📚 Work and Energy in GCSE Mathematics | GCSE 数学中的功和能量
In GCSE CCEA Mathematics, the topic of work and energy bridges physics and algebra. You will not be asked to recall scientific principles in isolation, but to apply straightforward formulas to calculate work done, kinetic energy and gravitational potential energy. Questions often involve unit conversions, substitution into equations and solving for an unknown variable. This revision guide covers every key formula you need to memorise, common problem types and strategies to avoid the usual pitfalls.
在 GCSE CCEA 数学课程中，功和能量是连接物理学与代数的一座桥梁。考试不会要求你单独背诵科学原理，而是需要运用简单的公式来计算功、动能和重力势能。题目常常涉及单位换算、代入公式以及求解未知量。这份复习指南涵盖了你需要牢记的每一个关键公式、常见题型以及避免常见错误的方法。
1. Understanding Work: The Product of Force and Distance | 理解功：力与距离的乘积
Work is done when a force moves an object in the direction of the force. In mathematical terms, if a constant force F acts on an object and causes it to move a distance d in the same direction as the force, then the work done W is given by the product of the magnitude of the force and the distance moved. Note that if the force is not parallel to the movement, only the component in the direction of movement is considered, but at GCSE CCEA you will typically be given parallel forces only.
当力使物体沿力的方向移动时，便做了功。从数学角度来说，如果一个恒定的力 F 作用在物体上，并使物体沿力的方向移动距离 d，那么所做的功 W 等于力的大小与移动距离的乘积。需要注意的是，如果力与运动方向不平行，则只考虑沿运动方向的分量，但在 CCEA 的 GCSE 考试中，通常只出现平行力的情况。
Work is a scalar quantity measured in joules (J). One joule is equal to one newton of force acting over one metre. The word ‘work’ in everyday language can mean effort, but in mathematics it always means the transfer of energy that results from a force acting over a distance. If there is no movement, no work is done, no matter how large the force.
功是一个标量，以焦耳（J）为单位。1 焦耳等于 1 牛顿的力作用 1 米的距离。日常语言中的“工作”可以指付出的努力，但在数学中，功始终指力作用于一段距离所引起的能量转移。如果没有移动，无论力有多大，都没有做功。
2. The Work Formula W = Fd and Units | 功的公式 W = Fd 与单位
The fundamental work equation you must know is W = F × d, where W is work done in joules (J), F is force in newtons (N), and d is distance moved in metres (m). Always check that distance is in metres and force is in newtons before substituting. If the question gives distance in centimetres or kilometres, convert to metres first. Similarly, mass may be given in grams or tonnes; convert to kilograms when needed for force calculations, as weight is a force.
你必须掌握的基本功公式是 W = F × d，其中 W 表示功（单位：焦耳 J），F 表示力（单位：牛顿 N），d 表示移动距离（单位：米 m）。在代入数值之前，一定要检查距离是否为米，力是否为牛顿。如果题目给出的距离是厘米或千米，请先换算成米。同样，质量可能以克或吨为单位；计算力时如果需要，请先换算成千克，因为重力本身就是力。
When the force is the weight of an object, you must first calculate the weight using W = mg, where m is mass in kg and g is acceleration due to gravity. On Earth, g is taken as 10 m/s² in GCSE CCEA exams unless otherwise stated. This is a simplification to make calculations manageable without a calculator. For example, a 5 kg mass has a weight of 5 × 10 = 50 N. If this weight is lifted vertically through 2 m, the work done against gravity is 50 × 2 = 100 J.
当力是物体的重力时，必须先使用 W = mg 计算重力，其中 m 是质量（kg），g 是重力加速度。在 GCSE CCEA 考试中，除非特别说明，地球上 g 取 10 m/s²。这一简化使计算无需计算器即可完成。例如，一个 5 kg 的物体重力为 5 × 10 = 50 N。如果将此物体竖直向上提升 2 m，克服重力所做的功为 50 × 2 = 100 J。
3. Work Done Against Gravity: Weight and Height | 克服重力做功：重力与高度
Lifting an object vertically means doing work against gravity. The force applied must equal the weight of the object if it is lifted at constant speed. The distance moved is the vertical height h. Therefore, the work done in lifting is W = mgh. This result is also the gain in gravitational potential energy, which we will look at in a later section. Typical questions ask: ‘A crate of mass 20 kg is lifted 3 m. Calculate the work done.’ Here, force = 20 × 10 = 200 N, distance = 3 m, so work = 200 × 3 = 600 J.
