节: 应用:工作与力量 | 6.8 应用:工作与力量

章节大纲

In physics, work is defined as force acting through a distance (or displacement), and is computed as the product of the component force and the distance. The force can be constant through the distance, or vary as the distance. How would you express the total work done by a force $\begin{aligned} F (x) \end{aligned}$ acting from position $\begin{aligned} x = a \end{aligned}$ to position $\begin{aligned} x = b \end{aligned}$ ?
::在物理学中,工作的定义是:通过距离(或偏移)发挥作用的力,并计算成组成部分力和距离的产物。力可以通过距离保持恒定,或因距离而变化。您将如何表达从位置 x=a 到位置 x=b的F(x) 部队完成的全部工作?

Work and Force
::工作与力量

Work in physics is defined as the product of a force and displacement. Force and displacement are vector quantities, and therefore have direction and magnitude . Work is computed using the components of force and displacement that are in the same direction. The product of these two quantities gives the work done by the force in the direction of displacement. Mathematically, we say
::物理工作被定义为力量和流离失所的产物; 武力和流离失所是病媒的数量,因此具有方向和规模; 工作是使用同一方向的武力和流离失所的成分来计算的; 这两部分的产物提供了部队在流离失所方面所做的工作。

$\begin{aligned} W = F d, \end{aligned}$
::W=Fd, W=Fd, W=Fd, W=Fd, W=Fd, W=Fd, W=Fd, W=Fd, W=Fd, W=Fd, W=Fd, W=Fd, W=Fd, W=Fd, W=Fd,

where $\begin{aligned} F \end{aligned}$ is the force and $\begin{aligned} d \end{aligned}$ is the displacement, both in the same direction. If the force is measured in Newtons and distance is in meters, then work is measured in the units of energy which is in joules (J).
::F是力量,d是流离失所,两者方向都相同。如果用牛顿测量力,距离以米计,那么用焦耳(J)的能量单位衡量工作。

Say you push an empty grocery cart with a force of 44 N for a distance of 12 meters. How much work is done by you (the force)?
::说你们用44N的力推一辆空的杂货车,距离12米。你们做了多少工作(部队)?

Assuming that all of the force is in the direction of motion, we can use the formula above,
::假设所有力量都朝着运动的方向前进, 我们可以使用上面的公式,

$\begin{aligned} W & = F d \\ = (44) (12) \\ = 528 J \end{aligned}$

::W=Fd=(44)(12)=528 J

The work done is 528 $\begin{aligned} J \end{aligned}$ .
::已完成的工作是528 J。

Now, say a librarian displaces a book from an upper shelf to a lower one. If the vertical distance between the two shelves is 0.5 meters, and the weight of the book is 5 Newtons, how much work is done by the librarian?
::如果图书馆员将一本书从上架移到下架。如果两个书架之间的垂直距离是0.5米,而书的重量是5牛顿,图书馆员做了多少工作?

In order to be able to lift the book and move it to its new position, the librarian must exert a force that is at least equal to the weight of the book. In addition, since the displacement is a vector quantity, then the direction must be taken into account. So, $\begin{aligned} d = - 0.5 m e t e r s . \end{aligned}$
::图书馆员必须施加至少与书重量相等的力量。此外,由于迁移是一个矢量,所以必须考虑到方向。所以, d0.5米。

Thus
::因此,

$\begin{aligned} W & = F d \\ = (5) (- 0.5) \\ W & = F d \\ = - 2.5 J \end{aligned}$

::W=Fd=(5)(--0.5)W=Fd=2.5 J

Here we say that the work is negative since there is a loss of gravitational potential energy rather than a gain in energy. If the book is lifted to a higher shelf, then the work is positive, since there will be a gain in the gravitational potential energy.
::我们在这里说,工程是负的,因为重力潜在能量的丧失而不是能量的增益。如果书被提升到更高的架子上,那么工程就是正的,因为重力潜在能量将会增加。

In general, if a force $\begin{aligned} F \end{aligned}$ is a function of position in a given direction, the work done by the force in the direction from position $\begin{aligned} a \end{aligned}$ to position $\begin{aligned} b \end{aligned}$ can be expressed as:
::一般来说,如果F部队是在某一方向上的位置函数,F部队在从a到b阵地的方向上所做的工作可表述为:

$\begin{aligned} W = \int_{a}^{b} F (x) d x . \end{aligned}$

::WabF(x)dx 。

A force of 10 Newtons is required to stretch a 20 cm spring to a length of 26 cm. Find the work done in stretching the spring from a length of 24 cm to 28 cm.
::需要10个牛顿的力量才能将春季的长度从20厘米伸展到26厘米。找到将春季的长度从24厘米伸展到28厘米所完成的工作。

