principle. But before we get to that we should define both work and energy. What is energy?

Energy is the ability to do work. That does not help much. What is the energy we are going

to be talking about in this video? It is going to be kinetic energy. So that is energy due

to motion. And to figure that out remember it is one-half mv squared, where m is the

mass and v is the velocity of the object. If the object is bigger, if it is faster it

is going to have more kinetic energy. Now what is work? Work is simply a force times

a distance. So applying a force for a given amount of distance.

And so what is neat is that these two are equivalent. They are always going to be equal.

And so if you need to calculate the kinetic energy, a really good way to do it is to simply

calculate the work. And so if we are looking at baseball as it is moving it clearly has

kinetic energy. Where did it come from? It came from work of the pitcher. So the pitcher

is applying a force to that baseball over a given distance. And the amount of that work

is going to be equivalent to the amount of energy that is given to that baseball. And

so there is a equivalence between these two. So let’s start with work. So let’s say

we are applying a force to this object right here. So we are going to apply a constant

force to the right. And then we are going to move it a given distance. So watch what

happens. You can see it started at rest and it is accelerating. I just stopped the simulation

right here. And so we applied a constant force over a given distance. And so we did work.

Now what is neat is that is equivalent to the amount of energy that was given to that

object. This object now has kinetic energy. Now if we wanted to figure out that kinetic

energy we could break it down into the final kinetic energy, which is one half mv squared

minus the initial kinetic energy. But what is neat is if we just know the force and the

distance we have already figured out the amount of kinetic energy that is being added. Now

let’s work qualitatively for a second. Have we added energy to that object or have we

taken it away? Well since we applied a force to the right over a given distance we know

that the work is going to be a positive value. How else could we figure this out? Well if

we look at its initial kinetic energy, since it had a velocity of zero then we know that

this is a zero value. And if we are taking a value of a velocity minus a zero value we

know this is going to be a positive value. And so you can look at either the force and

where the force is being added over time. Or the change in the velocity and that is

going to tell you is energy being added or is it being taken away. Let’s look at a

different scenario. Now we have an object that is moving towards the right, and so when

I start it is going to be moving towards the right. Well let’s say we are applying a

net force to the left. What is going to happen here? You can see it is going to slow down.

And so what is happening to the amount of work? Well we have a negative force times

a distance and so we are decreasing the amount of kinetic energy in that object. We are taking

that energy away, which makes sense because the object is coming to rest. If we wanted

to look at the velocities we could see now that we have a final velocity of zero, so

this is a zero value right here, minus that initial velocity, and so you can see that

we are going to have a negative energy. We have lost energy. We have lost kinetic energy.

Now it is not as simple as that because sometimes you are going to apply a force and it is not

going to be in the same direction as the motion. And so let’s say we are pulling on that

object. So you could say we have a rope tied on to it. And we are moving it in that direction.

Okay. Now since we have moved it in that direction we have to breakdown that force vector into

its two component vectors. We have this, we will call that the force parallel. So it is

in the direction of the motion. And then we have this which is going to be the force perpendicular.

Now what is interesting is that it is only this force that is acting in the direction

of that motion. And therefore it is only that force that is doing work. The other force

is not doing any work since the motion is not in that direction. And so lots of times

when you are solving problems it is not as simple as figuring out the work. You have

to break that force down into its component force. And I will show you an example of that.

And so let’s start simple. Let’s say we have a cart. And that cart is moving from

the left to the right. And we are applying a constant force of 7.1 newtons. So how would

you figure out the work on that cart? It is really simple. You just say work is equal

to the force parallel. Again it is going to be in the direction of that motion times the

distance. And since we are given a force of 7.1 newtons times the distance 0.18 meters

it is really simple to solve for that. So what is going to be our work? It is simply

going to be 1.3 joules. Now that is the amount of work that was done on the cart. What is

neat is it also shows us the amount of kinetic energy that was added to that cart. We do

not have to figure out the mass and the velocity and initial velocity. We know by the amount

of work being applied to it that that is going to tell us the amount of kinetic energy added.

Let’s look at a cart that is already moving towards the right. We are applying a net force

to the left. Watch what happens here. Again it is slowing down. So how do you figure out

the amount of work on that? Well again it is force parallel. In this case it is antiparallel.

It is in the opposite direction. And so we have to put that in as a negative value. And

so we are going to have negative 0.93 joules. In other words we are taking energy away.

We are doing negative work on that object. Now let’s move that force in a different

direction. This makes the problem a little bit harder. You can always pause the video

and try to work this out. But let’s say we are applying a 9.6 newton force in this

direction but it is 32 degrees up from parallel. And watch what happens to that object. So

it is moving in that direction. So if I want to figure out the amount of energy that is

added to that cart all I have to do is figure out the force, force parallel times the distance.

What makes this problem a little bit harder remember is that we have to target this, which

is going to be that force in the parallel direction. How do we do that? Well we just

use a little bit of trigonometry. Since you know this angle and you know the hypotenuse

we can figure out this adjacent. And so to figure out that it is simply going to be the

force times the distance times cosine of theta. What is cosine of theta? It is the cosine

of this angle right here. And so that is going to give me a force and I can always check

that. It should be a value that is going to be less than the hypotenuse or less than 9.6

newtons. So if I plug that in my calculator I can find the cosine of 32. And the cosine

of 32 is 0.848. So it is decreasing that amount of force. But it is the force that is actually

moving in the direction of that motion. And so now I just simply multiply all of those

values and I get 1.5 joules. Again I am solving for significant digits here. And so how much

work is done on the cart? 1.5 joules. How much energy has been added to the cart? 1.5

joules. And so you know there is that equivalence. The amount of work done on that cart is equal

to the amount of energy that that cart has gained, which is kinetic energy minus initial

energy. And so if you wanted to, if you were given the mass of the object you could plug

that in. Since that object is accelerating we could say that its initial velocity is

0. And so you could figure out its final velocity. And so again, work and energy are going to

be equivalent. How did it get the energy? We applied a certain amount of force over

a given distance. And so did you learn to make predictions about the changes in kinetic

energy of an object? Again, if it is moving faster it is going to have a higher amount

of kinetic energy. Can you use the force and the velocity to just qualitatively figure

out are we adding energy or are we taking it away? And then finally can you figure out

the amount of work on an object and therefore the amount of energy that is being added to

the object? I hope so. And I hope that was helpful.