📚 Force and Acceleration | 力与加速度的关系
Classical mechanics is built on the relationship between net force and acceleration. This article reviews the key concepts from an A-Level mathematics perspective, focusing on vector treatment, problem solving and differential calculus.
经典力学建立于合外力与加速度之间的关系之上。本文从 A-Level 数学的视角回顾核心概念，重点讨论矢量的处理方式、解题步骤以及微分计算。
1. Newton’s Second Law as a Mathematical Relationship | 数学关系下的牛顿第二定律
Newton’s second law states that the net force acting on an object equals the rate of change of its momentum. When mass is constant, this reduces to the familiar equation F = m a, where F is measured in newtons, m in kilograms and a in metres per second squared.
牛顿第二定律指出：物体所受的合外力等于其动量的变化率。当质量恒定时，该定律简化为熟悉的方程 F = m a，其中 F 的单位是牛顿，m 的单位是千克，a 的单位是米每二次方秒。
The equation is often written as:
F = m a
For A-Level mechanics, it is vital to treat F as the resultant force. If multiple forces act, the vector sum is used before applying the equation.
在 A-Level 力学中，必须将 F 视为合力。如果物体受多个力作用，应先求矢量和，再代入公式。
When mass changes with time, the more general form is used:
F = d(mv)/dt
This differential form is particularly useful in variable-mass problems such as rockets or conveyor belts.
这种微分形式在火箭或传送带等变质量问题中尤其有用。
2. Force and Acceleration are Vectors | 力与加速度的矢量性
Both force and acceleration are vector quantities. A common mistake is to substitute only magnitudes into F = m a without considering direction. In one dimension, positive and negative signs represent opposite directions.
力和加速度都是矢量。一个常见错误是只将大小代入 F = m a，而忽略方向。在一维问题中，正负号代表相反方向。
A ball thrown upwards at 10 m/s has acceleration directed downward. Taking upward as positive, the equation becomes:
以 10 m/s 竖直上抛的小球，其加速度方向向下。若取向上为正，则方程写为：
-mg = m (-9.8)
In two dimensions, forces are resolved into perpendicular components, typically horizontal and vertical. The second law is then applied separately along each axis.
在二维问题中，力通常分解为互相垂直的分量，通常是水平和竖直分量，然后沿每个方向分别应用牛顿第二定律。
For example, with components Fₓ and Fᵧ, the acceleration components are aₓ = Fₓ / m and aᵧ = Fᵧ / m. The resultant acceleration is found using Pythagoras’ theorem.
例如，已知分力 Fₓ 和 Fᵧ，则加速度分量为 aₓ = Fₓ / m 和 aᵧ = Fᵧ / m，合加速度由勾股定理求得。
3. Inertial Mass vs Gravitational Mass | 惯性质量与引力质量
The mass in F = m a is called inertial mass. It measures how strongly an object resists acceleration when a given force is applied. A larger inertial mass produces a smaller acceleration for the same force.
F = m a 中的质量称为惯性质量，它度量物体在给定力作用下抵抗加速度的程度。惯性质量越大，相同力产生的加速度越小。
Gravitational mass, on the other hand, determines the gravitational force acting on an object. In Newton’s law of gravitation, the force is proportional to the product of gravitational masses.
另一方面，引力质量决定物体所受的引力大小。在牛顿万有引力定律中，引力与引力质量的乘积成正比。
Precision experiments show that inertial mass and gravitational mass are equivalent to a high degree of accuracy. This equivalence is the starting point for Einstein’s general theory of relativity.
精确实验表明，惯性质量与引力质量在很大程度上是相等的。这一等价性是爱因斯坦广义相对论的出发点。
In most A-Level problems, mass is assumed constant, but it is important to remember that the symbol m refers to the same physical property in both dynamics and gravitation.
