Edexcel Physics: Work and Energy – Revision Essentials | 功与能量考点精讲

📚 Edexcel Physics: Work and Energy – Revision Essentials | 功与能量考点精讲

Work and energy are central concepts in Edexcel Physics, linking forces, motion, and the fundamental principle of conservation of energy. This article unpacks all the key definitions, equations, and typical exam scenarios, from calculating work done by a force to analysing efficiency and power. Each section pairs clear English explanations with precise Chinese translations, mirroring the bilingual approach that supports international learners. Let’s dive into the essentials that will help you master this topic and secure top marks.

功与能量是 Edexcel 物理的核心概念，将力、运动与能量守恒的基本原理联系在一起。本文深入剖析所有关键定义、方程和典型考题场景，从计算力所做的功到分析效率和功率。每个小节都配有清晰的英文解释和对应的中文译文，体现了支持国际学习者的双语教学方式。让我们一起掌握这些要点，帮助你在考试中取得高分。

1. Work Done by a Constant Force | 恒力做功

Work is done when a force causes an object to move in the direction of the force. For a constant force F acting on an object that undergoes a displacement s in the same direction as the force, the work done W is defined as W = F × s.

当力使物体沿力的方向发生位移时，力就做了功。如果一个恒力 F 作用在物体上，物体在该力方向上产生了位移 s，那么所做的功定义为 W = F × s。

If the force and displacement are not parallel, but act at an angle θ to each other, the work done is given by W = Fs cosθ.

如果力与位移不平行，而是互成角度 θ，那么所做的功由 W = Fs cosθ 给出。

Work is a scalar quantity with the unit joule (J); 1 J is the work done when a force of 1 N moves an object 1 m in the direction of the force.

功是标量，单位为焦耳 (J)；1 J 表示 1 N 的力使物体沿力的方向移动 1 m 所做的功。

When θ = 0°, cos0° = 1, giving maximum work W = Fs. When θ = 90°, cos90° = 0, meaning no work is done. A force applied perpendicular to the displacement, such as the normal reaction on a moving block, does zero work.

当 θ = 0° 时，cos0° = 1，做功最大 W = Fs。当 θ = 90° 时，cos90° = 0，表示不做功。垂直于位移的力，例如作用在运动物体上的法向反力，做功为零。

2. Work Done by a Varying Force | 变力做功

When the force is not constant, the work done cannot be calculated simply by multiplying force and displacement. Instead, the work done equals the area under a force–displacement graph.

当力不是恒力时，不能简单地用力乘以位移来计算功。此时，所做的功等于力–位移图线下的面积。

If the force varies linearly, as with a stretched spring obeying Hooke’s law (F = kx), the graph is a straight line through the origin, and the area underneath is a triangle: W = ½Fmax x or W = ½kx². This gives the work done in stretching or compressing the spring.

如果力呈线性变化，比如遵循胡克定律 (F = kx) 的拉伸弹簧，图线是一条通过原点的直线，下方的面积是一个三角形：W = ½Fmax x 或 W = ½kx²。这就是拉伸或压缩弹簧所做的功。

In general, you can estimate the area by counting squares or using trapezoidal integration. This concept is vital for understanding energy stored in elastic objects and for analysing varying forces in Edexcel Unit 1 and Unit 4.

一般来说，可以通过数格点或梯形积分来估算面积。这一概念对于理解弹性物体中储存的能量以及分析 Edexcel 单元1和单元4中的变力问题至关重要。

3. Kinetic Energy (KE) | 动能

Kinetic energy is the energy possessed by an object due to its motion. For an object of mass m moving at speed v, the kinetic energy is KE = ½mv².

动能是物体由于运动而具有的能量。对于质量为 m、速度为 v 的物体，动能是 KE = ½mv²。

Kinetic energy is a scalar, always positive, and measured in joules. The expression can be derived from the definition of work: a resultant force F acting over a distance s changes the speed from u to v, and the work done equals the change in KE, giving Fs = ½mv² – ½mu².

