📚 KS3 CIE Physics: Hooke's Law Problems and Worked Examples | KS3 CIE 物理：胡克定律问题与例题精讲
In KS3 physics, Hooke' meaning of the spring constant, the difference between extension and total length, and the full range of question types you may meet in CIE-style tests, with worked examples, graph reading skills and practice questions with answers.
在 KS3 物理中，胡克定律描述了弹簧或其他弹性物体在受力时如何伸长。本文解释这一定律本身、公式 F = kx、弹簧常数的含义、伸长量与总长度的区别，以及你在 CIE 风格测试中可能遇到的各类题型，并配有例题、图像分析技巧和带答案的练习题。
1. What Is Hooke's Law? | 什么是胡克定律？
Hooke's Law describes the behaviour of springs and many other elastic materials. It says that the extension of a spring is directly proportional to the force applied to it, provided the elastic limit is not exceeded. In other words, double the force and you double the extension.
胡克定律描述弹簧和许多其他弹性材料的行为。它说的是：只要不超过弹性极限，弹簧的伸长量与施加的力成正比。换句话说，力加倍，伸长量也加倍。
The law is named after Robert Hooke, a 17th-century English scientist who first studied the stretching of springs in a systematic way. It works well for metal springs, metal wires and some other elastic objects.
该定律以 17 世纪英国科学家罗伯特·胡克的名字命名，他最早系统地研究了弹簧的伸长现象。它适用于金属弹簧、金属丝和其他一些弹性物体。
ion. For rubber, the graph curves even at small forces, so F = kx is not a reliable model.
它并不适用于橡皮筋这类材料，因为橡皮筋的力-伸长量图像不是直线。对于橡皮筋，即使力很小图像也会弯曲，所以 F = kx 并不是可靠的模型。
The key word in the definition is 'directly proportional'. That means the two quantities always keep a constant ratio, and the graph must be a straight line passing through the origin.
定义中的关键词是'正比'。这意味着两个量始终保持恒定的比值，而且图像必须是一条通过原点的直线。
2. The Hooke's Law Equation: F = kx | 胡克定律公式 F = kx
The equation for Hooke' wtons (N), k is the spring constant in newtons per metre (N/m), and x is the extension in metres (m). Each symbol has a fixed unit, and examiners expect those units to appear in your working.
胡克定律的公式写作 F = kx，其中 F 是力，单位牛顿 (N)；k 是弹簧常数，单位牛顿每米 (N/m)；x 是伸长量，单位米 (m)。每个符号都有固定单位，阅卷老师希望在你的解题过程中看到这些单位。
F = k × x
ose the form that matches the quantity the question asks for, then substitute the numbers.
这个公式可以变形为三种有用的形式：F = kx、k = F ÷ x 和 x = F ÷ k。选择与题目所求量对应的形式，然后代入数值。
nto the formula. A spring stretched by 8 cm has x = 0.08 m, not x = 8 m and not x = 800 m.
代入公式前，务必先把长度换算成米。被拉长 8 cm 的弹簧，其 x = 0.08 m，而不是 x = 8 m，也不是 x = 800 m。
| Quantity | Symbol | Unit |
|---|---|---|
| Force | F | newton (N) |
| Spring constant | k | newton per metre (N/m) |
| Extension | x | metre (m) |
In problem solving, the units of force, spring constant and extension must be consistent. Lengths are normally given in metres, forces in newtons, and spring constants in newtons per metre.
在解题时，力、弹簧常数和伸长量的单位必须统一。通常长度用米，力用牛顿，弹簧常数用牛顿每米。
3. Understanding the Spring Constant k | 理解弹簧常数 k
The spring constant k tells you how stiff a spring is. A large k means the spring is stiff and needs a large force to produce a small extension. A small k means the spring is soft and stretches easily.
弹簧常数 k 表示弹簧的刚度。k 大意味着弹簧硬，需要较大的力才能产生较小的伸长量；k 小意味着弹簧软，容易被拉长。
(N/m). A spring with k = 50 N/m needs 50 N to stretch by 1 m, or 5 N to stretch by 0.1 m.
