
Chinese: 
大家好，我是安德森。这里是AP物理必备系列的第64个视频，有关功和能。
在生活中，能把能和功的关系成功表达出来的事件是撑杆跳。
现在女子撑杆跳的世界纪录保持者是Elena Isinbeava。这项纪录
是16英尺7英寸。那这是多高呢？如果我站在一颗篮球
边缘上，她将能够撑杆跳过我的头顶，这是非常疯狂的一半。
我相信跳过那个高度杆子一定感觉很棒吧。但它确实是一个物理学问题。
当选手从跑道远端跑来时，她增加了她自己的动能。她将那些动能存储
在杆的弹性势能中。然后她将那些能又转换成了她的
势能。因此杆在这里做了功。它在一定距离内施加
了一个力。而这些都将是相同的。她在跑道上所积累的能量

English: 
Hi. It’s Mr. Andersen and this is AP Physics
essentials video 64. It is on work and energy.
And a track event that gets at the relationship
between energy and work is the pole vault.
And the world record holder in the women’s
pole vault is Elena Isinbeava. And that record
is sixteen feet seven inches. So what does
that mean? If I were to stand on a basketball
rim she would be able to pole vault a half
of a foot over my head which is pretty crazy.
I am sure it feels amazing to get over that
bar. But it is really a physics question.
As she runs down the runway she is increasing
her kinetic energy. She is then storing that
in the elastic potential energy of the pole.
And then she is converting that into gravitational
potential energy. And so that pole is doing
work. It is applying a force over a given
distance. And those are all going to be equivalent.
The amount of energy she stores on the runway

English: 
is going to be the amount of potential energy
she has at the top. And so in a system we
can add or pull out mechanical energy. And
that could either be potential due to its
position or kinetic energy. But as we do that
we have to move that system a certain amount
of distance. There has to be displacement.
And that displacement is caused by an unequal
parallel force. In other words the parallel
force has to be in the direction of that displacement.
And so since we have a force and a displacement,
what are we talking about here? We now have
work. And work and that mechanical energy
that is added are going to be equivalent.
And a nice tool that we can use in physics
to look at how much work are we doing is to
look at a force versus displacement graph.
And so let’s say we apply a constant force,
a parallel force, and we get a certain displacement,
all we do is look at the area under that curve
and that is going to tell us how much work
we have done and therefore how much mechanical
energy we have added to that system. Now let’s
say it is not straight. Let’s say the amount

Chinese: 
也将是（等同于）她在空中时的势能。因此我们看到，在一个系统中，
我们可以添加或减少机械能。而且所改变的能量既可以是由位置变化产生的势能，
也可以是动能。但当我们做出机械能改变时，我们需要移动一定
的距离。这个过程中必须要有位移​​。而位移是由一个不平衡的
平行力造成。换言之，平行力必须是在该位移的方向。
好，那么现在我们有了力和位移。然后呢？我们做了功。
所做的功和增加的能量也将会是数值相等的。
然后要看所做功的多少，我们可以运用物理学中一个很好的工具：
力 - 位移图。好，现在假设我们一直在施加大小方向均不变的力，
平衡力。我们又得到了一定的位移。现在我们所要做的就是看图像曲线下的面积。
也就是那块面积告诉了我们做功的多少，还有因此增加到系统中的机械能。
现在，让我们假设情况不是这么直截了当的。假设力的数量是

Chinese: 
随时间变化的，我们做的功将仍然将是该曲线下方面积的量。
由于我们在AP物理并不使用微积分，我们只能将这块面积分割成
不同的部分。我们接着可以计算出功以及被增加入系统中
的能量。现在我来给大家举一个例子。
假设我们有一个质量很大的巴士，然后我们又在向它施加
12,000牛顿的力。现在施加的力将会使物体产生加速度。这个加速度恒定不变，使物体
产生48米的位移。那么现在我们想求这位移过程中对巴士所做的功。
这里的功W将等于衡力F乘以s。这里的s就是物体产生的位移。
所以要求功，我们只用知道衡力的大小和位移是多少就好。
现在我把这个例子里的两个数值代入看看：12000牛顿的衡力乘以
一个48米的位移。那么我们所做的功就是五十八万焦耳左右。
现在我们就正确地解决了这个问题。看来我们对巴士做了很多功啊！

