📚 GCSE WJEC Maths: Work and Energy Exam Focus | GCSE WJEC 数学：功和能量 考点精讲
In WJEC GCSE Mathematics you will regularly meet questions built around scientific formulas such as work done, gravitational potential energy and kinetic energy. These are not science tests: the context comes from physics, but the marks come from maths - substituting into a formula, rearranging it, converting units and interpreting a graph. This revision guide walks you through every mathematical technique you need so that you can tackle work and energy problems confidently in the exam hall.
在 WJEC GCSE 数学考试中，你会经常遇到以科学公式为背景的题目，例如做功、重力势能和动能。这并不是科学测试：情境来自物理，但分数来自数学——代入公式、变形公式、换算单位以及解读图像。本复习指南将带你逐一掌握所有必需的数学技巧，让你在考场上自信应对功和能量类问题。
1. Why Work and Energy Appear in a Maths Paper | 为什么功和能量会出现在数学卷中
WJEC uses real-life and scientific contexts to test pure mathematical skills. A question about lifting a box is really testing substitution, rounding and unit reasoning. The examiner is not checking whether you remember a physics lesson; the examiner is checking whether you can handle a formula accurately.
WJEC 常用真实生活与科学情境来考查纯数学能力。一道关于"搬起箱子"的题目，实际考查的是代入、取整和单位推理。考官并不是在检查你是否记得物理课内容，而是在检查你能否准确处理一个公式。
Most of these questions sit in the algebra, ratio and number strands, and they are usually worth 2 to 4 marks each. Method marks are available for a correct substitution or a correct rearrangement, even if the final arithmetic slips, so always show your working line by line.
这类题目大多属于代数、比与比例以及数的板块，通常每题 2 到 4 分。即使最后算错，正确的代入或正确的变形也能拿到方法分，所以一定要逐行写出过程。
- Substitution into a given formula - 将数值代入给定公式
- Rearranging a formula to make another letter the subject - 变形公式，换主项
- Converting between units (cm, m, km, g, kg) - 单位换算
- Reading values and gradients from a graph - 从图像中读数与求斜率
- Rounding answers to a required degree of accuracy - 按要求精度取整
2. Understanding Work Done | 理解做功
Work done is the energy transferred when a force moves an object through a distance in the direction of that force. The mathematical relationship is short and simple, which makes it a favourite for substitution questions.
做功是指力使物体沿力的方向移动一段距离时所传递的能量。它的数学关系式简短明了，因此非常受代入类题目青睐。
W = F × d
Here W is the work done measured in joules (J), F is the force measured in newtons (N), and d is the distance moved along the line of the force, measured in metres (m). Because 1 joule is defined as 1 newton-metre, multiplying a force in newtons by a distance in metres always gives an answer in joules.
其中 W 是做功，单位为焦耳 (J)；F 是力，单位为牛顿 (N)；d 是沿力的方向移动的距离，单位为米 (m)。因为 1 焦耳定义为 1 牛顿·米，所以用牛顿表示的力乘以用米表示的距离，结果一定是焦耳。
lso run backwards: if the work done is 84 J over 3.5 m, the force must be 84 ÷ 3.5 = 24 N.
示例：用 24 N 的力把箱子沿地面推动 3.5 m。W = 24 × 3.5 = 84 J。注意题目也可以反过来问：如果移动 3.5 m 做功 84 J，那么力一定是 84 ÷ 3.5 = 24 N。
3. Units and Conversions | 单位与换算
Unit errors are the single most common reason for losing marks in work and energy questions. The formula only works if the force is in newtons and the distance is in metres, so convert first, substitute second.
单位错误是功和能量类题目中最常见的失分原因。公式只有在力用牛顿、距离用米时才成立，所以先换算、后代入。
To convert centimetres to metres, divide by 100. To convert kilometres to metres, multiply by 1000. To convert grams to kilograms, divide by 1000. To convert minutes to seconds, multiply by 60. Writing the conversion as a separate line makes your method clear and protects your method mark.
厘米换米要除以 100；千米换米要乘以 1000；克换千克要除以 1000；分钟换秒要乘以 60。把换算单独写一行，能让方法更清晰，也能保住方法分。
| Quantity 量 | Symbol 符号 | Unit 单位 | Watch out 注意 |
| Force 力 | F | newton (N) | Not kg |
| Distance 距离 | d | metre (m) | cm must be ÷ 100 |
| Work done 做功 | W | joule (J) | 1 J = 1 N m |
| Mass 质量 | m | kilogram (kg) | g must be ÷ 1000 |
| Gravity 重力场强度 | g | N/kg | Use the value given, often 9.8 |
| Speed 速度 | v | m/s | km/h must be ÷ 3.6 |
4. Rearranging the Work Formula | 做功公式变形
duct, the two rearrangements follow immediately: divide by d to find the force, or divide by F to find the distance.