将物体竖直向上提升意味着克服重力做功。如果物体以恒定速度上升，所需的力等于物体的重力。移动的距离是竖直高度 h。因此，提升物体所做的功为 W = mgh。这个结果也是重力势能的增加量，我们将在后文讨论。典型的问题会问：“一个质量为 20 kg 的板条箱被提升了 3 m。计算所做的功。”这里，力 = 20 × 10 = 200 N，距离 = 3 m，所以功 = 200 × 3 = 600 J。
A common variant states that the object is pulled up a smooth slope. Even then, the work done against gravity is still mgh, because vertical height gained is the only distance relevant for the change in gravitational energy. If the slope length s is given and the vertical height is not, you may need to use trigonometry to find h = s sin θ, but only if the angle is known. At CCEA level, the height is usually stated directly or can be deducted from a diagram.
另一种常见变体是，物体被沿光滑斜面向上拉动。即便如此，克服重力所做的功依然是 mgh，因为只有获得的竖直高度才与重力势能的变化有关。如果给出了斜面长度 s 而未直接给出竖直高度，你可能需要利用三角关系 h = s sin θ 来求解，但这仅在已知角度时才需要。在 CCEA 阶段，高度通常会被直接给出，或可从示意图中推导出来。
4. Kinetic Energy: The Energy of Motion | 动能：运动的能量
Kinetic energy (E_k) is the energy an object possesses due to its motion. Any moving object has kinetic energy, and the amount depends on both its mass and the square of its speed. This quadratic relationship means that doubling the speed quadruples the kinetic energy, which is a favourite concept for exam questions. For instance, a car travelling at 15 m/s has far more kinetic energy than the same car at 10 m/s, and the braking distance will be much longer.
动能（E_k）是物体由于运动而具有的能量。任何运动的物体都有动能，大小取决于其质量和速度的平方。这种平方关系意味着速度增大为原来的两倍时，动能会增大为原来的四倍，这是考试中经常考察的概念。例如，一辆以 15 m/s 行驶的汽车比同一辆以 10 m/s 行驶的汽车具有大得多的动能，因此刹车距离也会长得多。
In mathematical problems, kinetic energy is often linked to work done. If a constant force accelerates an object from rest, the work done on the object equals its gain in kinetic energy, provided there is no friction. This is the work–energy principle, which allows you to connect Fd and ½mv². For example, a force of 200 N pushes a 50 kg cart over 10 m on a frictionless surface. The work done is 2000 J, so the final kinetic energy is 2000 J, and you can then find the final speed.
在数学问题中，动能常常与功联系在一起。如果没有摩擦力，一个恒力使物体从静止加速时，对物体所做的功等于其动能的增加量。这就是功-能原理，它将 Fd 和 ½mv² 联系起来了。例如，一个 200 N 的力在光滑表面上将一个 50 kg 的小车推动了 10 m。所做的功为 2000 J，因此小车的最终动能为 2000 J，进而可以求出最终速度。
5. The Kinetic Energy Formula E_k = ½mv² | 动能公式 E_k = ½mv²
The formula you must learn is E_k = ½ m v², where m is mass in kilograms (kg) and v is speed in metres per second (m/s). The kinetic energy is measured in joules (J). It is vital to square the speed before multiplying by half the mass. A common mistake is to compute ½ × m × v and then square the result, which is incorrect. Use brackets or do the square first: v², then multiply by m, then halve.
你必须掌握的公式是 E_k = ½ m v²，其中 m 是质量（千克 kg），v 是速度（米每秒 m/s）。动能的单位是焦耳（J）。计算时务必先将速度平方，再乘以质量的一半。一个常见错误是先计算 ½ × m × v 然后再平方，这是不正确的。建议使用括号或先算 v²，再乘以 m，最后除以 2。
Example: A car of mass 1200 kg travels at 15 m/s. Calculate its kinetic energy. v² = 225, so E_k = ½ × 1200 × 225 = 600 × 225 = 135,000 J. This can be written as 135 kJ. Notice that if the speed doubles to 30 m/s, v² = 900, and E_k = ½ × 1200 × 900 = 540,000 J, which is exactly four times the previous value. This demonstrates the squared relationship clearly.
