Hooke’s Law Problems | 胡克定律问题

📚 Hooke’s Law Problems | 胡克定律问题

Hooke’s law is a fundamental principle in physics that describes the behaviour of springs and other elastic materials. This article explores a range of typical problems involving force, extension, spring constant, energy storage, and combinations of springs. Each section is designed to strengthen your problem-solving skills and deepen your conceptual understanding for physics examinations.

胡克定律是物理学中描述弹簧及其他弹性材料行为的基本原理。本文探讨一系列涉及力、伸长量、弹簧常数、能量储存以及弹簧组合的典型问题。每个小节旨在强化你的解题技巧，并加深你对物理概念的理解，为考试做好准备。

1. Understanding Hooke’s Law | 理解胡克定律

Hooke’s law states that the extension of a spring is directly proportional to the force applied to it, provided the elastic limit is not exceeded. Mathematically, this is written as F = kx, where F is the force in newtons, k is the spring constant in N/m, and x is the extension in metres. The negative sign often seen in vector form simply indicates that the restoring force acts opposite to the displacement.

胡克定律指出，只要不超过弹性极限，弹簧的伸长量与施加的力成正比。数学表达式为 F = kx，其中 F 是力（牛顿），k 是弹性系数（N/m），x 是伸长量（米）。矢量形式中常见的负号仅表示恢复力的方向与位移相反。

In many exam problems, you will be asked to calculate the extension produced by a known force, or to find the spring constant from experimental data. Always ensure the spring is not stretched beyond its elastic limit, otherwise the relationship no longer holds and permanent deformation occurs.

在许多考试问题中，你会被要求根据已知力求伸长量，或根据实验数据求弹簧常数。务必确保弹簧没有被拉伸超过弹性极限，否则正比关系不再成立，会发生永久形变。

2. Calculating Spring Constant | 计算弹簧常数

The spring constant k is a measure of the stiffness of a spring. A high k value indicates a stiff spring that requires a large force for a small extension. To calculate k, rearrange Hooke’s law: k = F / x. For example, if a force of 10 N causes an extension of 0.02 m, then k = 10 / 0.02 = 500 N/m.

弹簧常数 k 是衡量弹簧刚度的量度。较高的 k 值代表弹簧较硬，需要较大的力才能产生较小的伸长。计算 k 时，可重新整理胡克定律：k = F / x。例如，若 10 N 的力产生 0.02 m 的伸长，则 k = 10 / 0.02 = 500 N/m。

Problems often provide a table of force and extension values. Plotting a graph with force on the y‑axis and extension on the x‑axis yields a straight line through the origin for an ideal spring. The gradient of this line equals the spring constant k. Always comment on whether the line passes through the origin as a check of proportional behaviour.

题目经常提供力与伸长量的数据表格。以力为纵轴、伸长量为横轴作图，对于理想弹簧可得一条通过原点的直线。该直线的斜率就是弹簧常数 k。注意检查直线是否通过原点，以验证正比关系。

3. Force-Extension Problems | 力与伸长量问题

A classic problem asks: ‘A spring of spring constant 250 N/m is stretched by 0.04 m. What force is required?’ Simply apply F = kx: F = 250 × 0.04 = 10 N. Alternatively, you might be given the force and the original and final lengths, requiring you to calculate the extension first: extension = final length – original length.

一个经典问题是：“一根弹簧的弹性系数为 250 N/m，被拉伸了 0.04 m，需要多大的力？”直接使用 F = kx：F = 250 × 0.04 = 10 N。有时题目给出力、原长和最终长度，你需先算伸长量：伸长量 = 最终长度 – 原长。

Watch out for units. The extension must be in metres. If given in centimetres, convert to metres by dividing by 100. Also, remember that the force calculated is the tension in the spring, which is equal to the weight of a hanging mass if the system is in equilibrium.

注意单位换算。伸长量必须用米。如果以厘米给出，需除以 100 转换为米。另外，计算出的力即为弹簧中的拉力，如果系统处于平衡，则该力等于悬挂重物的重力。

4. Elastic Limit and Permanent Deformation | 弹性极限与永久变形

Every spring has an elastic limit. If the applied force is too large, the spring undergoes plastic deformation and does not return to its original length when the force is removed. In force–extension graphs, the straight line stops and the curve flattens. Problems may ask you to identify the elastic limit from a graph and explain why Hooke’s law no longer applies beyond that point.

每根弹簧都有一个弹性极限。若施加的力过大，弹簧会发生塑性变形，撤去力后无法恢复原长。在力‑伸长量图中，直线部分会终止，曲线趋于平坦。题目可能要求你从图中找出弹性极限，并解释为何超过该点后胡克定律不再适用。

To solve such problems, locate the last point that lies on the straight‑line portion. The corresponding force is the maximum force for which the spring obeys Hooke’s law. Any extension beyond that point is partly permanent. Understanding this concept helps avoid misapplying the formula in real‑world contexts.

