1

concept

## Work Done by a Constant Force

6m

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Hey guys. So now that we understand kinetic energy in this video, I want to start talking about work. So we're gonna see the relationship between work and energy. And we're also going to see the relationship between work and forces. So I'm gonna show you what work is. And then we'll see how to calculate the work done by a constant force. Let's check this out. First of all, what is work? Well, imagine that we had a box at rest on a frictionless surface, right? So velocity is equal to zero. So we wanted kinetic energy. The kinetic energy is also equal to zero. And then you start pushing this box. So if you push it and it's frictionless, then the box is gonna start to move. It's gonna start to move to the right. Which means that now it has some velocity that's not equal to zero. And so therefore it has some kinetic energy that's not equal to zero as we well. So if the box gains some speed by you pushing it some velocity, it also gains some kinetic energy. Right? That kinetic energy is always associated with your speed or your motion. So where did that energy actually come from? And the whole idea is that the energy actually comes from you. It comes from the force that you were pushing on this box with. So remember that energies in physics are always transferring between objects. The sun gives energy to the plants and then a cow eats the plant that's transfer of energy, then you eat the cow. That's a transfer of energy. And then you push on the box. So the whole idea here is that work which is written by the symbol W. Is actually the quantity it is the amount of energy that you transfer between objects or that is transferred between objects. So you push the box and let's say it gains 10 jewels of kinetic energy. You do work on the box, that's what we say in physics, you do work on the box or objects do work on others. So because work is really just energy that's transferred then the unit that we'll use for this is gonna be the jewel. Just like you know just like energy. Alright. So let me show you the examples for work. How do we calculate or? So the equations for work. How do we calculate it? And I'm gonna show you the fancy version of this that you'll see in your textbooks, it's the magnitude of the force times the magnitude of the displacement times. Cosine theta. So a lot of textbooks will rewrite. This is a shortcut way and this is how we're gonna write it from now on FD. Cosine Theta. So really we have to know about these equations about this equation in particular. Is that this theta term. Right? This angle is the angle that is between the force and the displacement. Alright, so data is always the angle between F. And delta X. Or F. And D. So basically in these problems where you calculate works, you're gonna have to know three things, you're gonna have to know the magnitudes of both of the force and the displacement. And remember the magnitudes are always gonna be plugged in as positive numbers. And then you're gonna figure out the angle that is between those vectors. And that's what you plug into your cosine theta term. Let's check out some examples. So here we're gonna pull a two kg box that's at rest on a frictionless surface. So the idea is this is two. And then I'm gonna pull this box with a force of three newtons. And I want to calculate the work if I pull it through a distance of five m. So basically what happens is I want to figure out the work that's done by my force. So remember that the work is gonna be F. D. Cosine of theta. And I'm gonna plug in everything as positive. So I have the force which is three newtons. And if I'm pulling it to the right then therefore it goes through a displacement a delta X. Of five. So really what I what happens is I want to plug this in And I'm gonna plug in three times five. And now what's the cosine of the angle between those two vectors? Well, if these two vectors are parallel to each other, they point in the same direction, then therefore the cosine of this angle here, this angle is equal to zero. And so with the co sign Of zero is it's always equal to just one. So really all we do is we actually just multiply three times five and we get 15 jewels. All right. So that's the work that's done. Let's move on to the second example. Now we have a car to five kg cart here that's already moving to the right with some velocity. So the idea here is now I have this cart like this. Now this is five. The velocity of this card. The initial is already equal to 10 m per second. But there's gonna be some stopping force. I'm gonna call this F. S. And it's 100 newtons. Now I want to figure out the work that's done by the stopping force. I'm just gonna use the same equation. FD. Cosine Theta. So this is my work done is gonna be F. D. And this is gonna be the stopping force times D. Times the cosine of Theta. Alright, so I've got the stopping force. This is 100. Now, what about the distance or the displacement? Well if the card is already moving to the right but it's being pulled to the left, it's still undergoes a displacement. That's to the left here. This delta X. Is 2.5. So this is 100 times 2.5. And now what's the angle that is between these two vectors. Well we said that zero degrees was when they're parallel to each other but when they're anti parallel, when you have a force like this and a displacement like this, then the angle is actually equal to 180 degrees here. It was equal to zero. So what happens here is that when you plug in cosigner Theta equals 180. Your co sign term will always turn to negative one for anti parallel forces. Right? So really this just turns into 100 times 2.5 times negative one. And this is just 250 jewels but it is negative. All right. So, we can see here is that works can actually be positive or they can be negative. Remember that work is really just the amount of energy transfer work is the quantity of energy that any force can either give to or take away from an object. So work can either be plus or minus. And the simple rule to figure this out is to look at the force and figure out whether it helps or hurts the motion. If the force helps, meaning it goes along with the force or the the objects motion. Then the work that's done is going to be positive like we had over here, the force causes the car to move the box to move to the right force goes along with the motion. The work is positive. Now the first hurts or goes against the objects motion. The work is going to be negative and that's basically what we had in this situation here. The cart was already moving to the right. This is the motion, but the force points this way it hurts or takes away from that kinetic energy. And that's why we got a negative number. Alright, so that's it for this one, guys, let's move on.