示例：一辆质量为 1200 kg 的汽车以 15 m/s 的速度行驶。计算其动能。v² = 225，所以 E_k = ½ × 1200 × 225 = 600 × 225 = 135,000 J。这可以写作 135 kJ。请注意，若速度加倍至 30 m/s，则 v² = 900，E_k = ½ × 1200 × 900 = 540,000 J，恰好是之前数值的四倍。这清晰地展现了平方关系。
6. Gravitational Potential Energy: E_p = mgh | 重力势能：E_p = mgh
Gravitational potential energy (E_p) is the energy stored in an object due to its position above the ground. The formula is E_p = mgh, where m is mass (kg), g is gravitational field strength (10 N/kg) and h is height above a reference level (m). E_p is measured in joules. The reference level is usually the ground or the lab bench, but the question will make it clear. Because E_p depends linearly on height, lifting an object twice as high stores twice as much energy.
重力势能（E_p）是物体由于处于地面以上的位置而储存的能量。公式为 E_p = mgh，其中 m 是质量（kg），g 是重力场强度（10 N/kg），h 是相对于基准面的高度（m）。E_p 的单位是焦耳。基准面通常是地面或实验台，题意会明确说明。由于 E_p 随高度线性变化，将物体提升到两倍高度，储存的能量也变为两倍。
This formula is particularly powerful when combined with kinetic energy in conservation of energy problems. A freely falling object converts its gravitational potential energy into kinetic energy. Neglecting air resistance, the loss in E_p equals the gain in E_k. This yields mgh = ½mv², and by cancelling m, we get v² = 2gh, so v = √(2gh). This result allows you to find the speed of an object dropped from a height without knowing its mass.
将该公式与动能结合用于能量守恒问题时特别有用。一个自由下落的物体会将重力势能转化为动能。忽略空气阻力，则 E_p 的减少量等于 E_k 的增加量。由此得到 mgh = ½mv²，消去 m 可得 v² = 2gh，即 v = √(2gh)。利用这一结果，无需知道质量，就可以求出从某一高度落下的物体的速度。
7. Conservation of Energy and Work-Energy Principle | 能量守恒与功-能原理
In all CCEA problems, you may assume that energy is conserved unless friction or air resistance is mentioned. The total energy in a closed system remains constant, changing only from one form to another. The work–energy principle states that the net work done on an object is equal to its change in kinetic energy. For a car accelerating on a flat road, work done by the engine (minus any work against friction) equals the increase in ½mv².
在所有 CCEA 的问题中，除非题目提到了摩擦力或空气阻力，否则你都可以假定能量是守恒的。一个封闭系统中的总能量保持不变，只会从一种形式转化为另一种形式。功-能原理指出，对物体做的净功等于其动能的变化量。对于在平地上加速的汽车，发动机所做的功（减去克服摩擦力所做的功）等于 ½mv² 的增加量。
For a vehicle travelling up a slope, the work done by the engine is converted into both kinetic energy and gravitational potential energy. If the speed is constant, all work goes into gaining height. Always set up an energy equation: work input = change in E_k + change in E_p. That is the most structured method to solve complex problems. For example, a 2000 kg car drives at a steady 20 m/s up a hill gaining 50 m in height. Ignoring friction, the work done by the engine is mgh = 2000 × 10 × 50 = 1,000,000 J.
对于沿斜坡向上行驶的车辆，发动机所做的功同时转化为动能和重力势能。如果速度恒定，则所有功都用于增加高度。一定要建立能量方程：输入功 = 动能变化量 + 势能变化量。这是解决复杂问题最清晰的方法。例如，一辆 2000 kg 的汽车以恒定的 20 m/s 爬上一个山坡，高度增加了 50 m。忽略摩擦力，发动机所做的功为 mgh = 2000 × 10 × 50 = 1,000,000 J。
8. Solving Problems Involving Work, E_k and E_p | 解决涉及功、动能和势能的问题
Typical multi-step problems ask you to find a missing variable: force, distance, speed or height. The approach is: (1) identify what information is given and what is required; (2) convert all units to SI (metres, kilograms, seconds, newtons, joules); (3) write the relevant formula; (4) substitute carefully; (5) solve the equation. In energy conversion problems, equate mgh and ½mv², cancel m if it appears on both sides, and rearrange.
