📚 Work and Energy in IB AQA Mathematics | IB AQA 数学：功和能量 考点精讲
In the IB Mathematics curriculum, particularly within the mechanics components of both Analysis & Approaches and Applications & Interpretation, the concepts of work and energy form a bridge between pure calculus and real-world physical systems. The AQA-style approach emphasises modelling, variable forces, and the application of definite integration to compute work done by a force along a straight line. This article revisits the essential definitions, provides worked examples aligned with IB-style questions, and connects energy principles to kinematic outcomes. Understanding how to set up integrals for variable forces and how kinetic and potential energies translate into equations of motion is critical for success in both calculator and non-calculator papers.
在 IB 数学课程中，尤其是分析与方法（AA）和应用与解释（AI）的力学部分，功和能量的概念是纯微积分与现实物理系统之间的桥梁。AQA 风格的教学注重建模、变力以及运用定积分计算沿直线运动时力所做的功。本文将回顾基本定义，提供符合 IB 考题风格的例题，并将能量原理与运动学结果联系起来。掌握如何为变力建立积分表达式，以及如何将动能和势能转化为运动方程，对于在允许和不允许使用计算器的试卷中取得好成绩至关重要。
1. Defining Work for a Constant Force | 恒力做功的定义
When a constant force F acts on an object and moves it through a displacement s in the direction of the force, the work done W is given by the product W = F × s. If the force is applied at an angle θ to the displacement, only the component along the displacement contributes: W = F s cos θ. Work is a scalar quantity measured in joules (J). In one-dimensional motion along the x-axis, the expression simplifies to W = F Δx, where Δx is the change in position.
当恒力 F 作用在物体上，并使其沿力的方向发生位移 s 时，所做的功 W 由乘积 W = F × s 给出。如果力与位移之间的夹角为 θ，则只有沿位移方向的分量做功：W = F s cos θ。功是标量，单位为焦耳（J）。在沿 x 轴的一维运动中，公式简化为 W = F Δx，其中 Δx 是位置的改变量。
2. Work as the Integral of a Variable Force | 变力做功的积分表示
For a force that varies with position, F(x), the work done in moving an object from x = a to x = b is the definite integral of the force function with respect to displacement:
W = ∫ab F(x) dx
This is the most common IB examination scenario: a particle moves along a straight line while a force dependent on its coordinate acts. The area under the force–displacement graph also represents the work. Students should be comfortable evaluating such integrals analytically or using a GDC when the function is complicated. Negative work occurs if the force opposes the motion direction.
对于随位置变化的力 F(x)，将物体从 x = a 移动到 x = b 时所做的功是力函数关于位移的定积分：
W = ∫ab F(x) dx
这是 IB 考试中最常见的情景：一个质点沿直线运动，而作用力依赖于它的坐标。力–位移图下方的面积也表示功。学生应熟练地对这类积分进行解析求值，或在函数复杂时使用图形计算器。当力与运动方向相反时，功为负值。
3. Variable Force Example: Spring (Hooke’s Law) | 变力示例：弹簧（胡克定律）
A spring obeying Hooke’s law exerts a restoring force F = –k x, where k is the spring constant and x is the extension from natural length. The work done by the spring as it returns from extension A to its natural length (x = 0) is:
W = ∫A0 (–k x) dx = ½ k A²
Notice the sign convention: the spring does positive work when moving towards equilibrium. The work done on the spring to extend it by A is also ½ k A², but the applied force does positive work while the spring force does negative work. IB questions often ask to find the energy stored in an elastic string or spring as a function of extension.
