📚 Mechanics Formula Summary and Typical Applications | 力学公式归纳与典型应用
This article summarises the core formulas of A-level and equivalent mechanics courses, and demonstrates their typical applications through clear examples. It is designed to help you consolidate the key concepts and apply them confidently in exams.
本文归纳了 A-level 及同等课程中力学部分的核心公式，并通过清晰的典型例子演示其应用，旨在帮助你巩固关键概念，并在考试中自信地运用。
1. Kinematics with Constant Acceleration | 匀变速直线运动学
The following “suvat” equations apply when acceleration is constant. Use the sign convention consistently: choose a positive direction and stick to it.
以下“suvat”方程适用于加速度恒定情形。使用符号约定时需保持一致：选定正方向并坚持它。
v = u + at
s = ut + ½at²
v² = u² + 2as
s = ½(u + v)t
List all known variables and the unknown you need. Each of these four equations omits one variable, so choose the most direct one.
列出所有已知量和待求量。这四个方程各自缺一个变量，因此选择最直接的那个。
For vertical motion under gravity, set a = g (often taken as 9.81 m s⁻² downwards).
对于重力作用下的竖直运动，令 a = g（通常取 9.81 m s⁻²，方向向下）。
Example: A ball is thrown upward at 20 m s⁻¹. Find its maximum height.
例：小球以 20 m s⁻¹ 竖直上抛。求其最大高度。
At maximum height, v = 0. Use v² = u² + 2as with u = 20, a = −g = −9.81, v = 0:
在最大高度处，v = 0。使用 v² = u² + 2as，其中 u = 20，a = −g = −9.81，v = 0：
0 = 20² − 2 × 9.81 × s → s = 400 / 19.62 ≈ 20.4 m
2. Newton's Laws of Motion | 牛顿运动定律
Newton's laws form the foundation of classical mechanics. The second law is the most used formula.
牛顿定律构成了经典力学的基础。第二定律是最常用的公式。
F = ma
The net force on a body equals the product of its mass and acceleration.
物体所受合力等于其质量与加速度的乘积。
If the net force is zero, acceleration is zero: the body is at rest or moves with constant velocity.
若合力为零，则加速度为零：物体静止或做匀速直线运动。
Example: A 3.0 kg block is pulled across a rough horizontal surface with a horizontal force of 15 N. The frictional force is 6.0 N. Find the acceleration.
例：质量为 3.0 kg 的木块在水平粗糙面上被 15 N 的水平力拉动，摩擦力为 6.0 N。求加速度。
Fnet = 15 − 6 = 9 N → a = Fnet / m = 9 / 3 = 3.0 m s⁻²
3. Work, Energy and Power | 功、能量与功率
Work is done when a force moves an object. Energy changes are often the simplest way to solve motion problems.
当力使物体移动时做功。能量变化常常是解决运动问题的最简方法。
W = Fs cos θ
Ek = ½mv²
Ep = mgh
P = W / t = Fv
For a force along the direction of motion, θ = 0 so W = Fs.
当力与运动方向一致时，θ = 0，故 W = Fs。
The work-energy theorem: work done by net force equals change in kinetic energy.
动能定理：合力所做功等于动能的变化。
Power is the rate of doing work; for a constant force and velocity, P = Fv.
功率是做功的快慢；对于恒定力和恒定速度，P = Fv。
Example: A motor lifts a 200 kg load at constant speed 1.5 m s⁻¹. Find the useful power output.
例：电动机以 1.5 m s⁻¹ 的恒定速度提升 200 kg 的重物。求有用功率。
F = mg = 200 × 9.81 = 1962 N → P = Fv = 1962 × 1.5 ≈ 2943 W
4. Momentum and Impulse | 动量与冲量
Momentum is conserved in an isolated system during collisions and explosions.
在孤立系统中，碰撞和爆炸过程中动量守恒。
p = mv
Impulse = F Δt = Δp
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
For elastic collisions, kinetic energy is also conserved; for inelastic collisions it is not.
弹性碰撞中动能也守恒；非弹性碰撞中动能不守恒。
The impulse-momentum relation is useful when force varies with time and you are given total time.
冲量-动量关系在力随时间变化且已知总时间时非常有用。
Example: A 0.5 kg trolley moving at 2 m s⁻¹ collides and sticks to a stationary 1.0 kg trolley. Find their common velocity.