由于 k 代表产生 1 米伸长量所需的力，它的单位是牛顿每米 (N/m)。k = 50 N/m 的弹簧需要 50 N 才能拉长 1 m，或者 5 N 拉长 0.1 m。
For a given force, a stiff spring (large k) gives a small extension while a soft spring (small k) gives a large extension. This idea is often tested by asking you to compare two springs on the same graph.
在相同的力下，硬弹簧（k 大）伸长量小，软弹簧（k 小）伸长量大。这个思路经常通过让你比较同一图像上的两条弹簧线来考查。
Adding springs in parallel makes the combination stiffer, so the effective k increases. Adding springs in series makes it softer, so the effective k decreases.
把弹簧并联会使其更硬，等效 k 增大；把弹簧串联会使其更软，等效 k 减小。
4. Extension vs Total Length | 伸长量与总长度
The word 'extension' means the extra length produced by the force, not the total length of the spring. This is one of the most common places where marks are lost in CIE-style papers.
'伸长量'一词指的是力产生的额外长度，而不是弹簧的总长度。这是 CIE 风格试卷中最常见的丢分点之一。
tched length − original (natural) length. Both lengths must be measured in the same unit before subtracting.
你可以用一个简单的减法算出伸长量：伸长量 = 拉伸后长度 − 原长（自然长度）。相减之前，两个长度必须使用相同的单位。
x = L − L₀
n is 18 − 12 = 6 cm, which is 0.06 m. The total length 18 cm must never be used in F = kx.
例如，一根自然长度为 12 cm 的弹簧被拉到 18 cm，伸长量为 18 − 12 = 6 cm，即 0.06 m。绝不能把总长度 18 cm 代入 F = kx。
The same rule applies when a spring is compressed. The compression (a shortening) is also called x, and Hooke's Law still works for small compressions of a metal spring.
当弹簧被压缩时也是同样的规则。压缩量（缩短的长度）同样记作 x，对于金属弹簧的小幅度压缩，胡克定律依然成立。
5. Worked Example: Finding the Force | 例题：求力
he equation F = kx and check the units, which are already in metres and newtons per metre.
题目：一根弹簧的弹簧常数为 40 N/m，被拉长 0.15 m。求所施加的力。第一步是写出公式 F = kx 并检查单位，题目已给出米和牛顿每米，单位正确。
Notice how the metre units cancel, leaving newtons, which confirms the answer is a force.
代入数值得 F = 40 N/m × 0.15 m。答案：F = 6 N。注意米这个单位相互抵消，只剩下牛顿，说明结果确实是力。
Now extend the question: what force would double the extension to 0.30 m? Since force and extension are directly proportional, the force also doubles, giving 12 N.
再把题目延伸一步：如果伸长量加倍到 0.30 m，需要多大的力？由于力与伸长量成正比，力也加倍，为 12 N。
Always finish a calculation answer with the correct unit and, where relevant, a full sentence such as 'the force applied is 6 N'. Examiners award marks for the unit as well as the number.
计算题作答时一定要写正确的单位，必要时写完整句子，例如'所施加的力为 6 N'。阅卷时数字和单位都占分。
6. Worked Example: Finding the Spring Constant | 例题：求弹簧常数
culate the spring constant. Here the unknown is k, so rearrange the equation to k = F ÷ x.
题目：12 N 的力把弹簧拉长 0.30 m，求弹簧常数。这里未知量是 k，所以把公式变形为 k = F ÷ x。
N/m. A larger k would mean the same 12 N produced a smaller extension, so the spring would be stiffer.
代入得 k = 12 N ÷ 0.30 m = 40 N/m。如果 k 更大，同样的 12 N 只会产生更小的伸长量，说明弹簧更硬。
Some questions give you a table of force and extension readings and ask you to find k from the data. In that case, divide each force by its matching extension; if the law holds, every pair should give the same value of k.
有些题目给出力和伸长量的数据表，要求你从数据中求 k。此时用每一组力除以对应的伸长量；如果定律成立，每一组算出的 k 都应相同。
| Force F (N) | Extension x (m) | k = F ÷ x (N/m) |
|---|---|---|
| 2 | 0.05 | 40 |
| 4 | 0.10 | 40 |
| 6 | 0.15 | 40 |
The constant value of k in the third column is strong evidence that the spring obeys Hooke's Law over this range of forces.