遵循胡克定律的弹簧施加恢复力 F = –k x，其中 k 为劲度系数，x 为从原长算起的伸长量。弹簧从伸长量 A 回到原长（x = 0）时所做的功为：
W = ∫A0 (–k x) dx = ½ k A²
注意符号规则：弹簧向平衡位置运动时做正功。将弹簧拉伸 A 长度时，外力做功亦为 ½ k A²，但外力做正功，而弹力做负功。IB 考题经常要求求出弹性绳或弹簧中储存的能量与伸长量的函数关系。
4. Kinetic Energy and the Work–Energy Theorem | 动能与功–能定理
The kinetic energy (KE) of a particle of mass m moving with speed v is:
KE = ½ m v²
The work–energy theorem states that the net work done by all forces acting on a particle equals the change in its kinetic energy: Wnet = ΔKE = ½ m v₂² – ½ m v₁². This principle allows students to relate the integral of force directly to speed without solving differential equations of motion explicitly. It is especially powerful when combining variable forces with given initial conditions.
质量为 m、速度为 v 的质点，其动能（KE）为：
KE = ½ m v²
功–能定理指出，作用在质点上的所有力所做的净功等于其动能的变化量：Wnet = ΔKE = ½ m v₂² – ½ m v₁²。这一原理使学生能够直接将力的积分与速度关联起来，而无需显式求解运动微分方程。当涉及变力以及给定初始条件时，该定理尤为有效。
5. Gravitational Potential Energy | 重力势能
In a uniform gravitational field near the Earth’s surface, the change in gravitational potential energy (GPE) when a mass m is raised by a vertical height h is ΔGPE = m g h. If the reference level is chosen at h = 0, then GPE = m g h. In IB problems, the work done against gravity is often part of the energy balance: the net work done by all forces (excluding gravity when using GPE) leads to the change in mechanical energy. Always be consistent: treat gravity either as an external force doing work or via GPE, but never both simultaneously.
在地表附近的均匀引力场中，将质量 m 的物体竖直提升高度 h 时，重力势能（GPE）的改变量为 ΔGPE = m g h。若将参考平面选在 h = 0 处，则 GPE = m g h。在 IB 问题中，克服重力做功通常是能量平衡的一部分：所有力（当使用 GPE 时重力除外）的净功导致机械能的变化。务必保持一致：要么将重力看作做功的外力，要么通过 GPE 处理，切不可两者同时使用。
6. Conservation of Mechanical Energy | 机械能守恒
When the only forces doing work are conservative (such as gravity or an ideal spring), the total mechanical energy E = KE + PE remains constant. For a particle moving under gravity:
½ m v₁² + m g h₁ = ½ m v₂² + m g h₂
For a horizontal spring–mass system:
½ m v² + ½ k x² = constant
IB questions frequently ask to use energy conservation to find maximum speed, maximum compression, or the speed at a given position. Remember that if friction or an applied non-conservative force is present, the work done by these forces must be added to one side of the equation: Wnc = ΔE.
当仅保守力（如重力或理想弹簧弹力）做功时，系统的总机械能 E = KE + PE 保持不变。对于仅在重力作用下的质点：
½ m v₁² + m g h₁ = ½ m v₂² + m g h₂
对于水平弹簧–振子系统：
½ m v² + ½ k x² = 常数
IB 考题常要求利用能量守恒来求最大速度、最大压缩量或特定位置的速度。请记住，若存在摩擦或其他非保守外力，则需将这些力的功加到方程的一侧：Wnc = ΔE。
7. Power: Rate of Doing Work | 功率：做功的快慢
Power P is the rate at which work is done. For a constant force moving an object at a constant speed v in the direction of the force, P = F v. More generally, instantaneous power is the derivative of work with respect to time, P = dW/dt. In many IB problems, a vehicle engine provides a certain power, and students must apply P = F v to find the tractive force or the maximum speed against resistive forces. When the speed changes, integration may be required: the total work done is the integral of power over time, W = ∫ P dt.
功率 P 是做功的速率。对于以恒定速度 v 沿力的方向运动的恒力，P = F v。更一般地，瞬时功率是功对时间的导数，P = dW/dt。在许多 IB 问题中，车辆发动机提供某一功率，学生需运用 P = F v 求得牵引力或克服阻力时的最大速度。当速度变化时，可能需要进行积分：总功为功率对时间的积分，W = ∫ P dt。
8. Work Done by a Force in Vector Form (Two Dimensions) | 二维空间中力的功
When a constant force vector F = (Fx, Fy) moves a particle along a straight-line displacement vector s = (Δx, Δy), the work is the dot product: W = F · s = Fx Δx + Fy Δy. For a variable force in 2D, the work is the line integral ∫ F · dr. In IB, such calculations are often restricted to cases where the path is a straight line and the force varies only with one coordinate, making integration straightforward. Still, recognising the dot product formulation helps when forces are given as vectors.