English: 
of force varies over time. It is still going
to be the amount of area underneath that curve.
And since we are not using calculus in AP
Physics we could just break this into all
of these different parts. We could figure
out the amount of work and then the amount
of energy that has been added to that system.
And so let’s give you an example of a problem.
Let’s say we have a bus that has a huge
mass and we are applying a 12,000 newton force
to it. And that newton force gets an acceleration,
since it is constant and moves a displacement
of 48 meters. So if we want to figure out
how much work is done on that bus, work is
going to be equal to the force, F, times s.
s is going to be the displacement of the object.
And so all we really need to know is what
is that force and what is the displacement.
And so if I plug in those two values right
here, it is that 12,000 newton force, times
a 48 meter displacement, and so the amount
of work that we have done is going to be around
580,000 joules as we solve this for correct
significant digits. So we have done that much

English: 
work on the bus. And so I will put that right
up here. Now work and energy are going to
be equivalent. And so we have added energy
to that system. We have added kinetic energy
to that system. And so the amount of work
is going to equal the amount of kinetic energy
that we have added. And so how much kinetic
energy did we have at the beginning? Well
the bus was at rest so there was zero amount
of kinetic energy. And we have done work on
that bus and we have converted into all of
that kinetic energy that we have at the end.
And so what makes this nice is once I know
the amount of work I could figure out the
amount of kinetic energy assuming we did not
lose a lot to friction. I could therefore
even figure out the velocity of that bus.
So let’s do that. The amount of work, which
we calculated before, is equal to 1/2 m v
squared. Again that is that final kinetic
energy. We know the mass of the bus, so we
can just plug that in. It is 6,500 kilograms.
And then we can simply solve for v. We can
solve for that final velocity of the bus.
So if I do that, I get a velocity of around
13 meters per second. And so it started at

Chinese: 
那么我就把这功的数值放在这儿。我们知道功和能将是
等同的。因此我们算是在系统中增加了能量。我们往系统中增加了
动能。所以我们所做的功将等于我们所增加的动能的量。
所以一开始我们有多少动能呢？当巴士原地不动的时候，
它所有的动能为零。现在我们给了它衡力和位移向它做功了，
我们也已经转换了所有的动能。
这转换的好处在于：我一求出所做功的量，我就可以求出在假设没有能量
流失情况下的动能。因此，我甚至可以求出那辆巴士
的速度。现在让我们试试看。我们之前计算过的
所做的功，也等于1/2 MV平方。我们还需要最后求出的动能。
我们知道巴士的质量是6500千克，所以我们把m=6500kg代入。
接着我们就可以解决了V，也就是速度了。我们也可以解决巴士的终速度。
所以如果我这样做，我就得到了每秒约13米的速度。所以它从0开始，

Chinese: 
在每秒13米结束。因此在这个过程最后速度大约是每小时30英里。
因此我们能够使用功、能理论或是说功、能之间的关系来找出
我们做了功后在系统中增加的机械能的量。
现在，我给大家出一个类似的问题。这是个12000牛顿的力。但
与此前情况不同，我们假设一个巨人正在拉这辆巴士，形成一个38度角。
那现在你怎么解决这个问题呢？记得要运用之前讲过的概念，力必须要与
位移平行。所以现在你不得不用一些三角的知识。
你就这样画个图。这将是我们需要求出的
未知邻边x。所以现在你要用这个等式：cos38度=
x比上所施加的力。然后你就可以求出平行于位移的力。
接着你就可以求出你正做的功。所以我们再一次证明了
力 - 位移图对于这类问题的实用性。在图上，纵轴是我们施加的力，

