📚 Work and Energy in IB & CIE Mathematics: Key Exam Points | IB CIE 数学：功和能量 考点精讲
In IB and CIE Mathematics, the concepts of work and energy provide a rich context for applying vectors, calculus, and algebraic techniques to real-world problems. Whether calculating the work done by a force along a straight line using dot product, integrating a variable force over a distance, or using the work-energy theorem to solve motion problems, mastery of these ideas is crucial for high marks in mechanics sections.
在 IB 和 CIE 数学中，功和能量的概念为向量、微积分和代数技巧的实际应用提供了丰富的场景。无论是使用点积计算沿直线的做功，对变力进行距离积分，还是运用功能定理解决运动问题，掌握这些思想对于在力学部分取得高分至关重要。
1. Definition of Work as a Scalar Product | 功的标量积定义
In mathematics, work done by a constant force F acting on an object that undergoes a displacement d is defined as the scalar product (dot product) of the two vectors: W = F · d = |F||d| cos θ, where θ is the angle between the force and displacement vectors.
在数学中，一个恒力 F 作用于物体并使其发生位移 d 所做的功，定义为两个向量的标量积（点积）：W = F · d = |F||d| cos θ，其中 θ 是力与位移向量之间的夹角。
The unit of work is the joule (J). In pure mathematics problems, the vector components are often given, and you must compute the dot product directly: if F = (F₁, F₂, F₃) and d = (d₁, d₂, d₃), then W = F₁d₁ + F₂d₂ + F₃d₃.
功的单位是焦耳（J）。在纯数学问题中，通常会给出向量的分量，你必须直接计算点积：如果 F = (F₁, F₂, F₃) 且 d = (d₁, d₂, d₃)，则 W = F₁d₁ + F₂d₂ + F₃d₃。
W = F ⋅ d = F₁d₁ + F₂d₂ + F₃d₃
2. Work Done by a Constant Force: Two-Dimensional Cases | 恒力做功：二维情况
For problems in the xy-plane, a force F = ai + bj moving an object from point A to B with displacement d = Δxi + Δyj yields work W = aΔx + bΔy. When the path is straight and the force constant, this simple calculation is all that is needed. Exam questions often ask for the angle between force and displacement using the dot product formula.
对于 xy 平面上的问题，力 F = ai + bj 将物体从点 A 移动到点 B，位移为 d = Δxi + Δyj，所做的功为 W = aΔx + bΔy。只要路径是直线且力恒定，这样简单的计算就足够了。考试题经常会要求利用点积公式计算力与位移之间的夹角。
Example: A force F = (3, 4) N moves a particle from (1,2) to (5,6) (units in metres). The displacement is (4,4). Work = 3×4 + 4×4 = 12+16=28 J.
示例：力 F = (3, 4) N 将质点从 (1,2) 移动到 (5,6)（单位为米）。位移为 (4,4)。功 = 3×4 + 4×4 = 12+16=28 J。
3. Work Done by a Variable Force in One Dimension | 一维变力做功
When the force varies with position, F(x), work is found by integrating the force function with respect to displacement: W = ∫x₁x₂ F(x) dx. This is a direct application of definite integration. The integral represents the area under the force-displacement graph.
当力随位置变化 F(x) 时，功通过将力函数对位移积分求得：W = ∫x₁x₂ F(x) dx。这是定积分的直接应用。该积分表示力-位移图下的面积。
In IB and CIE, you may see forces following Hooke’s Law (F = –kx) or arbitrary polynomial functions. Always set up the integral with limits matching the starting and ending positions, and be careful with signs: if the force opposes the direction of motion, work can be negative.
在 IB 和 CIE 的考试中，你可能会遇到遵循胡克定律（F = –kx）的力，或任意多项式函数。务必设置好积分限，使其与起始和终点位置匹配，并注意符号：如果力与运动方向相反，功可能为负值。
W = ∫x₁x₂ F(x) dx
4. Work Along a Curved Path: Vector Line Integrals | 沿曲线路径做功：向量线积分
For higher-level syllabi (IB HL, CIE Further Maths), you might need to compute the work done by a variable force vector F(x,y,z) along a curve C. The work is given by the line integral: W = ∫C F · dr. This is parameterised using r(t) = (x(t), y(t), z(t)) and dr = (dx/dt, dy/dt, dz/dt) dt. Then W = ∫t₁t₂ F(r(t)) · r‘(t) dt.
对于更高层次的课程（IB HL、CIE 进阶数学），你可能需要计算一个变力向量 F(x,y,z) 沿曲线 C 所做的功。这个功用线积分表示：W = ∫C F · dr。可以使用参数化计算，设 r(t) = (x(t), y(t), z(t))，dr = (dx/dt, dy/dt, dz/dt) dt，于是 W = ∫t₁t₂ F(r(t)) · r‘(t) dt。
It’s essential to remember that if the force field is conservative, the line integral is path-independent and can be evaluated using a potential function. Checking for a conservative field involves showing that the curl of F is zero, or in two dimensions that ∂F₂/∂x = ∂F₁/∂y.
必须记住，如果力场是保守的，线积分与路径无关，并且可以利用势函数来计算。判断保守场的方法是证明 F 的旋度为零，或在二维情况下证明 ∂F₂/∂x = ∂F₁/∂y。
5. Kinetic Energy and the Work-Energy Theorem | 动能与功能定理
The work-energy theorem states that the net work done on an object equals its change in kinetic energy: Wnet = ΔEk = ½mv² − ½mu², where u and v are initial and final speeds. This principle is often used in mathematics to find velocity without directly integrating acceleration.
功能定理指出，合力对物体所做的净功等于其动能的变化量：Wnet = ΔEk = ½mv² − ½mu²，其中 u 和 v 分别为初速和末速。这一原理在数学中常被用来求解速度，而无需直接对加速度积分。
In examination problems, you might be given a force function, a mass, and an initial speed, then asked to find the speed at a certain position. You simply compute the work integral and equate it to ½mv² − ½mu².
在考试问题中，可能会给出一个力函数、质量和初速度，然后要求求出某一位置处的速度。你只需计算功的积分，并将其等于 ½mv² − ½mu² 即可。
Wnet =
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