It could be that it winds up getting really tall or it could be practically flat and then falling immediately.1449

It could be either one of these, I did not tell you how it is going to look.1455

So how could we figure out what it is going to be.1459

Remember, we know about how negative work works.1461

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.1466

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.1480

It has to go up by H, but then to be able to make it down, it has to also go down by H.1491

We have got a positive H and a negative H.1498

What happens if we look at the work done over the positive H and the negative H section?1500

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.1505

We only have to care about the up and down components.1515

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.1518

In this case, we only have to care about the up by H and down by H, and then h.1534

We already know that the amount of work done in the h is going to be mgh.1542

What about the amount of work done by H?1547

This one, g is negative number.1550

So, +H, if we go this way, they are actually going to have an angle, θ = 180 degrees.1558

This is going to give us, -mgH.1566

What about the direction where it goes down?1572

That is going to be θ = 0, because now they are going in the same direction.1574

So this is going to be, +mgH.1578

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.1583

-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.1592

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.1604

That is the important part, mgh.1626

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.1630

The two works go opposite to one another, so it just gets canceled out.1647

In the end, it is mgh, the end of our answer is going to be the exact same answer we got for example 2.1651

The work = mgh, which is going to be, 0.25 × 9.8 × 12 = 29.4 J, is what we got.1656

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.1671

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.1686

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.1701

Hope you enjoyed this, hope you got a good understanding of work.1708

This is going to be really useful in the next section when we talk about energy.1714