第三列中 k 值保持不变，这是该弹簧在此受力范围内遵守胡克定律的有力证据。
7. Worked Example: Finding the Extension | 例题：求伸长量
nd a 3 N weight is attached. Calculate the extension. Rearrange the equation to x = F ÷ k.
题目：一根 k = 20 N/m 的弹簧竖直悬挂，挂上 3 N 的重物，求伸长量。把公式变形为 x = F ÷ k。
5 m, which is 15 cm. The natural length of the spring is not needed unless the question also asks for the new total length.
代入得 x = 3 N ÷ 20 N/m = 0.15 m，即 15 cm。除非题目还要求新的总长度，否则弹簧的自然长度并不需要用到。
+ 15 cm = 35 cm. Mixing up 15 cm and 35 cm in the final answer is a very frequent mistake.
如果自然长度为 20 cm，新的总长度为 20 cm + 15 cm = 35 cm。在最终答案中把 15 cm 和 35 cm 弄混，是非常常见的错误。
re and after loading and asks for the force. First find x by subtraction, then use F = kx.
第二类题目给出加载前后的总长度，要求求力。先用减法求出 x，再用 F = kx 计算。
8. Force-Extension Graphs | 力-伸长量图像
A force-extension graph plots force F on the y-axis and extension x on the x-axis. For a spring obeying Hooke's Law, the graph is a straight line through the origin, which shows that F is directly proportional to x.
力-伸长量图像以力 F 为纵轴、伸长量 x 为横轴。对于遵守胡克定律的弹簧，图像是一条通过原点的直线，说明 F 与 x 成正比。
adient = change in F ÷ change in x. A steeper line means a stiffer spring with a larger k.
这条直线的斜率等于弹簧常数 k，因为斜率 = F 的变化量 ÷ x 的变化量。直线越陡，说明弹簧越硬，k 越大。
For a straight line through the origin, the area is a triangle: area = ½ × base × height.
力-伸长量图像下方的面积单位是 N × m = 焦耳 (J)，代表弹簧储存的能量。对于通过原点的直线，面积是一个三角形：面积 = ½ × 底 × 高。
Before you read a gradient, check the scale on both axes. Exams often give extensions in centimetres on the axis while the formula needs metres, so convert before calculating k.
读取斜率之前，先检查两个坐标轴的刻度。考试中横轴常以厘米给出伸长量，而公式需要米，因此计算 k 前要先换算。
9. The Elastic Limit and Limit of Proportionality | 弹性极限与正比极限
Beyond this point, extension is no longer proportional to force, so F = kx stops working.
正比极限是图像上直线结束、曲线开始的点。超过这一点，伸长量与力不再成正比，F = kx 不再适用。
The elastic limit is the point beyond which the spring does not return to its original length when the force is removed. It is usually very close to the limit of proportionality for a metal spring.
弹性极限是超过之后弹簧在撤去外力时不再回到原长的那个点。对于金属弹簧，它通常非常接近正比极限。
If a spring is stretched past its elastic limit, it becomes permanently deformed and remains longer than before. The graph on unloading no longer follows the same line back to the origin.
如果弹簧被拉伸超过弹性极限，它会发生永久形变，长度比原来更长。卸载时的图像不再沿同一条直线回到原点。
Exam answers should use the phrase 'the elastic limit has been exceeded' rather than vague wording such as 'the spring is broken', unless the question really is about a spring snapping.
考试作答时应使用'超过了弹性极限'这样的表述，而不要用'弹簧坏了'之类含糊的说法，除非题目确实在问弹簧被拉断。
10. Elastic Potential Energy Stored in a Spring | 弹簧储存的弹性势能
is done on it and energy is stored as elastic potential energy. For a spring obeying Hooke's Law, this energy can be calculated with the formula E = ½kx².
弹簧被拉长时，有外力对它做功，能量以弹性势能的形式储存起来。对于遵守胡克定律的弹簧，这份能量可以用公式 E = ½kx² 计算。
E = ½ × k × x²
The energy is measured in joules (J). Because x is squared, doubling the extension quadruples the stored energy, which is why the graph curves upwards when energy is plotted against extension.
能量的单位是焦耳 (J)。由于 x 是平方项，伸长量加倍会使储存的能量变为四倍，这也是能量对伸长量作图时曲线向上弯曲的原因。
k = 200 N/m is stretched by 0.05 m. Then E = ½ × 200 × 0.05² = ½ × 200 × 0.0025 = 0.25 J.