动能是标量，始终为正，单位是焦耳。这个公式可以从功的定义推导出来：合力 F 作用一段距离 s 使速度从 u 变为 v，所做的功等于动能的变化量，即 Fs = ½mv² – ½mu²。

In many exam questions, you must use KE calculations alongside momentum or potential energy. Remember that v is the instantaneous speed, and v² means the speed is squared; doubling the speed quadruples the kinetic energy.

在许多考题中，你需要结合动量或势能进行动能计算。记住 v 是瞬时速率，v² 表示速率平方；速率翻倍会使动能变为原来的四倍。

4. Gravitational Potential Energy (GPE) | 重力势能

Gravitational potential energy is the energy an object possesses due to its position in a gravitational field. Near the Earth’s surface, for an object of mass m raised through a vertical height h, the change in GPE is ΔGPE = mgh, where g is the gravitational field strength (9.81 N kg⁻¹ on Earth).

重力势能是物体因其在重力场中的位置而具有的能量。在地球表面附近，对于质量为 m 的物体，竖直升高 h，其重力势能的改变量为 ΔGPE = mgh，其中 g 是重力场强度（地球上为 9.81 N kg⁻¹）。

GPE is a relative quantity; you can choose any reference level as zero potential. Changes in GPE depend only on vertical height, not on the path taken. If an object falls from height h₁ to h₂, the decrease in GPE is mg(h₁ – h₂).

重力势能是一个相对量；你可以任意选择零势能参考面。重力势能的变化只取决于竖直高度，与路径无关。如果物体从高度 h₁ 落到 h₂，重力势能减少量为 mg(h₁ – h₂)。

In uniform gravitational fields, mgh is sufficient. For A-level, you may also encounter gravitational potential V = –GM/r for radial fields, but that appears in Unit 4/5 topics. Here we focus on mgh.

在均匀重力场中，使用 mgh 就足够了。在 A-level 阶段，你可能还会接触到径向场中的引力势 V = –GM/r，但这出现在单元4/5的内容中。此处我们重点讨论 mgh。

5. Elastic Potential Energy (EPE) | 弹性势能

When a spring or any elastic object is stretched or compressed, work is done against the restoring force, and energy is stored as elastic potential energy. For a spring that obeys Hooke’s law (F = kx), the elastic potential energy stored is EPE = ½kx², where k is the spring constant and x is the extension or compression.

当弹簧或任何弹性物体被拉伸或压缩时，克服回复力做功，能量以弹性势能的形式储存起来。对于遵循胡克定律 (F = kx) 的弹簧，储存的弹性势能为 EPE = ½kx²，其中 k 是弹簧劲度系数，x 是伸长量或压缩量。

This expression can be derived from the area under the force–extension graph, as the force increases linearly from 0 to F = kx. The area under the right-angled triangle gives ½ × base × height = ½ × x × kx = ½kx².

这个表达式可以从力–伸长量图线下的面积推导出来，因为力从 0 线性增加到 F = kx。直角三角形的面积是 ½ × 底 × 高 = ½ × x × kx = ½kx²。

Be careful to interpret k correctly. A stiffer spring has a larger k, so more work is needed to produce the same extension. Also, EPE is always positive, irrespective of whether x is extension or compression.

要注意正确理解 k 的含义。劲度系数越大的弹簧越硬，产生相同伸长量需要做的功更多。此外，无论 x 是伸长还是压缩，弹性势能总是正值。

6. The Principle of Conservation of Energy | 能量守恒原理

Energy cannot be created or destroyed; it can only be transferred from one form to another. In any isolated system, the total energy remains constant.

能量既不能凭空产生，也不能凭空消失，它只能从一种形式转化为另一种形式。在任何孤立系统中，总能量保持不变。

For example, as a pendulum swings, gravitational potential energy converts to kinetic energy and back again. In the presence of air resistance, some mechanical energy is transferred to thermal energy of the surroundings, but the total energy of the universe is unchanged.

例如，当单摆摆动时，重力势能转化为动能，再转化回来。在存在空气阻力的情况下，一部分机械能转化为周围环境的热能，但宇宙的总能量不变。

This principle is the foundation for solving many Edexcel problems. You often equate total initial energy (KE + GPE + EPE) to total final energy, accounting for any work done against friction or resistive forces, which is transferred to heat.