解答此类问题时，找到直线部分的最后一点，该点对应的力就是弹簧符合胡克定律的最大力。超出该点的伸长部分都是不可恢复的。理解这一概念有助于避免在实际情境中误用公式。

5. Springs in Series | 串联弹簧

When two springs are connected end‑to‑end, they are in series. The total extension is the sum of the individual extensions, and the same force acts through both springs. The effective spring constant k_eff for springs in series is given by 1/k_eff = 1/k₁ + 1/k₂. This means the combined spring is less stiff than either spring alone.

当两根弹簧首尾相连时，它们就是串联。总伸长量为各弹簧伸长量之和，且各弹簧受到的力相同。串联弹簧的有效弹性系数 k_eff 满足 1/k_eff = 1/k₁ + 1/k₂。这意味着组合弹簧的刚度比任一根单独的弹簧都要小。

A typical exam question provides two springs with constants 100 N/m and 200 N/m in series. Calculate k_eff: 1/k_eff = 1/100 + 1/200 = 3/200, so k_eff = 200/3 ≈ 66.7 N/m. Then use the effective constant to find the extension for a given load: x = F / k_eff.

典型的考试题给出两根弹性系数分别为 100 N/m 和 200 N/m 的弹簧串联。计算 k_eff: 1/k_eff = 1/100 + 1/200 = 3/200，因此 k_eff = 200/3 ≈ 66.7 N/m。然后用等效弹性系数求给定负载下的伸长量：x = F / k_eff。

6. Springs in Parallel | 并联弹簧

Springs in parallel are connected side by side so that they share the load and undergo the same extension. The effective spring constant for parallel springs is simply the sum of the individual constants: k_eff = k₁ + k₂. The combined spring is stiffer, so it requires a larger force to produce the same extension.

并联弹簧是指并排连接，共同承受负载且产生相同的伸长量。并联弹簧的有效弹性系数就是各弹簧常数之和：k_eff = k₁ + k₂。组合弹簧更硬，因此产生相同的伸长量需要更大的力。

If a 50 N/m spring is placed in parallel with a 150 N/m spring, k_eff = 200 N/m. If a 40 N load is hung from the combination, the extension is x = F / k_eff = 40 / 200 = 0.2 m. You can also deduce the force carried by each spring: F₁ = k₁ × x and F₂ = k₂ × x.

若将一根 50 N/m 的弹簧与一根 150 N/m 的弹簧并联，k_eff = 200 N/m。若在该组合下悬挂 40 N 的负载，伸长量 x = F / k_eff = 40 / 200 = 0.2 m。你还可以求出每根弹簧承受的力：F₁ = k₁ × x，F₂ = k₂ × x。

Configuration	有效弹性系数 keff	特点
串联 Series	1/keff = 1/k₁ + 1/k₂	总刚度变小
并联 Parallel	keff = k₁ + k₂	总刚度变大

7. Energy Stored in a Spring | 弹簧储存的能量

The work done in stretching or compressing a spring is stored as elastic potential energy. The formula for this energy is E = ½ k x², derived from the area under the force–extension graph. The energy is measured in joules when k is in N/m and x is in metres.

拉伸或压缩弹簧所做的功以弹性势能的形式储存起来。能量公式为 E = ½ k x²，源自力‑伸长量图线下的面积。当 k 的单位为 N/m、x 的单位为米时，能量以焦耳为单位。

Problems often involve calculating the energy stored and then linking it to kinetic energy or gravitational potential energy. For instance, a spring with k = 1000 N/m compressed by 0.1 m stores E = 0.5 × 1000 × (0.1)² = 5 J. When released, this energy can launch a projectile, so you can equate 5 J to ½ m v² to find the speed.

题目经常要求计算储存的能量，再将其与动能或重力势能联系起来。例如，一根 k = 1000 N/m 的弹簧被压缩 0.1 m，储存的能量为 E = 0.5 × 1000 × (0.1)² = 5 J。释放时，这部分能量可弹射物体，因此可令 5 J = ½ m v² 以求出速度。

8. Work Done and Elastic Potential Energy | 做功与弹性势能

A common trap is using the formula work = force × distance without considering that the force varies from zero to F as the spring stretches. The work done against the spring’s tension is the average force multiplied by the extension. The average force is (0 + F)/2 = F/2, so work = (F/2) × x. Since F = kx, work = (kx/2) × x = ½ k x².

常见的陷阱是直接使用功 = 力 × 距离，而忽略了弹簧拉伸过程中力从零逐渐增至 F。克服弹簧张力的功等于平均力乘以伸长量。平均力为 (0 + F)/2 = F/2，因此功 = (F/2) × x。因为 F = kx，所以功 = (kx/2) × x = ½ k x²。

This explains why the energy stored is not simply F × x but half of that product. Always use the ½ factor in energy calculations, or compute the area under the force–extension graph as a right‑angled triangle.