在大多数 A-Level 题目中，质量被假定为恒定值，但需要注意，m 在动力学和引力中都指同一个物理属性。
4. Solving Linear Problems | 直线运动问题求解
Straight-line problems are the simplest application of Newton’s second law. The steps are: draw a force diagram, choose a positive direction, resolve forces, and apply F = m a.
直线运动问题是牛顿第二定律最简单的应用。解题步骤如下：画受力图，选取正方向，分解力，然后代入 F = m a。
Consider a box of mass 5 kg pulled horizontally by a force of 20 N on a frictionless surface. The acceleration is simply:
考虑一个 5 kg 的箱子在光滑水平面上受到 20 N 的水平拉力。其加速度为：
a = 20 / 5 = 4 m/s²
If friction is included, the net force is the applied force minus the frictional force. Suppose friction is 6 N:
如果考虑摩擦力，则合力为拉力减去摩擦力。设摩擦力为 6 N：
F_net = 20 – 6 = 14 N
a = 14 / 5 = 2.8 m/s²
When the force is constant, the acceleration is constant, and the SUVAT equations can be used. When the force varies, calculus is required.
当力恒定时，加速度恒定，可以使用 SUVAT 运动学公式。当力变化时，则需要使用微积分。
5. Motion on Inclines | 斜面上的运动
Inclined plane problems require careful resolution of weight into components parallel and perpendicular to the plane.
斜面问题需要将重力分解为平行于斜面和垂直于斜面的两个分量。
For a block of mass m on a plane inclined at angle θ, the component of weight down the slope is:
对于质量为 m 的物块，在倾角为 θ 的斜面上，重力沿斜面向下的分量为：
m g sin θ
The component perpendicular to the plane is:
m g cos θ
If the plane is smooth, the acceleration down the plane is:
若斜面光滑，物块沿斜面下滑的加速度为：
a = g sin θ
If friction acts, the frictional force is μ R, where R = m g cos θ is the normal reaction. The net force is:
若存在摩擦，摩擦力为 μ R，其中 R = m g cos θ 是法向反力。合力为：
m g sin θ – μ m g cos θ = m a
Therefore the acceleration is:
a = g (sin θ – μ cos θ)
This formula is only valid for objects moving down the plane. If the object moves up the plane, friction reverses direction and the minus becomes a plus in the friction term.
该公式仅适用于物体沿斜面向下运动的情形。如果物体沿斜面向上运动，摩擦力的方向相反，摩擦力项前要加正号。
6. Connected Particles | 连接体
Connected particle problems involve two or more masses linked by a light, inextensible string. The tension has the same magnitude throughout the string if the pulley is light and smooth.
连接体问题涉及由轻质不可伸长的绳子连接的多个质量。若滑轮轻且光滑，绳中张力处处相等。
Consider two masses m₁ and m₂ connected by a string over a pulley. The heavier mass m₂ accelerates downward while m₁ accelerates upward. Taking the direction of motion as positive for each mass:
考虑两个质量 m₁ 和 m₂ 通过绳子绕过定滑轮连接。较重质量 m₂ 向下加速，m₁ 向上加速。对每个质量取运动方向为正：
For m₂:
对 m₂：
m₂ g – T = m₂ a
For m₁:
对 m₁：
T – m₁ g = m₁ a
Adding the two equations eliminates T:
两式相加消去 T：
(m₂ – m₁) g = (m₁ + m₂) a
a = (m₂ – m₁) g / (m₁ + m₂)
The tension can then be found by substituting back into either equation.
将 a 代回任一方程即可求出张力 T。
7. Non-Constant Acceleration and Force | 变加速度与变力
When the net force is not constant, acceleration is not constant, and the SUVAT equations cannot be used. Instead, we use the differential relationship:
当合力不恒定时，加速度也不恒定，此时不能使用 SUVAT 运动学公式，而应使用微分关系：
F(x) = m a = m (dv/dt)
Since v = dx/dt, we can write:
由于 v = dx/dt，可以写为：
a = v (dv/dx)
This form is especially useful when force is given as a function of displacement. Integrating both sides gives the work-energy relationship.
