IGCSE CCEA Maths: Work and Energy Revision | IGCSE CCEA 数学：功和能量考点精讲

📚 IGCSE CCEA Maths: Work and Energy Revision | IGCSE CCEA 数学：功和能量考点精讲

In IGCSE CCEA Mathematics, the topic of work and energy applies fundamental mathematical skills to real-world physical scenarios. This revision guide covers essential formulas, unit conversions, energy calculations, power, efficiency, and common problem-solving techniques. You will learn to set up equations, rearrange terms, and interpret graphs – all within a mathematical context that prepares you for both the exam and practical applications.

在 IGCSE CCEA 数学中，功与能量专题将基本数学技能应用到现实物理情境中。本考点精讲涵盖核心公式、单位换算、能量计算、功率、效率以及常见解题技巧。你将学习如何建立方程、移项变形并解读图像——所有这些内容都置于数学背景中，帮助你备战考试及实际应用。

1. Work: Definition and Basic Formula | 功：定义与基本公式

In mathematics, work is done when a constant force acts on an object and causes displacement in the direction of the force. The work done W is calculated as the product of the force F and the distance d moved in the direction of the force, provided the force is constant and motion is in a straight line.

在数学中，当一个恒力作用在物体上并使物体沿力的方向发生位移时，即做了功。所做的功 W 等于力 F 与沿力方向移动的距离 d 的乘积，前提是力恒定且运动沿直线方向。

W = F × d

This linear relationship allows direct proportionality problems: if the force doubles while distance remains the same, the work doubles. In CCEA exams, you may need to rearrange the formula to find an unknown force or distance.

这种线性关系支持正比例问题：若力加倍而距离不变，功也加倍。在 CCEA 考试中，你可能需要改写公式以求出未知力或距离。

Example: A constant force of 12 N pushes a box 5 m across a floor. The work done is W = 12 × 5 = 60 J.

示例：一个 12 N 的恒力推动箱子沿地面移动 5 m。所做的功为 W = 12 × 5 = 60 J。

2. Units of Work and Conversions | 功的单位与换算

The SI unit of work is the joule (J). One joule is defined as the work done when a force of 1 newton moves an object 1 metre in the direction of the force: 1 J = 1 N·m. In mathematical problems, you will often need to convert distances from centimetres to metres or from kilometres to metres before substituting into the formula.

功的国际单位是焦耳 (J)。1 焦耳定义为 1 牛顿的力使物体沿力的方向移动 1 米所做的功：1 J = 1 N·m。在数学问题中，经常需要先将距离从厘米换算为米或从千米换算为米，再代入公式。

Quantity (量)	SI Unit (国际单位)	Symbol (符号)
Force	newton	N
Distance	metre	m
Work	joule	J

Common conversions: 100 cm = 1 m, 1000 m = 1 km, so 1 cm = 0.01 m. Always check that your units are consistent to avoid calculation errors.

常见换算：100 cm = 1 m，1000 m = 1 km，故 1 cm = 0.01 m。始终检查单位是否一致，以免计算错误。

3. Work Done Against Gravity | 克服重力做的功

When an object is lifted vertically upwards at constant speed, the applied force must balance its weight. The work done against gravity depends on the mass m, gravitational field strength g, and the vertical height h. The formula used is:

当物体以恒定速度竖直向上提升时，施加的力必须与重力平衡。克服重力做的功取决于质量 m、引力场强度 g 和垂直高度 h。所用公式为：

W = mgh

Here g is approximately 9.8 m/s² or 10 m/s², depending on the exam question. The weight force is mg, and since the object is lifted through height h, the work done is weight × height. This is a direct application of W = F × d.

此处 g 约取 9.8 m/s² 或 10 m/s²，取决于考题。重力为 mg，因物体被提升高度 h，做的功等于重力乘以高度。这是 W = F × d 的直接应用。

For example, lifting a 2 kg mass through 3 m vertically (take g = 10 m/s²) gives W = 2 × 10 × 3 = 60 J. In CCEA problems, you may need to rearrange to find mass or height from known work.

