Work and Energy | 功和能量考点精讲

📚 Work and Energy | 功和能量考点精讲

In A-Level OCR Mathematics (Mechanics), work and energy provide a powerful alternative to Newton’s laws for solving problems involving forces and motion. This topic connects the concept of work done by a force to changes in kinetic energy, potential energy, and the principle of conservation of mechanical energy. Mastery of these ideas is essential for tackling a wide range of mechanics questions, from simple inclined planes to systems involving springs and power.

在A-Level OCR数学（力学）中，功和能量是解决力和运动问题的有力工具，可以替代牛顿定律。这个主题将力所做的功与动能、势能的变化以及机械能守恒原理联系起来。掌握这些概念对于解决从简单斜面到涉及弹簧和功率的系统等各种力学问题至关重要。

1. Definition of Work Done | 功的定义

Work is done when a force moves its point of application in the direction of the force. For a constant force F (in newtons) acting on an object that moves a distance s (in metres) along the line of action of the force, the work done W is given by W = F s. If the force is not entirely parallel to the displacement, we use the component of the force in the direction of the displacement: W = F s cos θ, where θ is the angle between the force and the displacement vector.

当力使其作用点沿力的方向移动时，力就做了功。对于作用在物体上的恒力 F（以牛顿为单位），物体沿力的作用线移动距离 s（以米为单位），所做的功 W = F s。如果力不完全平行于位移，我们使用力在位移方向上的分量：W = F s cos θ，其中 θ 是力与位移矢量之间的夹角。

The unit of work is the joule (J). 1 joule is the work done when a force of 1 newton moves through a distance of 1 metre in the direction of the force. Work is a scalar quantity, and it can be positive (when the force component is in the direction of motion) or negative (when it opposes motion, e.g., friction).

功的单位是焦耳（J）。1焦耳是1牛顿的力使物体沿力的方向移动1米所做的功。功是标量，可以是正的（当力的分量与运动方向相同时）或负的（当力阻碍运动时，例如摩擦力）。

Situation	Formula for Work Done
Force parallel to displacement	W = F s
Force at angle θ to displacement	W = F s cos θ
Work done against gravity (lifting)	W = m g h
Work done by a variable force	Area under force–distance graph

2. Kinetic Energy | 动能

The kinetic energy (KE) of an object of mass m moving with speed v is given by KE = ½ m v². Like work, kinetic energy is measured in joules. It is a scalar quantity that depends only on the mass and the magnitude of velocity, not on the direction of motion.

质量为 m、速度为 v 的物体的动能（KE）由公式 KE = ½ m v² 给出。和功一样，动能的单位是焦耳。它是一个标量，只取决于质量和速度的大小，与运动方向无关。

The work–energy principle states that the net work done on an object by all forces is equal to the change in its kinetic energy: W_net = ΔKE = ½ m v² − ½ m u², where u is the initial speed and v is the final speed. This principle is derived from Newton’s second law and is especially useful when forces are not constant or when the exact path is complicated.

功能原理指出，作用在物体上所有力的净功等于其动能的变化：W_net = ΔKE = ½ m v² − ½ m u²，其中 u 是初速度，v 是末速度。该原理由牛顿第二定律导出，当力不恒定或路径复杂时尤为有用。

In problems, always remember to include all forces when calculating net work, including friction, tension, and components of weight. If friction is present, the work done against friction is usually negative and reduces the kinetic energy.

在解题时，计算净功时务必包括所有力，如摩擦力、张力和重力的分量。如果存在摩擦，克服摩擦所做的功通常为负，会减少动能。

3. Gravitational Potential Energy | 重力势能

Gravitational potential energy (GPE) is the energy an object possesses due to its position in a gravitational field. For an object of mass m raised through a vertical height h near the Earth’s surface, the change in GPE is ΔGPE = m g h, where g is the acceleration due to gravity (9.8 m s⁻², unless stated otherwise). We usually define a reference level where GPE = 0, often the starting point or the lowest point.

