## 字幕表 動画を再生する

• Hi. It’s Mr. Andersen and this is AP Physics essentials video 51. It is on the work energy

• 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.

Hi. It’s Mr. Andersen and this is AP Physics essentials video 51. It is on the work energy

ワンタップで英和辞典検索 単語をクリックすると、意味が表示されます

# 仕事のエネルギー原理 (Work Energy Principle)

• 54 9
Cheng-Hong Liu に公開 2021 年 01 月 14 日