2

Problem

You pull a 5kg box vertically up with a constant 100N force for 2m. How much work do you do?

A

51 J

B

100 J

C

102 J

D

200 J

E

500 J

F

1000 J

3

example

## Work by a 2D Push/Pull

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Alright guys, so let's check this one out here. We've got this 10 kg suitcase, that's the mass of the box. And we're also pulling with some applied force of 50 newtons and we're pulling it across the floor. And it's gonna be pulled some distance of 20 m. But which way it's gonna be pulled even though we're pulling it upwards like this, what happens is it's just gonna move along the floor, right? It actually wouldn't fly up like this. So what's gonna happen is it's gonna travel some displacement here. Let me draw this a little bit bigger. And we know that this delta X. Or D. Doesn't matter which letter we use um is gonna be 20 m. So we want to do this first part here is we want to figure out the work that's done by the pool. Right? So your pole here. So the work that's done by F. Well remember the work that's done by any force is always gonna be F. Always that force times deke data. So the work of my F. Is gonna be F. D. Co sign of data. Remember that? This angle is the angle between F. And D. So let's check this out. I have the magnitude of the force. That's 50. I have my ex. Or my D. Right? It's just equal to 20 m. So I can just plug this into my work equation. This is gonna be 50 times one. Sorry, 20 times the cosine of the angle that's between these two vectors. We can take a look here, they're actually not parallel. Right? The force actually points this way but your displacement points along the floor like this. So this is the angle that we use between these vectors. And we know that this angle here from the problem is equal to the 37°. So we're just gonna plug this in 50 times 20 times the cosine of 37. You go ahead and work this out. What you're gonna get is you're gonna get 800 jewels. That's the answer to party. Alright so now in part B what we have to do is we're gonna have to decompose are forced into its components FX and Fy and then calculate the work done by each one of those components. So for part B we have the work is done by F. X. Remember what we said? The work done by any forces that force. So F. X. Times D. Cosine of theta. Now we actually don't have F. X. In our diagram just yet. So we're gonna do is we're gonna take this force and we're gonna break it down into its components like this. So this is gonna be my F. X. And I know to figure that out. I'm just gonna use F Times cosign of data here. So this is just gonna be 50 times the cosine of actually let me go ahead and write it somewhere else. So I've got that F. X. Is just equal F. Cosine Theta. So it's just equal to 50 times the cosine of 37. So if you go ahead and plug this in, you're gonna get 40. And the same thing is gonna happen with F. Y. Except now we're just gonna use sign, right? So F. X. F. Y. Is F. F. Sine theta. That's basically this vector like this, this has been my F. Y. And this is gonna be 50 times the sine of 37. So you're gonna get 30 here. So those are the components, it's just basically a 30 40 50 triangle. Which makes sense. So now we have to do is we have to calculate our work. So we're gonna take the magnitude of fx. Which we know this fx is really just equal to 40. Remember this is gonna be my 40. Now the distance that it travels is still going to be the 20. That doesn't change. But now, what's the cosine between the angle between? What's the angle between these two vectors? Well, now what happens is your FX. Remember actually points along the horizontal in the same direction as your displacement. So what's gonna happen here is you're gonna have 40 times 20. And the co signed between these two angles is actually gonna be between these two vectors is actually zero. Right? Because basically your fx points to the right and your distance also points to the right here. So what happens is you get a cosine of zero that we know that equals one if you plug this in, you're gonna get 800 jewels. So you get the same exact number. But and that's basically because the only four that's actually doing work on this object is this Fx here. That's what gives you the 800 jewels. One way you can also think about this is that we use F. D. Cosine theta When we calculated part A. And part B. What happens is you already kind of have a cosigner of data that is absorbed in this term here. Right? It's kind of wrapped up inside this Fx because really this is already F. Times the cosine of theta. So you're doing FD cosine theta here. But you're doing F. Times Cosine Theta times D. When you're doing it this way. So you're just gonna get the same exact number here. Alright, so to finish off the problem, we're gonna take a look at part C. This is gonna be the work that's done by F. Y. So this is gonna be my F. Y. Times D. Cosine of Theta. So we take a look we got F. Y. Remember this, that's just the 30 that we calculated over here times the displacement. Which is 20. But now what's the cosigner, what's the angle between these two vectors? Well basically with this F. Y. Is vertical. But we know our displacement actually points to the right like this. So the angle between these two is actually 90. And so therefore the cosine of 90 is gonna be zero and the work gets done is just basically gonna cancel out to zero jewels. That makes sense because there's a component of your force that's pulling vertically, but if you're traveling horizontally then there's no work that that force does. Alright, so that's it for this one. Guys, let me know if you have any questions.