重力势能（GPE）是物体因其在引力场中的位置而具有的能量。对于在地球表面附近被举升垂直高度 h、质量为 m 的物体，重力势能的变化为 ΔGPE = m g h，其中 g 是重力加速度（9.8 m s⁻²，除非另有说明）。我们通常定义势能为零的参考水平面，通常是起点或最低点。

The work done against gravity when lifting an object is stored as gravitational potential energy (assuming no other forces do work). Conversely, when an object falls, its GPE decreases and is converted into kinetic energy or work against resistive forces.

将物体举升时克服重力所做的功储存为重力势能（假设没有其他力做功）。反之，当物体下落时，其重力势能减少，转化为动能或克服阻力做功。

Note that GPE depends only on vertical height, not on the path taken. This makes energy calculations very convenient for inclined planes or curved paths.

注意，重力势能仅取决于垂直高度，而与物体经过的路径无关。这使得能量计算用于斜面或曲线路径时非常方便。

4. Elastic Potential Energy | 弹性势能

When a spring or an elastic string is stretched or compressed, it stores elastic potential energy (EPE). For a spring obeying Hooke’s Law, the applied force is proportional to the extension (or compression) x: F = k x, where k is the spring constant (stiffness). The work done in stretching the spring from its natural length by an extension x is ½ k x², and this is the elastic potential energy stored: EPE = ½ k x².

当弹簧或弹性绳被拉伸或压缩时，它会储存弹性势能（EPE）。对于遵循胡克定律的弹簧，施加的力与伸长量（或压缩量）x 成正比：F = k x，其中 k 是弹簧的劲度系数。将弹簧从原长拉伸 x 所做的功为 ½ k x²，这就是储存的弹性势能：EPE = ½ k x²。

This formula assumes the spring is not stretched beyond its elastic limit and that the natural length is the position of zero EPE. The unit of EPE is also the joule. In OCR exam problems, you may encounter light elastic strings and springs, and you must use the energy stored correctly in conservation of energy calculations.

该公式假设弹簧没有超出弹性极限，并且原长位置为零弹性势能。EPE的单位也是焦耳。在OCR考试题中，你可能遇到轻质弹性绳和弹簧，必须正确使用储存的能量进行能量守恒计算。

For an elastic string stretched to length L (natural length L₀), extension x = L − L₀.
弹性绳拉伸至长度L（原长L₀），伸长量x = L − L₀。
Hooke’s Law: T = (λ x)/L₀, where λ is the modulus of elasticity, and k = λ/L₀.
胡克定律：T = (λ x)/L₀，其中λ为弹性模量，k = λ/L₀。

5. The Principle of Conservation of Mechanical Energy | 机械能守恒原理

If no external forces (such as friction or air resistance) do work on a system, the total mechanical energy (KE + GPE + EPE) remains constant. This is the principle of conservation of mechanical energy. It applies strictly only when all forces are conservative (gravity, elastic forces) and no energy is dissipated as heat or sound.

如果没有外力（如摩擦或空气阻力）对系统做功，则总机械能（KE + GPE + EPE）保持恒定。这就是机械能守恒原理。严格来说，它仅在所有力都是保守力（重力、弹力）且没有能量以热或声音形式散失时才成立。

Typical A-Level problems involve an object sliding down a smooth slope, a pendulum, or a mass on a spring. You write an equation equating initial total mechanical energy to final total mechanical energy, and solve for the unknown speed, height, or extension.

典型的A-Level问题包括物体沿光滑斜面下滑、单摆或弹簧上的质量块。你会写出初始总机械能等于最终总机械能的方程，并求解未知的速度、高度或伸长量。

Example: A particle of mass m is released from rest at the top of a smooth quarter-circle track of radius r. Using conservation of energy: m g r = ½ m v² at the bottom → v = √(2 g r). This is much simpler than using circular motion dynamics.