这解释了为何储存的能量不是简单的 F × x，而是其乘积的一半。在能量计算中一定要使用 ½ 这个系数，或者计算力‑伸长量图下直角三角形的面积。

9. Combining Hooke’s Law with Newton’s Laws | 结合胡克定律与牛顿定律

Many problems combine Hooke’s law with Newton’s second law. For example, a mass hanging on a spring in equilibrium is stationary because the spring force balances the weight: kx = mg. From this, the extension can be found: x = mg / k. If the system is accelerating, write the net force equation: T – mg = ma, where T is the spring force.

许多题目将胡克定律与牛顿第二定律结合起来。例如，悬挂在弹簧上的质量块静止时，弹簧力与重力平衡：kx = mg。由此可以求出伸长量：x = mg / k。若系统在加速，则列出合力方程：T – mg = ma，其中 T 为弹簧力。

In a typical elevator problem, the spring scale reading changes when the elevator accelerates upward or downward. The apparent weight is m(g ± a), and the spring extension corresponds to this apparent force. This reinforces the idea that Hooke’s law relates force and extension regardless of whether the system is inertial.

在典型的电梯问题中，当电梯向上或向下加速时，弹簧秤的读数会变化。表观重量为 m(g ± a)，弹簧伸长量则对应这个表观力。这强化了一个观念：无论系统是否处于惯性系，胡克定律都只关联力与伸长量。

10. Experimental Determination of k | 实验测定弹簧常数 k

To find the spring constant experimentally, add known masses to a spring, measure the new length, and calculate the extension for each mass. Plot force (weight mg) against extension. If the points lie on a straight line through the origin, Hooke’s law is obeyed and the gradient equals k. Use the linear portion only if the graph curves at higher loads.

要通过实验求弹簧常数，先在弹簧上添加已知质量，测量新长度，计算每次质量的伸长量。以力（重量 mg）为纵轴、伸长量为横轴作图。如果数据点落在一条通过原点的直线上，说明弹簧服从胡克定律，且斜率为 k。如果图线在较高负载下弯曲，则仅使用线性部分。

A measurement question may ask you to analyse data with uncertainties. The gradient of the best‑fit line gives k, and you can determine the percentage uncertainty in k by considering the spread of possible gradients. Remember to include the gravitational field strength g = 9.81 N/kg when converting mass to weight.

测量类题目可能要求你分析带有不确定度的数据。最佳拟合线的斜率给出 k，你还可以通过考虑可能斜率的分布范围来确定 k 的百分比不确定度。注意在将质量转换为重量时使用引力场强 g = 9.81 N/kg。

11. Graphical Analysis | 图形分析

Force–extension graphs are central to Hooke’s law problems. A straight line through the origin confirms direct proportionality. If the graph does not pass through the origin, there may be a zero error (e.g., the spring was already extended). The gradient gives k only for the linear region. The area under the graph up to a given extension equals the work done or energy stored.

力‑伸长量图是胡克定律问题的核心。一条通过原点的直线可以确认正比关系。若图线不通过原点，则可能存在零点误差（例如弹簧已经被拉伸）。只有在线性区域内，斜率才代表 k。图线下至某一伸长量为止的面积等于所做的功或储存的能量。

Problems sometimes present a graph with two or more springs. By comparing gradients, you can deduce which spring is stiffer. A steeper line means a larger k. You may also be asked to sketch the graph for a spring that has been stretched beyond its elastic limit, showing a curved region followed by a plateau.

有时题目会给出多条弹簧的图线。通过比较斜率，你可以推断哪根弹簧更硬。线越陡峭意味着 k 越大。你还可能被要求画出超过弹性极限的弹簧的图形，其中会先出现一段弯曲区域，随后趋于平缓。

12. Problem-Solving Strategies | 解题策略

Start by identifying the quantities given and what is required. Convert all units to SI (metres, newtons, kilograms). Write down Hooke’s law and any energy formulas. For combined springs, determine whether they are in series or parallel and use the appropriate effective constant. Draw a free‑body diagram when forces are involved, and check if the elastic limit is mentioned.

解题时，先确定已知量和待求量。将所有单位转换为国际单位制（米、牛顿、千克）。列出胡克定律及相关的能量公式。对于弹簧组合，判断它们是串联还是并联，并使用相应的等效弹性系数。当涉及力时，画出受力图，并检查题目是否提到了弹性极限。

When solving energy problems, remember that the elastic potential energy formula is only valid for extensions within the elastic limit. Always double‑check whether the question asks for the work done to stretch the spring or the energy stored. If in doubt, calculate the area under the force–extension graph, which always gives the correct work regardless of linearity.

在解答能量问题时，请记住弹性势能公式仅适用于弹性极限内的伸长量。务必仔细核对题目是要求拉伸弹簧所做的功还是储存的能量。如有疑问，可计算力‑伸长量图下的面积，无论是否线性，这一方法总能给出正确的功。

Published by TutorHao | Physics Revision Series | aleveler.com

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