例：质量为 0.5 kg、速度为 2 m s⁻¹ 的小车与静止的 1.0 kg 小车相撞并粘在一起。求共同速度。
0.5 × 2 = (0.5 + 1.0)v → v = 1 / 1.5 ≈ 0.67 m s⁻¹
5. Circular Motion and Centripetal Force | 圆周运动与向心力
Uniform circular motion requires a net force directed toward the centre.
匀速圆周运动需要指向圆心的合力。
ac = v² / r = ω²r
Fc = mv² / r = mω²r
Angular speed ω is related to linear speed by v = rω.
角速度 ω 与线速度的关系为 v = rω。
Period T and frequency f: T = 2π/ω = 1/f.
周期 T 与频率 f：T = 2π/ω = 1/f。
Example: A 0.10 kg mass, attached to a 0.50 m string, moves in a horizontal circle at 3.0 m s⁻¹. Find the tension in the string.
例：质量为 0.10 kg 的小球系在 0.50 m 长的绳子上，以 3.0 m s⁻¹ 的速度在水平面上做圆周运动。求绳的拉力。
T = mv² / r = 0.10 × 9.0 / 0.50 = 1.8 N
6. Newton's Law of Gravitation | 万有引力定律
Gravitational force provides the centripetal force for planetary and satellite motion.
万有引力提供行星和卫星做圆周运动所需的向心力。
F = Gm₁m₂ / r²
g = GM / r²
G is the universal gravitational constant, approximately 6.67 × 10⁻¹¹ N m² kg⁻².
G 为万有引力常量，约为 6.67 × 10⁻¹¹ N m² kg⁻²。
For a satellite at radius r with speed v: GMm / r² = mv² / r, hence v = √GM/r.
对于半径 r 处的卫星：GMm / r² = mv² / r，因此 v = √GM/r。
Example: Calculate the speed of a satellite orbiting Earth at radius 7.0 × 10⁶ m. Use GM = 3.99 × 10¹⁴ m³ s⁻².
例：计算在轨道半径 7.0 × 10⁶ m 处绕地球运动的卫星速度。已知 GM = 3.99 × 10¹⁴ m³ s⁻²。
v = √GM/r = √3.99 × 10¹⁴ / 7.0 × 10⁶ ≈ 7.55 × 10³ m s⁻¹
7. Hooke's Law and Elastic Potential Energy | 胡克定律与弹性势能
For springs and other elastic materials within their limit of proportionality, extension is proportional to applied force.
对于弹簧及其他弹性材料，在比例极限内，伸长量与所受外力成正比。
F = kx
Eel = ½kx²
k is the spring constant, measured in N m⁻¹; x is the extension or compression from natural length.
k 为劲度系数，单位 N m⁻¹；x 是相对原长的伸长或压缩量。
The elastic potential energy stored is given by ½kx².
储存的弹性势能为 ½kx²。
Example: A spring has natural length 0.20 m. When a 4.0 N weight is hung on it, its length is 0.25 m. Find the spring constant and the stored energy.
例：一根弹簧原长 0.20 m。挂上 4.0 N 重物后长度为 0.25 m。求劲度系数和储存的弹性势能。
x = 0.05 m → k = F/x = 4.0 / 0.05 = 80 N m⁻¹
E = ½kx² = ½ × 80 × 0.05² = 0.10 J
8. Simple Harmonic Motion | 简谐运动
In simple harmonic motion (SHM), acceleration is proportional to displacement and directed toward the equilibrium point.
在简谐运动中，加速度与位移成正比，且方向始终指向平衡位置。
a = −ω²x
x = A cos(2πt / T)
vmax = Aω
For a mass-spring system: ω = √k/m.
对于弹簧振子系统：ω = √k/m。
For a simple pendulum: ω = √g/l.
对于单摆：ω = √g/l。
Example: A mass of 0.20 kg oscillates on a spring with k = 80 N m⁻¹. Find the angular frequency and the maximum speed when the amplitude is 0.10 m.