例子：质量为m的质点从半径为r的光滑四分之一圆弧轨道顶端从静止释放。利用能量守恒：在底部有 m g r = ½ m v² → v = √(2 g r)。这比使用圆周运动动力学简单得多。

6. Work Done by Variable Forces | 变力所做的功

When a force is not constant, the work done cannot be found simply by W = F s. Instead, if the force varies as a function of displacement, the work done is the area under the force–displacement graph. This is a key concept in Mechanics 2 module of OCR. In simple cases, the graph is a straight line (e.g., force with a spring where F = k x) and the area is a triangle or trapezium; for more complex functions, integration is used: W = ∫ F dx.

当力不恒定时，不能简单地用 W = F s 求功。相反，如果力随位移变化，所做的功就是力-位移图下的面积。这是OCR力学2模块中的一个关键概念。在简单情况下，图形是一条直线（例如，弹簧的力满足F = k x），面积是三角形或梯形；对于更复杂的函数，则使用积分：W = ∫ F dx。

In OCR exams, you might be given a force–displacement graph and asked to estimate the area using the trapezium rule or count squares, or you might be given a function for F(x) and required to integrate. Remember that the limits of integration correspond to the initial and final positions.

在OCR考试中，可能会给出一张力-位移图，要求你使用梯形法则或数方格来估算面积，或者给出一个力F(x)的函数并要求积分。请记住积分的上下限对应于初始和最终位置。

Force variation	Work done W
Constant F	F × distance
F = k x (spring)	½ k x² (for extension x from natural length)
General function F(x)	∫ F(x) dx from x_initial to x_final

7. Power | 功率

Power is the rate at which work is done or energy is transferred. The SI unit is the watt (W), where 1 W = 1 J s⁻¹. If a constant force F moves an object at a steady speed v in the direction of the force, the power developed is P = F v. This formula is derived from the definition of work: power = (work done)/time = (F × distance moved)/time = F v.

功率是做功或能量转移的速率。国际单位是瓦特（W），1 W = 1 J s⁻¹。如果一个恒力 F 使物体在力的方向上以恒定速度 v 运动，则产生的功率为 P = F v。该公式由功的定义导出：功率 = 功/时间 = (F × 移动距离)/时间 = F v。

If the force is not in the direction of motion, then P = F v cos θ. For a vehicle moving at speed v against resistances, the driving force F produced by the engine equals the total resistive force when speed is constant; then the power output is P = (resistive force) × v.

如果力不在运动方向上，则 P = F v cos θ。对于以速度v克服阻力行驶的车辆，当速度恒定时，发动机产生的驱动力 F 等于总阻力；那么输出功率为 P = (阻力) × v。

Power can also be expressed as the derivative of work with respect to time. In problems with varying speed, you may need to use P = F v to find the force at a given instant and then use Newton’s second law to find acceleration.

功率也可以表示为功对时间的导数。在变速问题中，你可能需要使用 P = F v 求出某一时刻的力，然后使用牛顿第二定律求加速度。

8. Work–Energy Principle with Non-Conservative Forces | 含非保守力的功能原理

When non-conservative forces such as friction or air resistance are present, mechanical energy is no longer conserved. The work done by these forces is equal to the change in the total mechanical energy. We write:

W_nc = ΔKE + ΔGPE + ΔEPE

where W_nc is the work done by non-conservative forces. Usually, W_nc is negative because friction opposes motion and removes energy from the system.

当存在非保守力（如摩擦力或空气阻力）时，机械能不再守恒。这些力所做的功等于总机械能的变化。我们写成：

W_nc = ΔKE + ΔGPE + ΔEPE

其中 W_nc 是非保守力所做的功。通常 W_nc 为负，因为摩擦阻碍运动并使系统能量减少。

In exam questions, you may be asked to find the work done against friction during a motion. Set up an energy equation comparing initial and final mechanical energies plus the work done against friction (which is positive when taken as “work against”).