4

example

## Work Done By Friction

4m

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Hey guys, so that's probably gonna be calculating the works done by a bunch of these forces, kinetic friction, weight and normal. So, let's go ahead and check out this problem here, we have the box that weighs 50 newtons, which means that your MG, which goes down, is already equal to 50 newtons. Don't assume that to mean 50 kg and then multiplied by nine point, exhibit at the wrong answer. The MG is already 50 which means that the normal force, right? If these are the only two forces that are acting in the vertical direction, have to cancel, that's our normal force. We actually have one last force to consider which is the force of friction. So we're pulling this block to the rights and friction is going to oppose that by acting to the left here. So this is going to be RFK, that's the friction force. Now, what we want to do this first part here is we want to calculate the work that is done by this kinetic friction. So, that's the work that's going to be done by F K. Well, remember if we're trying to figure the work done by F. K. That's going to be F. K times D. Co signed photo. So that's gonna be the equation for that. We have this data here. That's the angle between F. K. And D. All right. So, how do we figure out this force of friction will remember the force of friction is always going to be um U. K. Times the normal force. So, in UK times normal times D. Co sine of theta. All right. So, we have this coefficient of friction here. This mu K is 0.7. Now, we just have to figure out the normal force. Well, actually, we already know that it's equal to 50. So already start plugging in the work that's done by the friction force is going to be, this is going to be 0.7 times 50. That's the normal force. Yeah 50 or here. Now we got the distance right? So the box actually travels a horizontal distance of eight m. So basically we've got this distance over here this delta X. Or D. Is going to be eight. So that means that we're just gonna plug in eight. And to here and now we're just gonna look at the co sine of the angle. Remember the co sine of the angle of the angles between the force of friction and your displacement. Your friction force acts to the left but your display your displacement acts to the right. So here what happens is your force of friction of your F. K. Is this way. But your displacement is this way this is your delta X. Or your D. And so what that means here is that the angle is equal to degrees. So your co sign is going to be the coastline. 180. Remember what happens is that when you plug in the coastline of 180 you're gonna get a negative one always. So what this means is that you're gonna get, the friction force is equal to negative 280 jewels. Now what I want to point out here is that the friction force is always going to be negative. Your friction force is always going to oppose your direction of motion. We were going to the right in this example in a friction force are friction force pointing to the left. If it was reversed, if we were going to the left are friction force would actually point to the rights. So because friction always opposes, your motion is always in the opposite direction. Is V. It always does negative work. So, what you're always going to see when you calculate the work done by friction because you're gonna see F. D. Co sign and the number of him here is always going to be 100 and 80 degrees. They're always gonna be opposite of each other. So one way we can simplify this from now on is all right. That the work done by friction is just negative F. K. Times D. All right. That's a pretty simple way to figure that out. So, you're plucking those numbers and you'll get negative 280 jewels. All right, So, let's move on to part B. Now. Now we're going to calculate the work done by weight force. So, the work done by weight is really going to be the w work uh Sorry, the work done by gravity work done by MG. So here we're just gonna do MG. That's the weight force times D. Co sign of data. But we can quickly tell here that your MG is gonna point downwards And your displacement is going to point to the right. So this angle here is 90°.. So what happens here is you're gonna have MG. Which is 50 Times the displacement which is eight. But it doesn't matter because the coastline 90 is gonna wipe everything out and the whole entire thing is gonna become zero. So what happens is the work that's done by MG is going to be zero jewels. That makes sense. Your weight force always points down if you're going to the right that way force doesn't actually do any work on you. All right now I've got one last one last part here which is the way to work done by the normal force. So this is the work done by this end. So this is going to be our Normal force times D. Co sign of theta. And we're going to see is that this basically is going to be exactly the same as part B. Your normal force points up. Your displacement is to the right. So this angle here. So this angle here is 90 degrees. So you have your weight or start your normal 50. Your displacement is eight. But it doesn't matter because you're gonna have the coastline of 90 again. And so this whole thing is gonna go away. You're gonna have to work the sun is zero jewels. All right? So that's it for this one. Guys. Let me know if you have any questions.