典型的多步问题会要求你找出一个缺失的变量：力、距离、速度或高度。解题方法是：(1) 明确已知信息和未知量；(2) 将所有单位换算成 SI 单位（米、千克、秒、牛顿、焦耳）；(3) 写出相关公式；(4) 仔细代入；(5) 解方程。在能量转化问题中，将 mgh 与 ½mv² 联立，如果两边都有 m 就把它消去，然后整理方程。
Example: A 4 kg ball is dropped from a height of 20 m. Find its speed just before impact. (g = 10 m/s²). Loss in E_p = mgh = 4 × 10 × 20 = 800 J. This equals gain in E_k: ½ × 4 × v² = 800. So 2v² = 800, v² = 400, v = 20 m/s. Note that the mass cancelled out. If the question instead gave the work done to accelerate the ball, you would start from W = ½mv². Be flexible with the starting point.
示例：一个 4 kg 的球从 20 m 高处落下。求它落地瞬间的速度（g = 10 m/s²）。E_p 减少量 = mgh = 4 × 10 × 20 = 800 J。这等于 E_k 的增加量：½ × 4 × v² = 800。因此 2v² = 800，v² = 400，v = 20 m/s。注意质量已被消去。如果题目改为给出使球加速的功，你就应从 W = ½mv² 入手。请灵活选择出发点。
9. Power: The Rate of Doing Work | 功率：做功的快慢
Power (P) is the rate at which work is done or energy is transferred. The formula is P = W / t, where W is work in joules and t is time in seconds. The unit of power is the watt (W), where 1 W = 1 J/s. You may also see power expressed in kilowatts (kW). A motor that does 6000 J of work in 20 seconds has a power output of 6000 / 20 = 300 W. If the same work were done in 5 seconds, the power would be 1200 W, showing that a more powerful motor does the same job faster.
功率（P）是做功或能量转化的速率。公式为 P = W / t，其中 W 是功（焦耳），t 是时间（秒）。功率的单位是瓦特（W），1 W = 1 J/s。你也可能会看到以千瓦（kW）表示功率。一台电动机在 20 秒内做 6000 J 的功，其输出功率为 6000 / 20 = 300 W。如果同样的功在 5 秒内完成，功率则会达到 1200 W，这说明功率更大的电动机可以更快地完成同一工作。
Another useful version arises when a constant force moves an object at constant speed. The power can be expressed as P = F × v, where F is force and v is velocity. This is because work Fd divided by time t gives F × (d/t) = Fv. For example, a car engine exerting a force of 500 N at a constant speed of 30 m/s delivers a power of 500 × 30 = 15,000 W = 15 kW. This relationship is valuable when force and velocity are known directly.
当恒力使物体以恒定速度移动时，还会用到另一个形式的公式。功率可表示为 P = F × v，其中 F 是力，v 是速度。这是因为功 Fd 除以时间 t 得到 F × (d/t) = Fv。例如，一辆汽车的发动机在 500 N 的驱动力下以 30 m/s 的恒定速度行驶，其输出功率为 500 × 30 = 15,000 W = 15 kW。当力与速度已经直接已知时，这一关系式非常有用。
10. Typical CCEA Exam-Style Questions | 典型的 CCEA 考试题型
Question 1: A box of mass 15 kg is lifted vertically through 2.5 m. Calculate the work done. Solution: Force = weight = 15 × 10 = 150 N. Work = 150 × 2.5 = 375 J. Question 2: A cyclist of total mass 80 kg speeds up from 5 m/s to 10 m/s. Find the increase in kinetic energy. Initial E_k = ½ × 80 × 25 = 1000 J; final E_k = ½ × 80 × 100 = 4000 J; increase = 3000 J. These two styles cover the direct application of formulas.