在考试题中，可能会要求你求出运动过程中克服摩擦力所做的功。建立能量方程，将初始和最终的机械能与克服摩擦力所做的功相加（当视为“克服…的功”时取正值）。

Example: A sledge of mass 20 kg slides down a rough slope of length 10 m inclined at 30° to the horizontal. Initial speed is 0, final speed is 5 m s⁻¹. Find the work done against friction. Using energies: m g h = 20 × 9.8 × (10 sin30) = 980 J, final KE = ½ × 20 × 5² = 250 J. Work against friction = initial GPE − final KE = 980 − 250 = 730 J.

例子：一个质量为20 kg的雪橇沿粗糙斜面滑下，斜面长10 m，倾角30°。初速为0，末速为5 m s⁻¹。求克服摩擦力所做的功。利用能量：m g h = 20 × 9.8 × (10 sin30) = 980 J，末动能 = ½ × 20 × 5² = 250 J。克服摩擦力的功 = 初始重力势能 − 末动能 = 980 − 250 = 730 J。

9. Connected Particles and Energy | 连接体和能量

Energy methods can be used to solve problems involving two or more connected particles moving in a straight line, provided the string is inextensible and pulleys are smooth. The total work done by the external forces (weights, applied forces) minus work against resistances equals the change in total kinetic energy of the system. For each particle, consider its change in GPE and KE.

能量方法可用于解决涉及两个或多个沿直线运动的连接粒子的问题，前提是绳子不可伸长且滑轮光滑。外力（重力、施加的力）所做的总功减去克服阻力所做的功等于系统总动能的变化。对每个粒子，考虑其重力势能和动能的变化。

Typically, you write one energy equation for the whole system. For example, two masses m₁ and m₂ (m₁ > m₂) hanging over a smooth pulley: when m₁ descends a distance h, its loss of GPE is m₁ g h and m₂ gains m₂ g h. The net work by gravity is (m₁ − m₂) g h. This equals the increase in kinetic energy: ½ (m₁ + m₂) v². Thus v² = 2(m₁ − m₂) g h / (m₁ + m₂).

通常，你会为整个系统写一个能量方程。例如，两个质量 m₁ 和 m₂ (m₁ > m₂) 通过光滑滑轮悬挂：当 m₁ 下降距离 h 时，它损失的重力势能为 m₁ g h，而 m₂ 获得 m₂ g h。重力做的净功为 (m₁ − m₂) g h。这等于动能的增加：½ (m₁ + m₂) v²。因此 v² = 2(m₁ − m₂) g h / (m₁ + m₂)。

Be careful: the tension in the string does no net work on the system because it is an internal force and the string is inextensible. In energy equations for connected particles, tensions cancel out.

注意：绳子中的张力对系统不做净功，因为它是内力且绳子不可伸长。在连接体的能量方程中，张力相互抵消。

10. Using Energy to Find Maximum Height and Speed | 利用能量求最大高度和速度

Energy principles are extremely handy for finding maximum height reached by a projectile or an object on a spring. At the highest point, the vertical component of velocity is zero (for projectile motion) or the speed is zero (if the object comes momentarily to rest). Setting KE = 0 at the maximum height and equating initial energy to final GPE yields the maximum height directly, without needing equations of motion.

利用能量原理求抛射体或弹簧上的物体所能达到的最大高度非常方便。在最高点，速度的竖直分量为零（对于抛体运动）或速度为零（如果物体瞬间静止）。在最大高度处令 KE = 0，并将初始能量等于最终的重力势能，可直接得到最大高度，无需使用运动学方程。

For a particle on an elastic string bouncing, you can find the maximum height by equating initial elastic potential energy to gravitational potential energy gained, provided no energy is lost. Conceptual understanding: the initial EPE is converted entirely into GPE at the top of the motion.

对于在弹性绳上弹跳的粒子，如果无能量损失，可以通过将初始弹性势能等于获得的重力势能来求最大高度。概念理解：初始的弹性势能在运动最高点完全转化为重力势能。

Similarly, finding speed at a given position is straightforward: write total energy at initial position = total energy at the required position. This often bypasses having to consider acceleration and variable forces.