公式变形是这类题目的核心。因为 W = F × d 只是一个简单的乘积，两种变形立刻可得：除以 d 求力，或除以 F 求距离。
F = W ÷ d d = W ÷ F
0 × 5 = 100. Testing with simple values catches most slips before you commit to an answer.
检验变形是否正确的可靠方法，是代入简单数字。若 W = 100、d = 5，则 F 应为 20，因为 20 × 5 = 100。用简单数值检验，能在写出答案前抓住大多数疏漏。
and the fraction form is often easier to see when you need to cancel units or estimate a size.
你还需要熟练掌握把除法写成分数。写成 F = W / d 与上面完全等价，而当你需要约掉单位或估算大小时，分数形式往往更直观。
5. Gravitational Potential Energy | 重力势能
When an object is lifted, work is done against gravity, and the energy stored is called gravitational potential energy. In WJEC mathematics the formula is given to you on the page, so the skill being tested is correct substitution and careful use of the vertical height.
当物体被举高时，需要克服重力做功，储存的能量称为重力势能。在 WJEC 数学卷中该公式会直接给出，因此考查的技能是正确代入和谨慎使用垂直高度。
Ep = m × g × h
he energy is again in joules. A mass of 5 kg lifted through 12 m with g = 9.8 N/kg gives Ep = 5 × 9.8 × 12 = 588 J.
其中 m 是质量，单位千克；g 是重力场强度，单位牛顿每千克；h 是上升的垂直高度，单位米。能量同样以焦耳为单位。质量 5 kg 提升 12 m，取 g = 9.8 N/kg，则 Ep = 5 × 9.8 × 12 = 588 J。
mg × h. The two formulas are the same idea written in two ways, and explaining that link can earn a reasoning mark.
注意它与做功的联系。物体的重力为 mg，把它举高 h 意味着力 mg 作用了一段距离 h，所以 W = F × d = mg × h。两个公式其实是同一个思想的不同写法，能说明这种联系往往可以拿到推理分。
If a question gives a sloping ramp instead of a vertical lift, use the vertical height, not the length of the slope, unless the question explicitly asks about the force along the slope.
如果题目给的是斜面而不是竖直提升，要用垂直高度，而不是斜面长度，除非题目明确要求求沿斜面的力。
6. Kinetic Energy | 动能
Kinetic energy is the energy an object has because of its motion. The formula contains a square, which makes it richer for examination questions, including questions where the speed doubles and the energy changes by a factor of four.
动能是物体由于运动而具有的能量。公式中含有平方，因此更适合命题，包括"速度翻倍、能量变为四倍"这类问题。
Ek = ½ × m × v²
Here m is the mass in kilograms and v is the speed in metres per second, so the energy comes out in joules. A 0.4 kg ball moving at 15 m/s has Ek = ½ × 0.4 × 15² = ½ × 0.4 × 225 = 45 J.
其中 m 是质量，单位千克；v 是速度，单位米每秒，能量结果为焦耳。0.4 kg 的球以 15 m/s 运动时，Ek = ½ × 0.4 × 15² = ½ × 0.4 × 225 = 45 J。
Two order-of-operations points matter here. Square the speed before multiplying, and apply the factor of one half before or after the multiplication - but never forget it. Since v² is always positive, kinetic energy can never be negative, however the direction of travel is described.
这里有两个运算顺序要点。先平方速度再相乘，而二分之一因子在乘法之前或之后应用都可以，但绝不能漏掉。由于 v² 恒为正，无论运动方向如何描述，动能都不可能为负。
7. The Work-Energy Principle | 功与能原理
Work done and energy change are interchangeable quantities: the work done on an object equals its change in energy. In a mathematics paper this appears as an equation to solve rather than as an abstract statement.
做功与能量变化是可以互换的量：对物体做的功等于它能量的变化。在数学卷中，这通常表现为一道需要求解的方程，而不是一句抽象陈述。
W = ΔE F × d = ½ m v² − ½ m u²
Here u is the initial speed and v is the final speed. For example, a 600 kg car accelerating from 8 m/s to 20 m/s gains Ek 20 000 − 19 200 = 100 800 J. That 100 800 J is exactly the work done by the driving force.
其中 u 是初速度，v 是末速度。例如，600 kg 的汽车从 8 m/s 加速到 20 m/s，动能增加 Ek = ½ × 600 × 20² − ½ × 600 × 8² = 120 000 − 19 200 = 100 800 J。这 100 800 J 正是驱动力所做的功。
A useful shortcut is to factorise: ½m(v² − u²). It reduces the chance of an arithmetic slip on a non-calculator paper and keeps the numbers smaller for longer, which helps with rounding at the end.