Hooke’s Law states that when a spring is stretched $\begin{aligned} x \end{aligned}$ units beyond its natural length it pulls back with a force $\begin{aligned} F (x) = k x \end{aligned}$ , where $\begin{aligned} k \end{aligned}$ is called the spring constant or the stiffness constant. The work required to stretch the spring a length $\begin{aligned} x \end{aligned}$ is $\begin{aligned} W = \int_{a}^{b} F (x) d x \end{aligned}$ , where $\begin{aligned} a \end{aligned}$ is the initial displacement of the spring ( $\begin{aligned} a = 0 \end{aligned}$ if the spring is initially unstretched) and $\begin{aligned} b \end{aligned}$ is the final displacement.
::Hooke的法律规定,当弹簧被拉长为超过其自然长度的x单位时,它会用F(x)=kx的力拉回来, k被称作弹簧常数或僵硬常数。伸展弹簧所需的工作是WabF(x)dx, 其中a 是弹簧的初始移位(a=0, 如果弹簧最初没有伸展), b 是最后的移位。

The problem requires that we first determine $\begin{aligned} k \end{aligned}$ , the spring constant:
::这个问题要求我们首先确定 k, 春季常数:

$\begin{aligned} F (x) & = k x \\ 10 & = k (6) \\ k & = \frac{5}{3} N / m \end{aligned}$

::F(x)=kx10=k(6)k=53 N/m

The work to displace the spring from 24 cm to 28 cm can now be computed:
::将弹簧从24厘米移到28厘米的工作现在可以计算如下:

$\begin{aligned} W & = \int_{a}^{b} F (x) d x \\ = \int_{24}^{28} \frac{5}{3} x d x \\ = \frac{5}{3} {[\frac{x^{2}}{2}]}_{24}^{28} \\ = \frac{5}{3} [\frac{784}{2} - \frac{576}{2}] \\ W & = 173.3 J \end{aligned}$

::WabF(x)dx242853xx=53[x22]2428=53[7842-5762]W=173.3J

The work required is 173.3 J.
::所需工作为173.3 J。

Examples
::实例

Example 1
::例1

Earlier, you were asked how to express the total work done by a force $\begin{aligned} F (x) \end{aligned}$ acting from position $\begin{aligned} x = a \end{aligned}$ to position $\begin{aligned} x = b \end{aligned}$ . Based on the definition of work and knowledge of the meaning of the definite integral, work is expressed as
::早些时候,有人问您如何表达F(x)部队从x=a位置到x=a位置的F(x)部队完成的全部工作。

$\begin{aligned} W = \int_{a}^{b} F (x) d x \end{aligned}$

::WabF(x)dx

Note that the force in Newtons (N), over a distance in meters (m) results in work in joules (J);
::注意牛顿(N)的功率超过米(m),可产生焦耳(J);

Force in pounds (lbs), over a distance in feet (ft) results in work in ft-lbs.
::以磅(磅)为单位,跨足距离(ft)导致工作以ft-磅(ft-lbs)为单位。

Example 2
::例2

A bucket has an empty weight of 23 N. It is filled with sand of weight 80 N and attached to a rope of weight 5.1 N/m. Then it is lifted from the floor at a constant rate to a height 32 meters above the floor. While in flight, the bucket leaks sand grains at a constant rate, and by the time it reaches the top no sand is left in the bucket. Find the work done:
::桶的空重为23牛顿。桶装满了80牛顿的沙子,并附在5.1牛顿/米的绳子上。然后,桶以恒定速度从地板上抬起,升至地表高度32米以上。当桶在飞行时,桶以恒定的速度漏出沙粒,当桶到达顶部时,桶里没有沙子。找到完成的工作 :

by lifting the empty bucket;
::卸下空桶;

by lifting the sand alone;
::单凭沙子就抬起沙子;

by lifting the rope alone;
::单凭绳索举起绳索;

by the lifting the bucket, the sand, and the rope together.
::举起桶沙子绳子

The empty bucket . Since the bucket’s weight is constant, the worker must exert a force that is equal to the weight of the empty bucket. Thus
::空桶,由于桶重量不变,工人必须施加相当于空桶重量的力量。

$\begin{aligned} W_{1} & = F d \\ = (23) (+ 32) \\ W_{1} & = 736 J \end{aligned}$