类似地，求给定位置的速度也很直接：写出初始位置的总能量等于所求位置的总能量。这通常会绕过考虑加速度和变力的问题。

11. Common Misconceptions and Exam Tips | 常见误区与考试技巧

One common mistake is confusing work done by a force with work done against a force. When you lift an object, the work done against gravity is mgh (positive), but the work done by gravity is −mgh. Always read the question carefully: “work done against friction” is positive, while “work done by friction” is negative in the energy balance.

一个常见误区是混淆力所做的功与克服力所做的功。当你举起物体时，克服重力做的功是 mgh（正），但重力做的功是 −mgh。务必仔细读题：“克服摩擦力做的功”是正的，而在能量平衡中“摩擦力做的功”是负的。

Another pitfall is using energy conservation when non-conservative forces are present. Always account for work done by friction or air resistance explicitly using the extended work–energy principle, not by assuming KE + GPE = constant.

另一个陷阱是在存在非保守力时使用能量守恒。务必通过扩展的功能原理明确考虑摩擦或空气阻力所做的功，而不是假设 KE + GPE = 常数。

Units: Ensure consistent use of SI units (metres, kilograms, seconds, newtons, joules, watts). If g is taken as 9.8, use that value unless the question specifies 10. Precision is important in OCR mark schemes.

单位：确保一致使用国际单位制（米、千克、秒、牛顿、焦耳、瓦特）。如果 g 取9.8，除非题目指定用10，否则就用9.8。OCR的评分标准对精度很重视。

Finally, always draw a clear diagram and define your reference level for GPE. Label initial and final positions, and write out the energy equation carefully symbolically before substituting numbers.

最后，一定要画出清晰的示意图并定义重力势能的参考水平。标出初始和最终位置，在代入数字之前先用符号仔细写出能量方程。

12. Worked Example: Combined Energy Problem | 综合能量问题例题

A 3 kg block is released from rest on a rough slope inclined at 25° to the horizontal. It slides 2 m down the slope, then compresses a spring (k = 500 N m⁻¹) placed at the bottom. The coefficient of friction is 0.2. Find the maximum compression of the spring.

一个3 kg的物块从与水平面成25°角的粗糙斜面上从静止释放。它沿斜面滑下2 m，然后压缩放在底部的弹簧（k = 500 N m⁻¹）。摩擦系数为0.2。求弹簧的最大压缩量。

Solution: Let the maximum compression be x metres. The total distance moved along the slope is (2 + x) m. The vertical height lost is (2 + x) sin25°. Work done against friction is μ R × distance, where R = m g cos25°, so work against friction = 0.2 × 3 × 9.8 × cos25° × (2 + x). Energy balance:

Loss in GPE = Gain in EPE + Work done against friction

3 × 9.8 × (2 + x) sin25° = ½ × 500 × x² + 0.2 × 3 × 9.8 cos25° × (2 + x)

Simplify and solve the quadratic for x. Reject negative root. (Numerical working yields x ≈ 0.16 m.)

解：设最大压缩量为 x 米。沿斜面运动的总距离为 (2 + x) m。下降的垂直高度为 (2 + x) sin25°。克服摩擦力做的功为 μ R × 距离，其中 R = m g cos25°，因此克服摩擦力的功 = 0.2 × 3 × 9.8 × cos25° × (2 + x)。能量平衡：

失去的重力势能 = 获得的弹性势能 + 克服摩擦做的功

3 × 9.8 × (2 + x) sin25° = ½ × 500 × x² + 0.2 × 3 × 9.8 cos25° × (2 + x)

化简并解关于 x 的二次方程。舍去负根。（数值计算得 x ≈ 0.16 m。）

This example illustrates combining GPE, EPE, and work against friction in one equation. It is typical of the more challenging OCR mechanics questions.

这个例子展示了如何在一个方程中结合重力势能、弹性势能和克服摩擦做的功。这是OCR力学中较难问题的典型。

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Work and Energy | 功和能量 考点精讲