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保护能源
There are many conversions between potential and as the cars travel around a double looping roller coaster. Throughout the ride, however, there will always be the same total amount of . Play around with the simulation below to learn more about the transfer of energy in a roller coaster ride:
::汽车在绕着双环滚滚车行驶时,潜力和车轮之间有许多转换。然而,在整个旅程中,总数量总是相同的。玩弄下面的模拟游戏,以更多地了解过山车载运的能源转移情况:

Conservation of Energy
::保护能源

The law of conservation of energy states that within a closed system , energy can change form, but the total amount of energy is constant. Another way of expressing the law of conservation of energy is to say that energy can neither be created nor destroyed. An important part of using the conservation of energy is selecting the system. Just as in conservation of , energy is conserved only if the system is closed. In a closed system , objects may not enter or leave, and it is isolated from external forces so that no work can be done on the system.
::节能法则指出,在一个封闭的系统中,能源可以改变形式,但能源总量是不变的。另一种表达节能法则的方式是说,不能创造或摧毁能源。节能的一个重要部分是选择系统。正如在节能方面一样,只有在系统关闭的情况下才能节能。在一个封闭的系统中,物体不得进入或离开,它不得与外部力量分离,从而无法对系统进行任何工作。

In the analysis of the behavior of an object, you must make sure you have included everything in the system that is involved in the motion. For example, if you are considering a ball that is acted on by gravity, you must include the earth in your system. If considered by itself, one can tell that the of the ball is increasing as it falls, but only by including the earth in the system can you see that the increasing kinetic energy is balanced by an equivalent loss of . The sum of the kinetic energy and the potential energy of an object is often called the mechanical energy .
::在分析物体的行为时,您必须确定您已经将所有与运动有关的元素都包括在系统中。例如,如果您在考虑一个以重力作用的球,则必须将地球包括在您的系统中。如果自我考虑,人们可以发现球在下降时正在增加,但只有将地球纳入系统,你才能看到,不断增长的动能以相等的损耗来平衡。动能和物体的潜在能量的总和通常被称为机械能量。

Consider a box with a of 20.0 N sitting at rest on a shelf that is 2.00 m above the earth. The box has zero kinetic energy but it has potential energy related to its weight and the to the earth’s surface.
::考虑一个在地球上方2千米的架子上休息20.0N的盒子。盒子的动能为零,但与重量和地球表面有关的潜在能量。

$\begin{aligned} P E = m g h = (20.0 N) (2.00 m) = 40.0 J \end{aligned}$
::PE=mgh=(20.0N)(2.00米)=40.0J

If the box slides off the shelf, the only force acting on the box is the force of gravity and so the box falls. We can calculate the of the box when it strikes the ground by several methods. We can calculate the speed directly using the formula $\begin{aligned} v {_{f}}^{2} = 2 a d \end{aligned}$ . We can also find the final by setting the kinetic energy at the bottom of the fall equal to the potential energy at the top, $\begin{aligned} K E = P E \end{aligned}$ , thus $\begin{aligned} \frac{1}{2} m v^{2} = m g h \end{aligned}$ . When reduced, we see that $\begin{aligned} v {_{f}}^{2} = 2 g h \end{aligned}$ . Note that these formulas are essentially the same; when gravity is the and the height is the distance, they are the same equation.
::如果盒子从架子上滑动, 盒子上的唯一作用力就是引力, 使盒子掉落。当盒子以几种方法撞击地面时, 我们可以计算出它的速度。我们可以使用公式 vf2=2ad 直接计算速度。我们也可以通过将跌落底部的动能与顶部的潜在能量等值来找到最后的能量, KE=PE, 也就是12 mv2=mgh 。当减小时, 我们可以看到 vf2=2gh 。注意这些公式基本上是一样的; 当重力和高度是距离时, 它们就是相同的方程。

$\begin{aligned} v = \sqrt{(2) (9.80 m / s^{2}) (2.00 m)} = 6.26 m / s \end{aligned}$
::v=(2)9.80米/秒2(2.00米)=6.26米/秒