这一原理是解决许多 Edexcel 题目的基础。你通常会将初态的总能量（动能+重力势能+弹性势能）与末态的总能量等同起来，并计入克服摩擦或阻力所做的功，这部分能量转化为了内能。

7. The Work–Energy Theorem | 功能定理

The work–energy theorem states that the net work done on an object is equal to the change in its kinetic energy: W_net = ΔKE = ½mv² – ½mu².

功能定理指出，作用在物体上的净功等于其动能的变化量：W_net = ΔKE = ½mv² – ½mu²。

This theorem is valid even when potential energy changes are involved, provided you define the net work as the work done by all forces except conservative forces (like gravity or elastic forces), or you can directly apply energy conservation. In practice, for a block sliding down a slope with friction, the net work is the sum of work done by the component of weight along the slope and work done by friction; this net work equals the change in KE.

即使涉及势能变化，该定理也成立，只要将净功定义为除保守力（如重力或弹力）以外所有力所做的功，或者你也可以直接应用能量守恒。在实践中，对于一个沿斜面有摩擦地滑下的物块，净功是重力沿斜面分力所做的功和摩擦力所做的功的代数和；这个净功等于动能的变化量。

You can use the work–energy theorem to find the speed of an object without dealing with time and acceleration explicitly, which is especially useful in multi-step problems.

你可以利用功能定理求出物体的速度，而不必显式地处理时间和加速度，这在多步骤问题中特别有用。

8. Power | 功率

Power is the rate at which work is done or energy is transferred. The average power P = W/t or P = ΔE/t, where W is the work done (or energy transferred) in a time interval t. The unit of power is the watt (W), where 1 W = 1 J s⁻¹.

功率是做功或能量转移的速率。平均功率 P = W/t 或 P = ΔE/t，其中 W 是在时间间隔 t 内所做的功（或转移的能量）。功率的单位是瓦特 (W)，1 W = 1 J s⁻¹。

For a constant force acting on an object moving at constant speed v, the instantaneous power can also be written as P = Fv. This follows from P = W/t = (Fs)/t = Fv. If the force and velocity are not in the same direction, use P = Fv cosθ.

对于作用在匀速 v 运动的物体上的恒力，瞬时功率也可以写作 P = Fv。这由 P = W/t = (Fs)/t = Fv 得出。如果力和速度方向不同，则用 P = Fv cosθ。

A typical exam question involves a car engine providing a driving force to overcome resistive forces at a certain speed; the maximum power rating determines the maximum possible speed when the net driving force is balanced.

常见的考题涉及汽车发动机提供驱动力，以某一速度克服阻力；当净驱动力平衡时，额定最大功率决定了最大可能速度。

9. Efficiency | 效率

Efficiency is a measure of how much of the total energy input is converted into useful energy output. It is defined as efficiency = (useful energy output) / (total energy input), or equivalently, efficiency = (useful power output) / (total power input).

效率是衡量总输入能量中有多少转化为有用输出能量的量度。其定义为效率 = (有用输出能量) / (总输入能量)，或者等效地，效率 = (有用输出功率) / (总输入功率)。

Efficiency can be expressed as a decimal or a percentage (multiply by 100%). Since some energy is always transferred to unwanted forms (e.g. thermal energy due to friction), efficiency is always less than 1 (or less than 100%).

效率可以用小数或百分数（乘以100%）表示。由于总有部分能量转化为无用形式（例如因摩擦产生的热能），效率总是小于1（或小于100%）。

In Edexcel physics, you may need to calculate efficiency from Sankey diagrams or from given values of work done against friction and total work input. Always treat energy losses as the difference between input and useful output.

