Mechanics Formula Summary and Typical Applications | 力学公式归纳与典型应用

📚 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。

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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

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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。

F_net = 15 − 6 = 9 N → a = F_net / m = 9 / 3 = 3.0 m s⁻²

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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

E_k = ½mv²

E_p = 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

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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

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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。

a_c = v² / r = ω²r

F_c = 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

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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

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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

E_el = 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

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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)

v_max = 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

v_max = Aω = 0.10 × 20 = 2.0 m s⁻¹

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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 = u_x t

Vertical: y = u_y 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。

u_y = 18 sin 35° ≈ 10.3 m s⁻¹ → t = u_y / g ≈ 10.3 / 9.81 ≈ 1.05 s

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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

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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, F_max = μ_s N; for kinetic friction, F = μ_k N.

  最大静摩擦力 F_max = 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

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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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