::W1=Fd=(23)(+32)W1=736 J

The sand alone . The weight of the sand is decreasing at a constant rate from 80 N to 0 N over the 32 meter lift. When the bucket is at $\begin{aligned} x \end{aligned}$ meters above the floor, the sand weighs
::在32米的电梯上,沙的重量在以恒定速度从80牛升降至0牛顿,在32米的电梯上,沙的重量从80牛降为0牛顿。

$\begin{aligned} F (x) & = [Original weight of sand] [Proportion left at elevation x] \\ = 80 (\frac{32 - x}{32}) \\ F (x) & = [Original weight of sand] [Proportion left at elevation x] \\ = 80 (1 - \frac{x}{32}) \\ F (x) & = 80 - 2.5 x N \end{aligned}$

::F(x)=[沙子的原件重量][高纬度时留下的距离]=80(32-x3232)F(x)=[沙子的原件重量][高纬度时留下的距离]=80(1-x32)F(x)=80-2.5xN

The graph of $\begin{aligned} F (x) = 80 - 2.5 x N \end{aligned}$ represents the variation of the force with height $\begin{aligned} x \end{aligned}$ . The work done corresponds to computing the area under the force graph.
::F(x) = 80- 2.5 x N 的图形表示力与高x的变异。已完成的工作与在力图下计算面积相对应。

Thus the work done is
::因此,所完成的工作就是:

$\begin{aligned} W_{2} & = \int_{a}^{b} F (x) d x \\ = \int_{0}^{32} [80 - 2.5 x] d x \\ = {[80 x - \frac{2.5}{2} x^{2}]}_{0}^{32} \\ W_{2} & = 1280 J \end{aligned}$

::W2abF(x)dx*%032[80-2.5 x]dx=[80x-2.52x2x2,032W2=1280J

The rope alone . Since the weight of the rope is 5.1 N/m and the height is 32 meters, the total weight of the rope from the floor to a height of 32 meters is
$\begin{aligned} (5.1) (32) = 163.2 N . \end{aligned}$

::由于绳索的重量为5.1牛顿/米,高度为32米,从地板到32米高度的绳索的总重量为5.1牛顿/米(32)=163.2牛顿。

But since the worker is constantly pulling the rope, the rope’s length is decreasing at a constant rate and thus the rope’s weight is also decreasing as the bucket is being lifted. So at $\begin{aligned} x \end{aligned}$ meters, the length of the rope that remains to be lifted is $\begin{aligned} (32 - x) \end{aligned}$ meters which corresponds to remaining weight (force) of the rope to be lifted of $\begin{aligned} F (x) = (5.1) (32 - x) N \end{aligned}$ . Thus the work done to lift the weight of the rope is
::但是,由于工人不断拉绳子,绳子的长度正在以不变的速度递减,因此绳子的重量也随着桶的抬举而下降。因此,在x米时,有待吊起的绳子长度是(32-x)米,相当于F(x)=(5.1)(32-x)N所要抬起的绳子的剩余重量(力)。

$\begin{aligned} W_{3} & = \int_{a}^{b} F (x) d x \\ = \int_{0}^{32} 5.1 [32 - x] d x \\ = {[163.2 x - \frac{5.1}{2} x^{2}]}_{0}^{32} \\ W_{3} & = 2611.2 J \end{aligned}$

::W3abF(x)dx=0325.1[32-x]dx=[163.2 x-5.12x2]032W3=2611.2]

The bucket, the sand, and the rope together . Here we are asked to sum all the work done on the empty bucket, the sand, and the rope. Thus
::桶子,沙子,和绳子一起。在这里,我们被要求总结在空桶、沙子和绳子上完成的所有工作。

$\begin{aligned} W_{t o t a l} & = W_{1} + W_{2} + W_{3} \\ = 736 + 1280 + 2611.2 \\ = 4627.2 J \\ W_{t o t a l} & = 2611.2 J . \end{aligned}$

::W1+W2+W3=736+1280+2611.2=4627.2 JW总计=2611.2 J。

The total work done to lift the bucket is 2611.2 J.
::提水桶的总工作量为2611.2 J。

Review
::回顾

A particle moves along the $\begin{aligned} x \end{aligned}$ -axis by a force $\begin{aligned} F (x) = \frac{1}{x^{2} + 1} \end{aligned}$ . If the particle has already moved a distance of 10 meters from the origin, what is the work done by the force?
::如果粒子已经移动了距离原产地10米的距离,那么该粒子所完成的工作是什么?