Examples
::实例

Example 1
::例1

Suppose a cannon is sitting on top of a 50.0 m high hill and a 5.00 kg cannon ball is fired with a velocity of 30.0 m/s at some unknown angle. What is the velocity of the cannon ball when it strikes the earth?
::假设一个大炮座落在50米高山顶上,而一个5.00公斤大炮炮弹以30.0米/秒的速度以某种未知的角度发射。

Since the angle at which the cannon ball is fired is unknown, we cannot use the usual equations from . However, at the moment the cannon ball is fired, it has a certain $\begin{aligned} K E \end{aligned}$ due to the mass of the ball and its speed and it has a certain $\begin{aligned} P E \end{aligned}$ due to its mass and it height above the earth. Those two quantities of energy can be calculated. When the ball returns to the earth, its $\begin{aligned} P E \end{aligned}$ will be zero. Therefore, its $\begin{aligned} K E \end{aligned}$ at that point must account for the total of its original $\begin{aligned} K E + P E \end{aligned}$ . T
::由于发射大炮球的角不明,我们无法使用来自......的通常方程式。然而,在发射大炮球时,由于球的重量和速度,它有一定的KE,由于球的重量和速度,它有一定的PE,由于其质量和高度高于地球,可以计算出这两批能量。当球返回地球时,它的PE将是零。因此,在那个时候,它的KE必须说明其原来的KE+PE的总数。T

$\begin{aligned} E_{TOTAL} = K E + P E = \frac{1}{2} m v^{2} + m g h \end{aligned}$
::ETOTAL=KE+PE=12 mv2+mgh

$\begin{aligned} = (\frac{1}{2}) (5.00 k g) (30.0 m / s)^{2} + (5.00 k g) (9.80 m / s^{2}) (50.0 m) \end{aligned}$
::= (12) (5.00公斤)(30.0米/秒)2+(5.00公斤)(9.80米/秒)(50.0米)

$\begin{aligned} = 2250 J + 2450 J = 4700 J \end{aligned}$
::=2250 J+2450 J=4700 J

$\begin{aligned} \frac{1}{2} m v {_{f}}^{2} = 4700 J \end{aligned}$
::12 mvf2=4700 J

$\begin{aligned} v_{f} = \sqrt{\frac{(2) (4700 J)}{5.00 k g}} = 43.4 m / s \end{aligned}$
::vf=(2)(2)-4700 J)5.00公斤=43.4米/秒

Example 2
::例2

A 2.00 g bullet moving at 705 m/s strikes a 0.250 kg block of wood at rest on a frictionless surface. The bullet sticks in the wood and the combined mass moves slowly down the table.
::以705米/秒移动的A 2.00克子弹击中无摩擦表面休息的0.250公斤木块,木材中的子弹棒和综合质量缓慢地从桌子下移。

(a) What is the $\begin{aligned} K E \end{aligned}$ of the bullet before the collision?
:a) 碰撞前子弹的KE是什么?

(b) What is the speed of the combination after the collision?
:b) 碰撞后组合的速度如何?

(c) How much $\begin{aligned} K E \end{aligned}$ was lost in the collision?
:c) 碰撞中损失了多少KE?

(a) $\begin{aligned} K E_{BULLET} = \frac{1}{2} m v^{2} = (\frac{1}{2}) (0.00200 k g) (705 m / s)^{2} = 497 J \end{aligned}$
:a) KEBULLET=12 mv2=(12)(0.00200公斤)(705 m/s)2=497 J

(b) $\begin{aligned} m_{B} v_{B} + m_{W} v_{W} = (m_{B + W}) (v_{B + W}) \end{aligned}$
:b) mBvB+mWvW=(mB+W)(vB+W)

$\begin{aligned} (0.00200 k g) (705 m / s) + (0.250 k g) (0 m / s) = (0.252 k g) (V) \end{aligned}$
:0.0000公斤(705米/秒)+(0.250公斤)(0米/秒)=(0.252公斤)(V)

$\begin{aligned} (1.41 k g m / s) = (0.252 k g) (V) \end{aligned}$
:1.41公斤m/s)=(0.252公斤)(V)

$\begin{aligned} V = 5.60 m / s \end{aligned}$
::V=5.60米/秒

(c) $\begin{aligned} K E_{COMBINATION} = \frac{1}{2} m v^{2} = (\frac{1}{2}) (0.252 k g) (5.60 m / s)^{2} = 3.95 J \end{aligned}$
:c) Kconmodiation=12 mv2=(12)(0.252公斤)(5.60 m/s)2=3.95 J