问题 1：一个质量为 15 kg 的箱子被竖直向上提升 2.5 m。计算所做的功。解：力 = 重力 = 15 × 10 = 150 N。功 = 150 × 2.5 = 375 J。问题 2：一个总质量为 80 kg 的自行车手速度从 5 m/s 提升到 10 m/s。求动能的增加量。初始 E_k = ½ × 80 × 25 = 1000 J；最终 E_k = ½ × 80 × 100 = 4000 J；增加量 = 3000 J。这两种题型涵盖了公式的直接应用。
A more integrated question: A parcel of mass 2 kg slides from rest down a smooth ramp of vertical height 1.8 m. Find its speed at the bottom. Energy conversion: mgh = ½mv² → v = √(2gh) = √(2 × 10 × 1.8) = √36 = 6 m/s. If friction were present, the work done against friction would reduce the kinetic energy; the energy lost to friction would be given, and the final speed would be lower. Expect questions that mix work and friction: total work – work against friction = kinetic energy gain.
一个综合性更强的问题：一个质量为 2 kg 的包裹从静止沿光滑斜面滑下，斜面竖直高度为 1.8 m。求它到达底端时的速度。能量转化：mgh = ½mv² → v = √(2gh) = √(2 × 10 × 1.8) = √36 = 6 m/s。如果存在摩擦力，克服摩擦力所做的功会减小动能；题目会给出因摩擦损失的能量，最终速度会相应变小。预计会出现功与摩擦力相结合的题目：总功 – 克服摩擦的功 = 动能增加量。
11. Common Mistakes and How to Avoid Them | 常见错误与避免方法
1. Unit confusion: Always convert to metres, kilograms and seconds. A distance given as 25 cm must become 0.25 m. Mass given in grams must be divided by 1000. Force given in kN must be multiplied by 1000. Write out the conversion explicitly before substituting into the formula. 2. Forgetting to square v: With E_k = ½mv², compute v² first, then multiply. It is wrong to do ½mv and square the result. Use a calculator sequence: v², then multiply by mass, then divide by 2. This sequence avoids errors.
1. 单位混淆： 务必换算为米、千克和秒。25 cm 的距离应转换为 0.25 m。以克为单位的质量必须除以 1000。以千牛（kN）为单位的力必须乘以 1000。在代入公式之前，请明确写出换算过程。2. 忘记对 v 平方： 对于 E_k = ½mv²，应先计算 v²，再相乘。绝不能先算 ½mv 后再平方。可以按以下顺序使用计算器：v²，然后乘以质量，再除以 2。这个顺序可以避免错误。
3. Misapplying the work formula: W = Fd only applies when force and displacement are in the same direction. If a person carries a weight horizontally, the distance moved horizontally is not used in calculating work against gravity, because the force of gravity is vertical. In that case, only vertical displacement matters. 4. Weight versus mass: Do not use mass in kilograms as the force unless you have multiplied by g first. Weight is a force, mass is not. Using m instead of mg will give answers that are a factor of 10 out – a costly mistake.
3. 误用功的公式： W = Fd 仅当力与位移方向相同时才适用。如果某人水平地提着重量，水平移动的距离不能用于计算克服重力所做的功，因为重力的方向是竖直的。此时只有竖直位移才作数。4. 重力与质量混淆： 不要直接用千克为单位的质量作为力，除非你已经先乘以了 g。重力是力，质量则不是。用 m 代替 mg 会导致答案相差 10 倍——这是一个代价高昂的错误。
12. Summary and Key Formulae | 总结与关键公式
You need to be fluent with these four equations and their units:
你需要熟练运用以下四个方程及其单位：
| Work done | W = Fd (J) | F in N, d in m |
| Kinetic energy | E_k = ½ m v² (J) | m in kg, v in m/s |
| Gravitational potential energy | E_p = m g h (J) | g = 10 N/kg, h in m |
| Power | P = W / t (W) | W in J, t in s |
Remember that energy problems often require you to link two formulas. The work–energy principle W = ΔE_k, and conservation of mechanical energy give you mgh = ½mv². These relationships allow you to find speed from height, or force from change in kinetic energy. Work systematically, show your substitutions clearly, and check that your final answer has the correct units.
请记住，能量问题通常需要你联系两个公式。功-能原理 W = ΔE_k 和机械能守恒 mgh = ½mv² 都可使用。这些关系式帮助你从高度求速度，或从动能变化求力。解题时要条理清晰，明确写出代入过程，并检查最终答案的单位是否正确。
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