A force of $\begin{aligned} c o s (\frac{π x}{2}) \end{aligned}$ acts on an object when it is $\begin{aligned} x \end{aligned}$ meters away from the origin. How much work is done by this force in moving the object from $\begin{aligned} x = 1 \end{aligned}$ to $\begin{aligned} x = 5 \end{aligned}$ meters?
::COs(xx2) 力在物体离原点有x米距离时作用。该力在将物体从 x=1 移动到 x=5米方面做了多少工作?

In physics, if the force on an object varies with distance then work done by the force is defined as $\begin{aligned} W = \int_{a}^{b} F (r) d r \end{aligned}$ . If a satellite of mass $\begin{aligned} m \end{aligned}$ is to be launched into space, then the force experienced by the satellite during and after launch is $\begin{aligned} F (r) = G \frac{m M}{r^{2}} \end{aligned}$ , where $\begin{aligned} M = 6 \times 10^{24} k g \end{aligned}$ is the mass of the Earth and $\begin{aligned} G = 6.67 \times 10^{- 11} \frac{N m^{2}}{k g^{2}} \end{aligned}$ is the Universal Gravitational Constant. If the mass of the satellite is 1000 kg and we wish to lift it to an altitude of 35,780 km above the Earth’s surface, how much work is needed to lift it? (Assume radius of Earth is 6370 km).
::在物理学方面,如果一个物体上的力因距离而异,那么由该力量完成的工作就定义为WabF(r)dr。如果将质量m卫星射入空间,那么卫星在发射期间和发射后所经历的力是F(r)=GmMMR2,其中M=6x1024千克是地球质量,而G=6.67×10-11 Nm2kg2是普遍重力常数。如果卫星质量为1000千克,我们希望将其升至地球表面35,780公里的高度,则需要做多少工作才能升升到地球表面? (地球半径为6370公里)。

Hooke’s Law states that when a spring is stretched $\begin{aligned} x \end{aligned}$ units beyond its natural length it pulls back with a force $\begin{aligned} F (x) = k x \end{aligned}$ , where $\begin{aligned} k \end{aligned}$ is called the spring constant or the stiffness constant. To calculate the work required to stretch the spring a length $\begin{aligned} x \end{aligned}$ we use $\begin{aligned} W = \int_{a}^{b} F (x) d x \end{aligned}$ , where $\begin{aligned} a \end{aligned}$ is the initial displacement of the spring $\begin{aligned} (a = 0) \end{aligned}$ if the spring is initially unstretched) and $\begin{aligned} b \end{aligned}$ is the final displacement. A force of 5 N is exerted on a spring and stretches it 1 meter beyond its natural length.

Find the spring constant $\begin{aligned} k \end{aligned}$
::查找春季常数 k

How much work is required to stretch the spring 1.8 m beyond its natural length?
::要将春季1.8米长超过其自然长度,需要做多少工作?

::Hooke 的法律规定,当弹簧被拉长x 单位超过其自然长度时,它会用F(x)=kx的功率拉回, k被称为弹簧常数或僵硬常数。要计算延展弹簧所需的工作长度x我们使用WabF(x)dx, 如果弹簧最初没有伸展, 弹簧的初始置换(a=0) 和双是最后的置换。 5 N 的力被施加在弹簧上, 将弹簧拉长超过其自然长度1米。找到弹簧常数 k 需要多少工作才能使弹簧超过其自然长度1.8 m ?

When a force of 30 N is applied to a spring, it stretches it from a length of 12 cm to 15 cm. How much work will be done in stretching the spring from 12 cm to 20 cm? (Hint: read the first part of problem #4 above.)
::当对春季施用30牛顿的力时,其长度从12厘米延伸至15厘米。将春季从12厘米延伸至20厘米需要做多少工作? (提示:请读上面问题4的第一部分。 )

An elevator of weight 600 lbs is supported by a cable 15 feet long whose weight is 18 pounds per linear ft. Find the work done in lifting the elevator 15 meters.
::重600磅的电梯由一条15英尺长的电缆支撑,其重量为每直线英尺18磅。

A worker on the second floor of a house that is being painted, lifts a can of paint (with a rope of negligible weight) a distance of 12 feet. The worker is a little careless, however, and the paint sloshes out at a constant rate as the can is raised. The can and paint weigh 6 lbs when the worker starts lifting, and 3 lbs once it gets to the scaffolding. How much work was done in lifting the can of paint?
::在一个被油漆的房子的二楼,一个工人拿起一罐油漆(绳子可忽略不计)12英尺的距离。然而,工人有点粗心,油漆在罐子抬高时会以恒定速度脱落。当工人开始举起时,罐子和油漆的重量为6磅,到达脚手架时为3磅。在提升油漆罐方面做了多少工作?

