by Patrick Ford

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.

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