例如，将 2 kg 物体竖直提升 3 m（取 g = 10 m/s²）得 W = 2 × 10 × 3 = 60 J。CCEA 题目中，可能需要通过已知功反求质量或高度。

4. Work Done on an Inclined Plane | 斜面上做的功

Mathematical questions often feature objects moving along an inclined plane. If a constant force is applied parallel to the plane, the distance moved along the plane is s. The work done is simply W = F × s. When the force is used to raise the object against gravity, the useful work output is still mgh, where h is the vertical height gained.

数学题目常出现物体沿斜面运动的情景。若恒力平行于斜面施加，沿斜面移动的距离为 s。所做的功就是 W = F × s。当该力用于提升物体克服重力时，有用功输出仍为 mgh，其中 h 是获得的垂直高度。

h = s × sin θ

Here θ is the angle between the incline and the horizontal. Thus the work done against gravity can also be expressed as W = mg × s × sin θ. This links trigonometry with energy calculations.

其中 θ 为斜面与水平面的夹角。因此克服重力做的功也可表示为 W = mg × s × sin θ。这将三角学与能量计算联系起来。

In CCEA Mathematics, you may be given the slope length and vertical rise without the angle, so you can use similar triangles or Pythagoras’ theorem to find the required values.

在 CCEA 数学中，可能给出斜面长度和垂直升高而不给角度，此时可用相似三角形或勾股定理求出所需数值。

5. Kinetic Energy | 动能

Kinetic energy is the energy an object possesses due to its motion. The kinetic energy Eₖ of an object of mass m moving at speed v is given by:

动能是物体因运动而具有的能量。质量为 m、速度为 v 的物体的动能 Eₖ 由下式给出：

Eₖ = ½ m v²

Notice the squared relationship: if speed doubles, kinetic energy quadruples. This is a non-linear relationship frequently tested in proportionality questions. You must be able to substitute correctly and solve for v or m.

注意平方关系：若速度加倍，动能变为原来的四倍。这是一种非线性关系，常在比例问题中考查。考生需能正确代入并求解 v 或 m。

Example: A car of mass 800 kg is travelling at 15 m/s. Its kinetic energy is Eₖ = ½ × 800 × (15)² = ½ × 800 × 225 = 90,000 J. The answer may be expressed in standard form: 9.0 × 10⁴ J.

示例：一辆 800 kg 的汽车以 15 m/s 行驶。其动能为 Eₖ = ½ × 800 × (15)² = ½ × 800 × 225 = 90,000 J。答案可用科学记数法表示为 9.0 × 10⁴ J。

6. Gravitational Potential Energy | 重力势能

Gravitational potential energy Eₚ is the energy stored in an object due to its position above the ground. The formula is identical in structure to work done against gravity:

重力势能 Eₚ 是物体因位于地面以上而储存的能量。公式结构与克服重力做的功相同：

Eₚ = mgh

In energy conversion problems, a common scenario is an object falling from a height: the loss in Eₚ equals the gain in Eₖ, assuming no air resistance. This gives the equation mgh = ½ mv², which simplifies to v = √(2gh).

在能量转换问题中，常见情景是从高处下落的物体：假设无空气阻力，重力势能的减少等于动能的增加。由此得方程 mgh = ½ mv²，化简后 v = √(2gh)。

Cancelling mass m shows that the final speed depends only on the height and g, not on mass. Such algebraic manipulation is a key skill in CCEA Mathematics.

约去质量 m 表明，末速度只取决于高度与 g，而与质量无关。这种代数变形是 CCEA 数学的关键技能。

7. Conservation of Energy and Work-Energy Principle | 能量守恒与功能原理

The principle of conservation of energy states that energy cannot be created or destroyed, only transferred from one form to another. In a closed system with no external work done, total mechanical energy (kinetic + potential) remains constant.

能量守恒原理指出，能量不能被创造或消灭，只能从一种形式转化为另一种形式。在没有外力做功的封闭系统中，总机械能（动能 + 势能）保持不变。

The work-energy principle is particularly useful: the net work done on an object equals its change in kinetic energy.

功能原理尤其有用：合力对物体做的功等于其动能的变化量。

W_net = ΔEₖ = ½ m v² – ½ m u²

Here u is initial speed and v is final speed. This principle can be applied even when non-conservative forces like friction are present, because the net work includes the work done by these forces.