例题：一根 k = 200 N/m 的弹簧被拉长 0.05 m。则 E = ½ × 200 × 0.05² = ½ × 200 × 0.0025 = 0.25 J。
.05 m and height F = kx = 10 N, so area = ½ × 0.05 × 10 = 0.25 J. Both methods must agree.
用图像下的面积也能得到同样的结果。可以试试：三角形底为 0.05 m，高为 F = kx = 10 N，所以面积 = ½ × 0.05 × 10 = 0.25 J。两种方法结果必然一致。
11. Common Mistakes and Exam Technique | 常见错误与应试技巧
The most common mistake is using the total length of the spring instead of the extension. Always ask yourself: 'has the question given me a length or a change in length?'
最常见的错误是把弹簧的总长度当作伸长量。每次都要问自己：'题目给我的是一个长度，还是一个长度的变化量？'
= 1 m, and that a mass of 1 kg produces a weight of about 10 N on Earth using g = 10 N/kg.
第二常见的错误是没有把厘米换算成米，或没有把克换算成牛顿。记住 100 cm = 1 m，并且在地球上取 g = 10 N/kg 时，1 kg 的质量产生的重力约为 10 N。
Other frequent errors include forgetting the unit on the final answer, rounding too early in a multi-step calculation, and assuming a spring still obeys Hooke's Law after the limit of proportionality.
其他常见错误包括：最终答案漏写单位；在分步计算中过早取近似值；以及在超过正比极限后仍假定弹簧遵守胡克定律。
A good exam habit is to write the equation, substitute the numbers with units, then state the answer with its unit. This three-step layout usually earns full marks even if the arithmetic slips.
良好的应试习惯是：写出公式、代入带单位的数值、然后写出带单位的答案。这个三步结构即使算术出错，通常也能拿到大部分分数。
12. Practice Questions with Answers | 练习题与答案
Try these five questions without looking at the answers first. They cover all the main Hooke's Law question types: finding F, finding k, finding x, reading a graph and calculating stored energy.
先不看答案，试着完成下面五道题。它们覆盖了胡克定律的主要题型：求 F、求 k、求 x、读图像，以及计算储存的能量。
- Q1. A spring of k = 25 N/m is stretched by 0.20 m. Find the force. | 一根 k = 25 N/m 的弹簧被拉长 0.20 m，求力。
- Q2. A force of 9 N stretches a spring by 0.45 m. Find k. | 9 N 的力把弹簧拉长 0.45 m，求 k。
- Q3. A spring of k = 60 N/m carries a 1.2 N weight. Find the extension in cm. | 一根 k = 60 N/m 的弹簧挂上 1.2 N 的重物，求伸长量（厘米）。
- Q4. A spring of natural length 15 cm is stretched to 21 cm by a 4.8 N force. Find k. | 一根自然长度 15 cm 的弹簧被 4.8 N 的力拉到 21 cm，求 k。
- Q5. A spring of k = 300 N/m is stretched by 0.02 m. Find the energy stored. | 一根 k = 300 N/m 的弹簧被拉长 0.02 m，求储存的能量。
n = 21 − 15 = 6 cm = 0.06 m, so k = 4.8 ÷ 0.06 = 80 N/m. Q5: E = ½ × 300 × 0.02² = 0.06 J.
答案。Q1：F = 25 × 0.20 = 5 N。Q2：k = 9 ÷ 0.45 = 20 N/m。Q3：x = 1.2 ÷ 60 = 0.02 m = 2 cm。Q4：伸长量 = 21 − 15 = 6 cm = 0.06 m，故 k = 4.8 ÷ 0.06 = 80 N/m。Q5：E = ½ × 300 × 0.02² = 0.06 J。
If you answered all five correctly, you have covered the essential Hooke's Law skills needed for KS3 and for the start of CIE IGCSE Physics. Practise drawing the force-extension graph as well, since graph questions carry a large share of the marks.
如果这五题你全部答对，说明你已经掌握了 KS3 以及 CIE IGCSE 物理入门阶段所需的胡克定律核心技能。同时也要练习绘制力-伸长量图像，因为图像类题目占分比重很大。
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