当恒力矢量 F = (Fx, Fy) 使质点发生直线位移矢量 s = (Δx, Δy) 时，功为点乘：W = F · s = Fx Δx + Fy Δy。对于二维变力，功为线积分 ∫ F · dr。在 IB 中，此类计算通常局限于路径为直线且力仅随一个坐标变化的情形，这使得积分较为简单。不过，理解点乘形式有助于处理以矢量形式给出的力。
9. Typical IB Question: Work from a Velocity–Displacement Graph | 典型 IB 考题：通过速度–位移图像求功
A common IB problem gives a graph of v² against displacement x. Since the gradient of this graph relates to acceleration via v² = u² + 2 a x, the work–energy theorem can be applied: the change in kinetic energy ½ m (v₂² – v₁²) equals the area under an equivalent force–displacement graph multiplied by m/2. Students might be asked to determine the net force or the work done directly from the gradient of the v²–x line. This technique reinforces the link between kinematics and energy arguments.
一种常见的 IB 问题是给出 v² 关于位移 x 的图像。由于该图像的斜率通过 v² = u² + 2 a x 与加速度相关联，可应用功–能定理：动能的变化量 ½ m (v₂² – v₁²) 等于等效力–位移图下的面积乘以 m/2。学生可能需要直接从 v²–x 直线的斜率求出合力或所做的功。这种方法强化了运动学与能量论述之间的联系。
10. Work–Energy with Resistive Forces | 考虑阻力的功–能问题
Realistic IB modelling tasks often include resistive forces proportional to speed (FR = –k v) or to the square of speed. The net work is then Wengine + Wresistance = ΔKE. Since resistive forces are non-conservative, energy is dissipated as heat. Students must carefully set up integrals for the work done by resistance over a distance, using the chain rule or expressing v as a function of x if possible. For instance, if resistance is –k v and the vehicle moves from x₁ to x₂, the work done by resistance is –k ∫ v dx = –k ∫ v (dx/dt) dt = –k ∫ v² dt, which could require knowledge of velocity as a function of time.
IB 课程中实际的建模任务常常涉及与速度成正比的阻力（FR = –k v）或与速度平方成正比的阻力。此时净功为 W发动机 + W阻力 = ΔKE。由于阻力是非保守力，能量会以热的形式耗散。学生必须仔细建立阻力在一段距离上所做功的积分，利用链式法则或在可能时将 v 表示为 x 的函数。例如，若阻力为 –k v 且车辆从 x₁ 移动到 x₂，阻力所做的功为 –k ∫ v dx = –k ∫ v (dx/dt) dt = –k ∫ v² dt，这可能需要知道速度作为时间的函数。
11. Summary of Key Formulas and Units | 关键公式与单位汇总
The core relationships for revision:
| Concept | Formula | Units |
| Work (constant force) | W = F s cos θ | J (N·m) |
| Work (variable 1D) | W = ∫ F(x) dx | J |
| Kinetic energy | KE = ½ m v² | J |
| Gravitational PE | GPE = m g h | J |
| Elastic PE | EPE = ½ k x² | J |
| Power | P = F v, P = dW/dt | W (J/s) |
Always check that your integration limits correspond to the physical situation and that resistive work is correctly signed (negative when opposing motion). Practise interchanging between energy and kinematic descriptions, and be comfortable with both GDC and analytic integration methods.
复习时需要掌握的核心关系如下表所示。请始终检查积分限是否与物理情景相符，并确保阻力做功的符号正确（阻碍运动时为负）。多加练习能量描述与运动学描述之间的相互转换，并熟练运用图形计算器和解析积分两种方法。
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