5

concept

## Work Done By Gravity

6m

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Hey guys, so now that we understand how to calculate works done by constant forces in this video, we're going to take a look at how a specific force which is the force of gravity. So I'm gonna show you how to calculate the work that is done by gravity. So the whole idea here is that because gravity is a force, that's the MG that we always plug into our free body diagrams, then that means that gravity can do work. So, just remember a quick definition of work is that it's the transfer of energy. And the way we figure out whether it's positive or negative is it's positive when the force goes along with the motion, whether it helps your motion and it's negative work whenever the force points against your motion. Right? So, we're really just going to take a look at two different scenarios here. Gravity acts vertically. Right? You're Ngos points down. So we're gonna look at two different situations when objects are falling, going down or when they're rising and going up. So let's just get to the example here. Right? So in this first example we have a 5.1 kg book that's falling from a two m bookshelf. We've got this diagram here to help us out with this and we've got some information about the speed right before it hits the ground. And in this first part here, we want to calculate the work that is done on the book, bye gravity. So we want to figure out W. G. That's how we're really that's how we're going to write the work done by gravity. So how do we do this? Remember when we calculate works? You're always just gonna start from F. D. Co signed data. You need to know the force, the distance and the angle between those vectors. So let's check this out when you have the book that's falling downwards. The only force that's acting on it is your MG. Is that that force of gravity that pulls things downwards. So your FD cosign data really just becomes MG. So your distance. Now if your book is falling downwards then it's just going to undergo displacement of delta Y. And it's gonna go downwards like this right now. Some books are also going to use delta H just in case you see that letter, they really just mean the same exact thing. So your distance in F. D. Cosign Theta really just becomes delta Y. Now what about the angle between those two vectors? Well they both point downwards, they're basically parallel which means that the angle between them is just zero degrees or theta equals zero. And we know that this term just becomes one. So your work is really just W. Equals MG times delta Y. And because your force, your MG points along your direction of motion, then gravity is going to do positive work. So in part we just do MG times delta Y. And we've got all of our numbers right? We've got the mass, which is 5.1, we've got 9.8 for G. And then the displacement, the delta Y. Is really just the two m that it falls. And notice how we plugged in. Everything is positive. So you're gonna get a work that is 100 jewels. And that's the first part. So let's look at the part B. Now, part B asks us to calculate the kinetic energy right before it hits the ground. So in part B we want the kinetic energy and this is gonna be the kinetic energy final really. Um So if you take a look at our diagram here, what happens is that one of the book is on the bookshelf, like here are kinetic energy is actually equal to zero, right? It's not moving. And then we know that gravity is going to pull this thing downwards and then it's gonna have some velocity right before it hits the ground. So we want to figure out the kinetic energy here at this point, we'll remember that kinetic energy is really just one half M. V. Squared and we've got those numbers. So kinetic energy is just going to be one half times the mass of 51 Let me go and write that again. So you've got one half Times 5.1 times the 6.26 squared. And you go ahead and plug this in and you're gonna get a kinetic energy of 100 jewels. So notice how we get the same exact number here and that is no coincidence. Remember that work is really just a transfer of energy. So when the force of gravity does 100 jewels of work on the box, then the box basically gains 100. So the work is equal to 100 jewels. And so therefore the box gains 100 joules of kinetic energy. All right, so let's move on to the second part now, which is when objects are rising or going up now, we're gonna take a look at a problem here in which Iraq is thrown vertically upwards now and we want to calculate the work and energy. And the basic idea here is that the set up of these problems is gonna be the same. Your gravitational force still points downwards. But now because this block has a velocity that's up, your displacement vector is actually going to point up now instead of down. So really when you go through your FD cosign theta, what happens is that your force is still gonna be MG, your distance is still gonna be delta. Why? Right, those are positives. But now the angle between those two things is not going to be zero because your MG points down and your delta Y. Points up so your co sign now is going to be 180 we know this becomes negative one. So really what happens? I'm sorry. So really what happens here is that your work that is done by gravity just becomes negative MG times delta. Why? Because your force and your motion point in opposite directions, gravity is actually going to do negative work when things are going up. Let's take a look at this example. Now we've got a rock that we're throwing vertically upwards, right? So it's got some velocity like this. We know this initial velocity is 15 and now it happens, it's gonna rise to a maximum height of 11.5. That's basically what my delta Y. Is. So this delta Y. Is equal to 11 0.5. So in this first part here, we want to calculate now the initial kinetic energy K. Initial. So really we're gonna do one half M. V. Initial times of starting the initial square. Right? So we have one half times the two kg rock times the 15 m per second squared. And you end up with a kinetic energy of 225 jewels. That's what you start off with here at the bottom. So we have K. E. Initial is to 25. So now what happens is we want to calculate the work now in part B. So we want to calculate the work that is done on the rock by the force of gravity. That's W. G. And when you have objects that are going up really, this is just gonna be negative MG times delta. Why? Because again your force and motion are opposite. So you're W. G. Is just gonna be negative. This is going to be uh 5.1. This is gonna be too for your mass. This is gonna be 9.8 and now you're gonna have a delta Y. It says that the maximum height is 11.5. So this is your delta Y. Like this, right? And so if you go ahead and calculate this, you're gonna get negative 225 jewels. So notice here again how we get the same exact numbers. And the idea here is that you have all this kinetic energy when you first throw the rock upwards. But remember that as it goes up, the force of gravity is doing negative work on the rocks when it gets to the top here and it finally stops, the kinetic energy is equal to zero. It's basically lost all of its energy because of work. All right, so that's it for this one. Guys, let me know if you have any questions.