English: 
0. Ends up at 13 meters per second. So it
is going around 30 miles per hour at the end.
And so we are able to use this work-energy
theorem or this work-energy relationship to
figure out once we have work we could figure
out the amount of mechanical energy that has
been added. Now let’s give you a similar
problem. This is a 12,000 newton force. But
let’s say a giant is pulling this bus. And
it is at a 38 degree angle. So how would you
solve this one? Remember going back to that
early concept map, that force has to be parallel
to the displacement. And so you would have
to use a little bit of trigonometry. So you
would set it up like this. So this is going
to be our adjacent side. So this is the x
value that we are looking for here. And so
you are going to use the cosine of 38 equal
to that x over that force. And then you would
figure out the force parallel to that movement.
And that is going to tell you how much work
is being done. And so again a really nice
tool to get at this is a force versus displacement
graph. And so what we have is force on the

English: 
side and then displacement along the bottom.
So let’s say we have this object, a 3 kilogram
block. And we are going to apply the following
forces to it. Let’s say we apply a constant
force to the right, a 30 newton force for
the first 4 meters. And then we lessen that
force down to 0 over the next three meters.
And so if I want to calculate how much work
is being done on that block, I am going to
figure out the area under the curve. And so
I would start with this area right here. So
how much work is being done? Well you just
figure out the area of this block right here.
And so it is a force of 30 newtons. What is
our displacement? We would read that across
the bottom. It is going to be 4 meters. And
so the amount of work that we are doing is
going to 128 joules. It is just base times
height. And so that is how much energy that
would block would have at the end. What kind
of energy is it going to be? It is going to
be kinetic energy. Now watch what happens
after that. We are lessening the force down
to zero. And so how do I figure out the area
there? Well that is simply a triangle. And
so I could figure out what my base is. It

Chinese: 
而横轴是产生的位移。假设我们现在有一个3千克的大块物体。
我们将对它施加图上所显示的力。所以我们施加4秒钟一个方向
始终向右的30牛顿的衡力。然后，我们在接下来的3米位移
中将力减小到0。那么现在如果我要计算做功的多少，
我就要求出曲线下的区域。所以我就从现在标绿得这一块
开始。那这一块做的功是多少呢？你只用把这块的面积
求出来就好啦。好，那么这里有30牛顿的力。位移又是
多少呢？我们从横轴度数：4米。那么现在我们就求出了要做
的功：128焦耳。这里就仅仅是底乘高。
所以这就是物体这段路程结尾将拥有的能量。这能量是
什么类型呢？这些能量将会是动能。现在我们看看之后会
发生什么。我们正在把施加的力慢慢减小到0。那我该怎么
求这里黄色的面积呢？这是一个简单的三角形。所以我可以看到

Chinese: 
这里3米的底边。那力是多大呢？如果我们说这一整个是一个矩形，那
单独一个三角形就是1/2底乘高。它仍然是30乘以位移3。
所以这就是90焦耳。但是由于我们是1/2底乘高，真正的答案
就会是焦耳。所以你总可以计算力 - 位移图曲线下的面积。
这个面积不仅告诉你所做的功的量，还会告诉你
添加到系统中的能量。好，那么你学会如何对机械能变化
做出预测了吗？还是那句话，我们必须确保所施加的力
与位移平行。最后一个问题是：你能成功地把能量守恒定律
运用在这里吗？在整个过程中，我们并没有丢失能量。你又能成功地
运用功能关系吗？我希望你可以哟。希望这个视频可以帮助到大家！

English: 
is going to be 3 meters. What is the force?
Well, if we say this is a rectangle it is
1/2 base time height. It would still be 30
on this side times the displacement of 3.
So that would be 90 joules. But since we are
1/2 base times height I am going to get 45
joules. And so you can always calculate the
area under a force displacement graph. That
is going to tell you not only the amount of
work that has been done but the amount of
energy that has been added to that system.
And so did you learn to make predictions about
changes in mechanical energy? Again, we have
to make sure that that force is a parallel
force to displacement. And then finally could
you apply the concepts of conservation of
energy? We are not losing energy. And this
whole idea of work energy theorem to figure
out for example the velocity of that bus at
the end? I hope so. And I hope that was helpful.