$\begin{aligned} K E_{LOST} = K E_{BEFORE} - K E_{AFTER} = 497 J - 4 J = 493 J \end{aligned}$
::KELST=KEEFORE-KEAFFTER=497 J-4 J=493 J

Further Reading
::继续阅读

Conversion
::改划

Summary
::摘要

In a closed system, energy may change forms but the total amount of energy is constant.
::在封闭的系统中,能源可能改变形式,但能源总量保持不变。

Review
::回顾

A 15.0 kg chunk of ice falls off the top of an iceberg. If the chunk of ice falls 8.00 m to the surface of the water,

what is the kinetic energy of the chunk of ice when its hits the water, and
::当冰块撞击到水面时冰块的动能是什么?

what is its velocity?
::它的速度是多少?

::冰块从冰山顶上掉下来1500公斤的冰块。如果冰块从水面上掉落8米,那么冰块撞击水时的动能是什么? 其速度是多少?

An 85.0 kg cart is rolling along a level road at 9.00 m/s. The cart encounters a hill and coasts up the hill.

Assuming the movement is frictionless, at what vertical height will the cart come to rest?
::假设这个运动是无摩擦的,

Do you need to know the mass of the cart to solve this problem?
::您需要知道车的重量才能解决这个问题吗 ?

::一辆85公斤的马车正在沿9:00米/秒的一条水平公路行驶。该马车遇到山坡和山上海岸。假设该马车没有摩擦, 该马车会在哪个垂直高度歇息? 您需要知道该马车的重量才能解决这个问题吗 ?

A circus performer swings down from a platform on a rope tied to the top of a tent in a pendulum-like swing. The performer’s feet touch the ground 9.00 m below where the rope is tied. How fast is the performer moving at the bottom of the arc?
::马戏团表演者从平台上摔下来,绳子绑在像钟摆一样的钟摆的帐篷顶部。表演者的脚碰到绳子绑在的下方9:00米处的地面。表演者在弧底移动的速度有多快?

A skier starts from rest at the top of a 45.0 m hill, coasts down a $\begin{aligned} 30^{\circ} \end{aligned}$ slope into a valley, and continues up to the top of a 40.0 m hill. Both hill heights are measured from the valley floor. Assume the skier puts no effort into the motion (always coasting) and there is no friction.

How fast will the skier be moving on the valley floor?
::滑雪者在山谷地板上移动的速度有多快?

How fast will the skier be moving on the top of the 40.0 m hill?
::滑雪者在40米山顶移动的速度有多快?

::滑雪车从45米高山顶休息开始,向下向30千瓦高的山顶靠岸,一直持续到40米高的山顶。两座山高都是从山谷地底测量的。假设滑雪车不费劲地运动(一直靠岸),也没有摩擦。滑雪车在山谷地上移动的速度有多快?滑雪车在40米高的山顶移动的速度有多快?

A 2.00 kg ball is thrown upward at some unknown angle from the top of a 20.0 m high building. If the initial magnitude of the velocity of the ball is 20.0 m/s, what is the magnitutde of the final velocity when it strikes the ground? Ignore air resistance.
::2.00千克球从高楼20.0米的顶部向上投出一个未知角度。如果球速度的初始大小为20.0米/秒,那么最后速度撞击地面时的放大度是多少?忽略空气阻力。

If a 2.00 kg ball is thrown straight upward with a KE of 500 J, what maximum height will it reach? Neglect air resistance.
::如果一个2.00公斤的球被直向上扔上500焦耳的KE,它能达到的最高高度是多少?忽略空气阻力。

Explore More
::探索更多

Use this resource to answer the questions that follow.
::使用此资源回答下面的问题。

What happens when one ball is pulled up to one side and released?
::当一个球被拉到一边释放出来时会怎么样?

What h appens when three balls are pulled up to one side and released?
::当三个球被拉到一边放出来时会怎么样?

What happens when two balls are pulled out from each side and released?
::当两个球从两边拉出来然后释放出来时会怎么样?

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