一个实用技巧是提取公因式：½m(v² − u²)。这能降低非计算器试卷上算错的风险，并让数字在更长时间内保持较小，有助于最后的取整。
Questions on efficiency also appear: useful energy output divided by total energy input, expressed as a decimal or a percentage. Typical values here are 0.35 or 35%, and the arithmetic is simple division once the energy values are correct.
效率类题目也会出现：有用能量输出除以总能量输入，用小数或百分数表示。常见结果是 0.35 或 35%，只要能量数值正确，运算就是简单的除法。
8. Worked Calculations and Substitution Skills | 计算示范与代入技巧
A disciplined four-step method will keep you accurate on every work and energy question: write the formula, convert the units, substitute with brackets, then evaluate and round.
一套规范的四步法能让你在每道功和能量题上都保持准确：写公式、换单位、带括号代入、计算并取整。
one by the pushing force and the gravitational potential energy gained, with g = 9.8 N/kg.
计算示例：质量为 30 kg 的箱子被 250 N 的力沿斜面推上 4 m，同时竖直上升 1.2 m。取 g = 9.8 N/kg，求推力所做的功以及获得的重力势能。
- Work done by the push: W = 250 × 4 = 1000 J - 推力做功：W = 250 × 4 = 1000 J
- Potential energy gained: Ep = 30 × 9.8 × 1.2 = 352.8 J - 获得的重力势能：Ep = 30 × 9.8 × 1.2 = 352.8 J
- The difference, 1000 − 352.8 = 647.2 J, is energy transferred to friction and heating - 差值 1000 − 352.8 = 647.2 J 是传递给摩擦与发热的能量
Notice that the question asked for two separate quantities using two different distances: 4 m along the slope for the work done, and 1.2 m of vertical rise for the potential energy. Reading which distance belongs to which formula is a genuine exam skill.
注意题目用两个不同距离求两个不同的量：沿斜面的 4 m 用于做功，竖直上升的 1.2 m 用于势能。判断哪个距离属于哪个公式，是一项真正的考试技能。
9. Rounding, Significant Figures and Calculator Skills | 取整、有效数字与计算器
Never round in the middle of a calculation. Carry full accuracy through your working and round only at the final line, otherwise small errors grow into answer-mark losses.
千万不要在计算过程中取整。全程保持完整精度，只在最后一行取整，否则小误差会累积成整分丢失。
The instruction "give your answer to 3 significant figures" means three digits counted from the first non-zero digit. So 100 800 J becomes 101 000 J at 3 s.f., and 352.8 J becomes 353 J. If no rounding instruction is given, an exact answer or a sensibly rounded one is accepted.
题目要求"保留 3 位有效数字"，指的是从第一个非零数字起数三位。因此 100 800 J 取 3 位有效数字为 101 000 J，352.8 J 则为 353 J。若题目未给取整要求，精确值或合理取整均可接受。
round the squared terms, or use the fraction key, so the calculator squares only the speeds and not the whole expression. Use brackets around every numerator and denominator in a rearranged formula.
计算器操作同样重要。输入 ½ × 600 × (20² − 8²) 时要在平方项外加括号，或使用分数键，确保计算器只对速度平方，而不是对整个式子平方。在变形公式中，分子和分母都要加括号。
Finally, check the units of your answer against the question. If the question asks for kilojoules, divide the joule value by 1000; 86 400 J becomes 86.4 kJ.
最后，把答案单位与题目要求对照。若题目要求千焦，就把焦耳值除以 1000；86 400 J 即 86.4 kJ。
10. Reading Force-Distance and Energy Graphs | 解读力-距离图与能量图
Graph questions turn a formula into a shape. On a force-distance graph the work done is the area under the line, and on an energy-height graph the straight-line gradient is mg.
图像题把公式变成了图形。在力-距离图上，做功等于图线下的面积；在能量-高度图上，直线斜率等于 mg。
30 × 4 = 60 J. For a trapezium-shaped region, W = ½ × (a + b) × h, where a and b are the two parallel force values.
力恒定时，面积是矩形，W = F × d 与前面一致。若力在 4 m 内从 0 均匀增加到 30 N，面积是三角形：W = ½ × 30 × 4 = 60 J。若区域是梯形，则 W = ½ × (a + b) × h，其中 a 与 b 是两条平行的力值。
ight. The gradient equals mg, so a gradient of 49 means mg = 49 and, if g = 9.8 N/kg, the mass is 5 kg.
在重力势能对高度的图像上，图线是过原点的直线，因为能量与高度成正比。斜率等于 mg，若斜率为 49，则 mg = 49；若取 g = 9.8 N/kg，质量就是 5 kg。
Always state the coordinate values you read from the axes before using them, and check whether the axes are in centimetres, grams or kilojoules - graph scales are a
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