A tank is in the shape of a hemisphere with radius 5 feet. The tank contains a liquid that weights 30 lbs/cubic foot, and this liquid almost fills the tanks, to a height 1 foot from the top of the tank. You need to evacuate the tank, removing the liquid to a height 2 feet above the top of the tank. Find the work done in emptying the tank.
::罐体的形状是半径为5英尺的半球形。罐体中含有的液体重量为30磅/立方英尺,该液体几乎填满罐体,从罐体顶部到高度1英尺。您需要撤离罐体,将液体移到罐体顶部的高度2英尺。找到清除罐体的工作。

A tank has the shape of a right rectangular pyramid, with the base a square of length 4 feet, and the height 6 feet. The tank is filled to the top with oil weighing 30 lbs per cubic foot. Find the work done in pumping the oil to a position 1 foot above the top of the tank.
::罐体的形状是右长方形金字塔,底座为4英尺方形,高度为6英尺。罐体顶部装满油,每立方英尺重30磅。找到将油泵到罐体顶部1英尺处的工作。

A tank has the shape of an inverted right rectangular pyramid, with the top a square of length 6 feet, and the height 9 feet. The tank is only partially filled, to a position 3 feet from the top of the tank. If the tank is filled with a liquid weighing 40 lb per cubic foot, find the work done in pumping all of the oil to the rim of the tank.
::罐体的形状是倒向右方形的矩形金字塔,顶部为平方6英尺,高度为9英尺。罐体只有部分填充,从罐体顶部排到3英尺处。如果罐体装满液体,每立方英尺重40磅,请找到将所有石油泵到罐体边缘的工作。

A trough has a height of 4 feet, is 10 feet long, and has trapezoidal sides. The top base for the trapezoidal side is 5 feet, the bottom base is 3 feet, and the height is 4 feet. If the trough is filled to a height of 3 feet with a liquid of density 40 lbs per cubic foot, find the work done lifting all of the liquid to the top of the trough.
::槽口的高度为4英尺, 高度为10英尺, 并且有诱杀性边缘。诱杀性边缘的顶部底部是5英尺, 底部底部底部是3英尺, 高度是4英尺。如果槽口的高度是3英尺, 每立方英尺有密度为40磅的液体, 找到将所有液体抬到槽口顶部的工作。

A recent storm has filled the crawlspace beneath a house with a mud that needs to be removed. A worker gets down in the crawl space with a bucket (negligible weight) that has holes in it, and fills it with mud weighing 4 lbs. He lifts the bucket, but the mud pours out the holes in the bucket at a constant rate. If the worker lifts the mud to a height of 3 feet, but only 2 pounds of the stuff remains in the bucket, find the work done in lifting the mud.
::最近的暴风雨把一个房子下面的爬行空间填满了需要清除的泥土。工人在爬行空间倒下时有一个桶子(可忽略的重量),里面有洞,满满满了泥土。他抬起桶,但泥块以恒定的速度从桶里倒出洞。如果工人把泥土抬高到3英尺高,但只有2磅的东西留在桶里,找到在抬泥土时完成的工作。

A force of $\begin{aligned} x \sin (x^{2}) + 2 \end{aligned}$ is applied to a block, where $\begin{aligned} x \end{aligned}$ is the distance from the block’s starting position and is measured in meters. How much force is required to move the block ten meters?
::xsin(x2)+2的力对一个区块适用,其中x是区块起步位置的距离,以米计测量。移动区块10米需要多少力?

A force of $\begin{aligned} \cos (\sin^{2} (x)) \end{aligned}$ is applied to a block, where $\begin{aligned} x \end{aligned}$ is the distance from the block’s starting position and is measured in meters. How much work is required to move the block five meters?
::Cos(sin2(x))力应用于一个区块,该区块的距离是x与区块起始位置的距离,并以米计测量。移动区块5米需要多少工作?

A force of $\begin{aligned} 2 x^{3} + x^{2} - x + 1 \end{aligned}$ is applied to a block, where $\begin{aligned} x \end{aligned}$ is the distance from the block’s starting position and is measured in meters. How much work is required to move the block two meters?
::2x3+x2+x2-x+1的功率适用于一个区块,该区块的距离为x与区块起始位置的距离,并以米计测量。移动区块两米需要多少工作?

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

章节大纲

Work and Force ::工作与力量

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Review ::回顾

Review (Answers) ::回顾(答复)

Work and Force
::工作与力量

Examples
::实例

Example 1
::例1

Example 2
::例2

Review
::回顾

Review (Answers)
::回顾(答复)