6

Problem

You push a 3kg box against a wall for a distance of 2m with a force of 40N that makes a 53° angle with the horizontal, as shown. Calculate the work done by gravity.

A

58.8 J

B

-58.8 J

C

- 160 J

D

-80 J

7

example

## Work Done on Curvy (Not straight) Paths

2m

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All right guys, let's check out this problem here, we have a hiker who starts at the bottom of the mountain. So basically I'm gonna call this point A. Like this and they're trying to get to the top of this mountain, so I'm gonna call this point B. But instead of taking a straight path like up like this and therefore kind of looking at an inclined plane, what's actually happening is this hiker is going to take an irregular path with varying speeds and also inclination. So what does it look like? It's kind of like a squiggly path to get up to the top here of this mountain. We want to do is we want to figure out the work that is done by gravity on the hiker during this entire hike. So basically that's what I'm going to draw a little line like this and we know that this height here of the mountain, this is going to be my delta Y. Is going to be 1000 m. But the path that I'm taking from A to B is actually going to be sort of not straight, it's gonna be curvy like this. So how do I calculate the work that's done by gravity? Well remember that the equation is just negative MG. Times delta Y. And we have to do is we have to figure out a direction first of positive, So because I'm going from the bottom to the top, I'm going to call the upward direction positive like this. So what this means is that our delta Y. Is actually positive 1000 here. So if we get to the equation W. G equals negative MG delta. Why we run into a problem here? Because you might think that as we go along this path here, the work that's done is going to be sort of changing and you're actually right, it is at any point along the path right here for this hiker, we know that the MG is gonna point downwards. However, what are the really cool things about the work that's done by gravity is from the equation, we can tell that the work done only really just depends on delta, why it's only depending on the change in height here. So the work that's done by gravity depends only on the change in your heights. And it doesn't actually cause it doesn't actually depend on the path that's taken. So you wouldn't be able to solve this problem if you didn't know this. So this is actually called path independence. You might see that in your textbooks. It basically just means that it doesn't matter what path you take. All that matters is that you started today and ended up B and the difference between those two is m. So the work done by gravity is just gonna be negative. We have 75 times 9.8 And then we're gonna multiply this by positive 1000. So as we as we should expect to work done by gravity is gonna be negative and it's going to be 735,000 jewels. That's the work that is done by gravity. Alright, so again, only just depends on the height. It doesn't actually matter the path that you take here. We'll talk about that more in a later video. So that's it for this one guys.

Additional resources for Intro to Calculating Work

PRACTICE PROBLEMS AND ACTIVITIES (6)

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