在 Edexcel 物理中，你可能需要根据桑基图或给出的克服摩擦做功和总输入功的数值来计算效率。始终将能量损耗视为输入与有用输出之间的差值。

10. Key Formulas Summary Table | 核心公式汇总表

Quantity \| 量	Equation \| 方程式	Symbol meaning \| 符号含义
Work (constant force) \| 恒力做功	W = Fs cosθ	F = force, s = displacement, θ = angle between F and s
Kinetic energy \| 动能	KE = ½mv²	m = mass, v = speed
Gravitational PE change \| 重力势能变化	ΔGPE = mgh	h = vertical height change, g = 9.81 N kg⁻¹
Elastic PE \| 弹性势能	EPE = ½kx²	k = spring constant, x = extension/compression
Work–energy theorem \| 功能定理	W_net = ΔKE	W_net = net work done by all forces (or by non-conservative forces depending on definition)
Power \| 功率	P = W/t ; P = Fv	v = velocity, t = time
Efficiency \| 效率	η = useful output / total input	η may be expressed as percentage

11. Common Exam Pitfalls and Tips | 常见考试陷阱与技巧

A frequent mistake is to confuse the angle θ in the work formula. Remember that θ is the angle between the force vector and the displacement vector, not the angle with the horizontal. Always draw a clear vector diagram.

一个常见错误是混淆功公式中的角度 θ。记住 θ 是力矢量与位移矢量之间的夹角，而不是与水平面的夹角。一定要画出清晰的矢量图。

When using conservation of energy, be meticulous about identifying the system and accounting for all energy transformations. If friction is present, include the work done against friction as energy transferred to thermal energy, which will reduce the mechanical energy.

在使用能量守恒时，要仔细确定系统并考虑所有能量转化。如果存在摩擦，要将克服摩擦做的功计入，这部分能量转化为内能，会减少机械能。

In power problems, distinguish between average power and instantaneous power. For a car accelerating with constant power, v is not constant, so P = Fv gives the instantaneous relationship, and F = P/v decreases as speed increases.

在功率问题中，要区分平均功率和瞬时功率。对于以恒定功率加速的汽车，v 不是常数，因此 P = Fv 给出瞬时关系，驱动力 F = P/v 随着速度增加而减小。

Verify that your answer has correct units. If you calculate work in N·m, that is joules. If you obtain a negative work, it means the force opposes the displacement (e.g., friction) or the component of force is in the opposite direction.

验证你的答案单位是否正确。如果你以 N·m 为单位计算功，那就是焦耳。如果得到负功，则表示力与位移方向相反（例如摩擦力），或力的分量方向相反。

12. Worked Example – A Block on an Incline | 例题 – 斜面上的物块

A 2.0 kg block slides from rest down a smooth incline of length 3.0 m, inclined at 30° to the horizontal. Find its speed at the bottom using energy considerations. (g = 9.81 m s⁻²)

一个 2.0 kg 的物块从静止开始沿光滑斜面滑下，斜面长 3.0 m，与水平面成 30° 角。用能量法求它到达底部的速度。（g = 9.81 m s⁻²）

The vertical height descended is h = 3.0 m × sin30° = 1.5 m. Loss in GPE = mgh = 2.0 × 9.81 × 1.5 = 29.43 J. Since the incline is smooth, no work is done against friction, so all GPE converts to KE. Hence, ½mv² = 29.43 J, so v = √(2 × 29.43 / 2.0) = √29.43 ≈ 5.42 m s⁻¹.

下降的竖直高度为 h = 3.0 m × sin30° = 1.5 m。减少的重力势能 = mgh = 2.0 × 9.81 × 1.5 = 29.43 J。由于斜面光滑，没有克服摩擦做功，所以全部重力势能转化为动能。因此，½mv² = 29.43 J，故 v = √(2 × 29.43 / 2.0) = √29.43 ≈ 5.42 m s⁻¹。

This simple example illustrates the power of the energy approach compared to using equations of motion. Always check if the inclined plane is smooth or rough – if rough, you would need to subtract work done against friction from the loss in GPE before calculating KE.

这个简单的例子展示了能量法相比运动学方程的便捷之处。始终检查斜面是光滑还是粗糙——如果是粗糙的，你需要在计算动能之前从减少的重力势能中减去克服摩擦做的功。

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Edexcel Physics: Work and Energy – Revision Essentials | 功与能量 考点精讲