例：质量 0.20 kg 的物体在劲度系数 k = 80 N m⁻¹ 的弹簧上振动。求角频率以及振幅为 0.10 m 时的最大速度。
ω = √k/m = √80 / 0.20 = √400 = 20 rad s⁻¹
vmax = Aω = 0.10 × 20 = 2.0 m s⁻¹
9. Projectile Motion | 抛体运动
Projectile motion is the superposition of uniform horizontal velocity and constant vertical acceleration.
抛体运动是水平匀速运动和竖直匀加速运动的合成。
Horizontal: x = ux t
Vertical: y = uy t − ½gt²
The horizontal component of velocity remains constant, ignoring air resistance.
忽略空气阻力时，水平速度分量保持不变。
At maximum height, the vertical velocity component is zero.
在最高点处，竖直速度分量为零。
Example: A ball is kicked at 18 m s⁻¹ at 35° above the horizontal. Find the time to reach maximum height.
例：足球以 18 m s⁻¹ 的初速度、35° 仰角被踢出。求达到最大高度所需时间。
uy = 18 sin 35° ≈ 10.3 m s⁻¹ → t = uy / g ≈ 10.3 / 9.81 ≈ 1.05 s
10. Rigid Body Equilibrium and Moments | 刚体平衡与力矩
For a body in equilibrium, both the resultant force and the resultant moment about any point must be zero.
物体处于平衡状态时，合力和对任意点的合力矩都为零。
Sum of clockwise moments = Sum of anticlockwise moments
Moment = F × d
d is the perpendicular distance from the line of action of the force to the pivot.
d 是力的作用线到支点的垂直距离。
This principle is used to solve problems involving seesaws, levers and balanced beams.
该原理用于解决涉及跷跷板、杠杆和平衡梁的问题。
Example: A 5.0 m uniform beam of weight 200 N is pivoted at one end. A 300 N weight is placed 2.0 m from the pivot. What force is needed downward at the other end to balance?
例：一根 5.0 m 的均匀横梁重 200 N，一端为支点。在距支点 2.0 m 处放置一个 300 N 的重物。为使梁平衡，在另一端需向下施加多大的力？
Assume beam weight acts at its centre, 2.5 m from pivot. Take moments about the pivot:
假设横梁重力作用在其中心，即距支点 2.5 m 处。对支点取矩：
F × 5.0 = 300 × 2.0 + 200 × 2.5 = 600 + 500 = 1100
F = 1100 / 5.0 = 220 N
11. Friction and the Basic Kinetic Model | 摩擦力与基本动力学模型
Friction is a contact force that opposes relative motion (or attempted motion).
摩擦力是一种接触力，总是阻碍相对运动（或相对运动趋势）。
F_friction ≤ μ N
For limiting static friction, Fmax = μ_s N; for kinetic friction, F = μ_k N.
最大静摩擦力 Fmax = μ_s N；动摩擦力 F = μ_k N。
N is the normal reaction force, which may be less than mg if there is an additional upward force.
N 为法向支持力；若存在额外的竖直向上的力，N 可能小于 mg。
Example: A 25 kg crate is pushed across a floor with μ_k = 0.30. What horizontal force is needed to keep it moving at constant speed?
例：质量 25 kg 的木箱在地板上滑动，μ_k = 0.30。需要多大的水平力才能使其匀速运动？
N = mg = 25 × 9.81 = 245 N → F = μ_k N = 0.30 × 245 ≈ 74 N
12. Practical Strategy for Solving Mechanics Problems | 解力学问题的实用策略
When facing any mechanics problem, follow a structured approach to avoid missing key information.
面对任何力学问题时，采用结构化步骤可以避免遗漏关键信息。
Step 1: Draw a clear free-body diagram, labelling all forces.
第一步：画出清晰的受力分析图，标出所有力。
Step 2: Choose a coordinate system and positive direction.
第二步：选择坐标系和正方向。
Step 3: Apply Newton's second law along each axis, or use energy/momentum if time is not involved.
第三步：沿每个轴应用牛顿第二定律；若不涉及时间，可用能量或动量。
Step 4: Check units and whether your answer has a reasonable order of magnitude.
第四步：检查单位以及答案的数量级是否合理。
By mastering these formulas and practising the typical examples above, you will develop a reliable toolkit for mechanics questions in your exam.
通过掌握上述公式并练习这些典型例题，你将建立一套可靠的解题工具包，从容应对考试中的力学问题。
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