这种形式在力为位移的函数时尤其有用。两边积分可得功能关系。
For example, if a particle of mass 2 kg experiences a force F = 6 x, then:
例如，质量为 2 kg 的质点受到力 F = 6 x，则：
2 v (dv/dx) = 6 x
v dv = 3 x dx
Integrating from x₀ to x gives:
从 x₀ 到 x 积分得到：
½ v² – ½ v₀² = (3/2)(x² – x₀²)
The acceleration is the time derivative of velocity, so if F is a function of time, we may integrate a = F(t)/m directly with respect to time.
加速度是速度对时间的导数，因此若 F 是时间的函数，可以直接对 a = F(t)/m 关于时间积分。
8. Friction and Drag Forces | 摩擦与阻力的力学关系
Friction between two surfaces in contact can be modelled as F ≤ μ R, where μ is the coefficient of static friction. When the object is sliding, kinetic friction is often approximated as F = μₖ R.
接触表面之间的摩擦力可建模为 F ≤ μ R，其中 μ 是静摩擦系数。当物体滑动时，动摩擦力通常近似为 F = μₖ R。
The direction of friction always opposes the relative motion or the tendency to move. This must be checked carefully when setting up equations.
摩擦力的方向总是阻碍相对运动或相对运动趋势。建立方程时必须仔细判断。
Drag force in a fluid is often modelled as proportional to speed:
流体中的阻力通常建模为与速度成正比：
F_drag = k v
For an object falling vertically under gravity and drag, the equation of motion is:
对于在重力和阻力作用下竖直下落的物体，运动方程为：
m g – k v = m (dv/dt)
The terminal velocity is found by setting dv/dt = 0:
终末速度由令 dv/dt = 0 求得：
v_T = m g / k
This differential equation can be solved by separation of variables to give v as a function of time.
该微分方程可分离变量求解，得到 v 关于时间的函数。
9. Simple Harmonic Motion as an Example | 简谐运动示例
When the net force is proportional to displacement and directed towards an equilibrium point, the motion is simple harmonic. The equation of motion is:
当合力与位移成正比且方向指向平衡位置时，物体做简谐运动。运动方程为：
m (d²x/dt²) = -k x
Rewriting with acceleration a = d²x/dt²:
用加速度 a = d²x/dt² 改写：
a = -(k/m) x
The quantity k/m is the square of the angular frequency:
其中 k/m 是角频率的平方：
ω² = k/m
The general solution is x = A sin(ω t) + B cos(ω t), showing that acceleration is always opposite to displacement and proportional to it.
通解为 x = A sin(ω t) + B cos(ω t)，表明加速度始终与位移方向相反且与位移成正比。
This example demonstrates how force and acceleration relationships translate into second-order differential equations.
这个例子展示了力与加速度的关系如何转化为二阶微分方程。
10. Momentum and the Impulse-Momentum Link | 动量与冲量
Newton’s second law can also be expressed as the impulse-momentum theorem. The impulse of a force equals the change in momentum.
牛顿第二定律也可以表达为冲量-动量定理。力的冲量等于动量的变化量。
For a constant force:
对于恒力：
F Δt = m v – m u
This is the integral form of F = m a. It is especially useful when force varies with time, because the area under a force-time graph gives the impulse.
这是 F = m a 的积分形式。当力随时间变化时特别有用，因为力-时间图像下的面积等于冲量。
Dividing both sides by Δt and taking the limit gives back the original differential form:
两边除以 Δt 并取极限，即可回到原来的微分形式：
F = dp/dt
Understanding the link between force, acceleration, momentum and impulse allows students to choose the most efficient method for a given problem.
理解力、加速度、动量与冲量之间的联系，可以帮助学生在面对具体题目时选择最有效的方法。
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