Guys, so now that we understand kinetic energy, in this video, I want to start talking about work. So we're going to see the relationship between work and energy and we're also going to see the relationship between work and forces. I'm going to 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 the velocity is equal to 0. So if we wanted kinetic energy, the kinetic energy is also equal to 0, and then you start pushing this box. So if you push it and it's frictionless, then the box is going to start to move. It's going to start to move to the right which means that now it has some velocity that's not equal to 0 and so, therefore, it has some kinetic energy that's not equal to 0 as 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 a transfer of energy, then you eat the cow, that's a transfer of energy, and then you push 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 joules 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 going to be the joule 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 going to 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 θ. So a lot of textbooks will rewrite this as a shortcut way, and this is how we're going to write it from now on, fdcosθ. So really, what we have to know about these equations, about this equation in particular, angle. θ is always the angle between f and Δx or f and d. So, basically, in these problems where you calculate work, you're going to have to know 3 things. You're going to have to know the magnitudes of both the force and the displacement. And remember, the magnitudes are always going to be plugged in as positive numbers. And then you're going to figure out the angle that is between those vectors. And that's what you plug into your cosine θ term.

Let's check out some examples. So here, we're going to pull a 2 kilogram box that's at rest on a frictionless surface. So the idea is this is 2, and then I'm going to pull this box with a force of 3 Newtons. And I want to calculate the work if I pull it through a distance of 5 meters. So basically, what happens is I want to figure out the work that's done by my force. So remember that the work is going to be fdcosθ, and I'm going to plug in everything as positive. So I have the force, which is 3 newtons. And if I'm pulling it to the right, then, therefore, it goes through a displacement, a Δx of 5. So really what I what happens is I'm going to plug this in and then I'm going to plug in 3 times 5. And now what's the cosine of this 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 0. And so with the cosine of 0 is, it's always equal to just 1. So, really, all we do is we actually just multiply 3 times 5 and we get 15 joules.

Alright. So that's the work that's done. Let's move on to the second example. Now we have a cart, a 5 kilogram 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 5. The velocity of this cart, the initial, is already equal to 10 meters per second, but there's going to be some stopping force. I'm going to call this fs, and it's a 100 Newtons. Now I want to figure out the work that's done by the stopping force. I'm just going to use the same equation, fdcosθ. So this is my work done is going to be fd, and this is going to be the stopping force times d times the cosine of θ. Alright? So I've got the stopping force. This is 100. Now what about the distance or the displacement? Well, if the cart is already moving to the right, but it's being pulled to the left, it still undergoes a displacement that's to the left here. This Δx is 2.5. So this is a 100 times 2.5. And now, what's the angle that is between these two vectors? Well, we said that 0 degrees was when they're parallel to each other. But when they're antiparallel, 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 0. So what happens here is that when you plug in cosine or θ equals a 180, your cosine term will always turn to negative one for antiparallel forces. Right? So, really, this just turns into a 100 times 2.5 times negative one, and this is just 250 joules, but it is negative.

Alright? So we can see here is that work 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 object's motion, then the work that's done is going to be positive, like we had over here. The force causes the cart to move or the box to move to the right. The force goes along with the motion. The work is positive. Now if the force hurts or goes against the object's motion, the work is going to be negative. 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.