其中 u 为初速度，v 为末速度。该原理即使存在摩擦力等非保守力时也能应用，因为合功包含了这些力所做的功。

For example, if a braking force does negative work on a car, the kinetic energy decreases. You can set up an equation to find the braking distance.

例如，若刹车力对汽车做负功，动能减少。可建立方程求出刹车距离。

8. Power: Rate of Doing Work | 功率：做功的快慢

Power is the rate at which work is done or energy is transferred. In mathematics, power P is calculated as work done W divided by the time t taken.

功率是做功或能量转移的速率。在数学中，功率 P 等于做的功 W 除以所用时间 t。

P = W / t

The unit of power is the watt (W), where 1 W = 1 J/s. Larger units like kilowatt (kW) are often used; 1 kW = 1000 W. You will need to convert between these units in calculations.

功率单位是瓦特 (W)，1 W = 1 J/s。更大单位如千瓦 (kW) 经常使用；1 kW = 1000 W。计算中需在这些单位之间转换。

Another useful form arises when a constant force moves an object at constant speed v: P = F × v. This is derived from P = (F × d) / t = F × (d / t) = F × v.

另一个实用形式是当恒力以恒定速度 v 移动物体时：P = F × v。推导自 P = (F × d) / t = F × (d / t) = F × v。

Example: A motor lifts a 50 kg mass vertically at 2 m/s. The force required equals the weight, 50 × 10 = 500 N (using g = 10 m/s²). The power output is P = 500 × 2 = 1000 W = 1 kW.

示例：电动机以 2 m/s 竖直提升 50 kg 物体。所需力等于重力，50 × 10 = 500 N（取 g = 10 m/s²）。输出功率为 P = 500 × 2 = 1000 W = 1 kW。

9. Efficiency | 效率

Efficiency measures how much of the input energy or work is converted into useful output. It is expressed as a percentage, so efficiency makes heavy use of ratio and proportion concepts in mathematics.

效率衡量输入能量或功中有多少转化为有用的输出。它用百分数表示，因此效率在数学中大量运用比和比例概念。

Efficiency = (Useful output work / Total input work) × 100%

Energy cannot be destroyed, but some input is always wasted, usually as heat due to friction. In exam questions, you might be given the total energy input and the useful work done, and asked to find the efficiency or the energy wasted.

能量无法被消灭，但总有一部分输入被浪费，通常因摩擦以热的形式散失。考题中，可能给出总输入能量和有用功，要求计算效率或浪费的能量。

Example: A machine receives 500 J of energy and does 350 J of useful work. Efficiency = (350 / 500) × 100% = 70%. The wasted energy is 500 – 350 = 150 J.

示例：一台机器接收 500 J 能量，做 350 J 有用功。效率 = (350 / 500) × 100% = 70%。浪费的能量为 500 – 350 = 150 J。

10. Graphical Analysis and Problem-Solving Strategies | 图形分析与解题策略

In CCEA Mathematics, force–distance graphs provide a visual method for calculating work. The work done by a varying force can be found as the area under the force–distance graph. For a constant force, this area is simply a rectangle; for a force that changes linearly, the area is a triangle or trapezium.

在 CCEA 数学中，力-距离图提供了一种计算功的直观方法。变力做的功可通过力-距离图下的面积求得。对于恒力，该面积就是一个矩形；对于线性变化的力，面积为三角形或梯形。

Work done = Area under F–d graph

Similarly, power–time graphs can be used to find total energy transferred, where energy = area under P–t graph. These graphical problems test your ability to apply geometric area formulas in a physical context.

类似地，功率-时间图可用于求总传递能量，即能量 = P–t 图下的面积。这类图形问题考查你在物理情境中应用几何面积公式的能力。

Problem-solving tips: always identify the known quantities and the required unknown, choose the appropriate formula, check that units are consistent, and where multiple steps are involved, consider using the conservation of energy or work-energy principle to link stages.

解题技巧：始终明确已知量和所求未知量，选择合适的公式，检查单位一致，若涉及多个步骤，可考虑用能量守恒或功能原理将各阶段联系起来。

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IGCSE CCEA Maths: Work and Energy Revision | IGCSE CCEA 数学：功和能量 考点精讲