📚 Complete Physics for Cambridge Secondary Workbook: Formula Derivations | 剑桥中学物理练习册公式推导
Understanding how physics formulas are derived not only boosts your problem‑solving confidence but also builds a deeper grasp of the underlying principles. In this article, we walk through the key derivations commonly found in the Cambridge Secondary Physics curriculum, step by step.
理解物理公式的推导过程不仅能增强你解题的信心，还能让你更深入地掌握基本原理。本文将一步步梳理剑桥中学物理课程中常见的核心公式推导。
1. Deriving v² = u² + 2as | 推导 v² = u² + 2as
at² yields s = u((v − u)/a) + ½a((v − u)/a)². Simplifying, s = (uv − u²)/a + (v² − 2uv + u²)/(2a). Multiplying through by 2a gives 2as = 2uv − 2u² + v² − 2uv + u², which simplifies to 2as = v² − u². Therefore, v² = u² + 2as.
我们从加速度的定义出发：a = (v − u) / t。变形得 t = (v − u) / a。将其代入位移公式 s = ut + ½at²，得到 s = u((v − u)/a) + ½a((v − u)/a)²。化简后 s = (uv − u²)/a + (v² − 2uv + u²)/(2a)。两边同乘 2a 得 2as = 2uv − 2u² + v² − 2uv + u²，整理得 2as = v² − u²。因此 v² = u² + 2as。
2. Deriving F = ma from Experimental Evidence | 由实验证据推导 F = ma
nd law is built on two proportionalities: acceleration is proportional to the net force (a ∝ F) and inversely proportional to mass (a ∝ 1/m). Combining these gives a ∝ ning the unit of force such that 1 N accelerates 1 kg by 1 m s⁻², k becomes 1. Hence the familiar F = ma.
牛顿第二定律建立在两个比例关系上：加速度与合力成正比（a ∝ F），与质量成反比（a ∝ 1/m）。合并后得到 a ∝ F/m。引入比例常数 k，可写成 F = kma。通过定义 1 N 的力使 1 kg 物体产生 1 m s⁻² 的加速度，令 k = 1，便得到熟悉的 F = ma。
3. Deriving Kinetic Energy Eₖ = ½mv² | 推导动能 Eₖ = ½mv²
a distance s. The work done is W = Fs. Using F = ma and the motion equation v² = 0² + 2as ⇒ we substitute to get W = ma × (v²/(2a)) = ½mv². This work is stored as kinetic energy, so Eₖ = ½mv².
设一个恒定的合力 F 使物体从静止加速，移动距离 s。做功 W = Fs。利用 F = ma 和运动学公式 v² = 0² + 2as ⇒ s = v²/(2a)，代入得 W = ma × (v²/(2a)) = ½mv²。这部分功转化为动能，因此 Eₖ = ½mv²。
4. Deriving Gravitational Potential Energy Eₚ = mgh | 推导重力势能 Eₚ = mgh
gravity is W = force × distance = mg × h. This work is stored as gravitational potential energy, so Eₚ = mgh. The formula assumes a uniform gravitational field near the Earth’s surface.
以恒定速度将质量为 m 的物体竖直提升高度 h，需要施加等于其重力 mg 的向上力。克服重力所做的功为 W = 力 × 距离 = mg × h。这部分功被储存为重力势能，因此 Eₚ = mgh。该公式适用于地球表面附近的匀强重力场。
5. Deriving Power in Terms of Velocity P = Fv | 推导功率与速度的关系 P = Fv
time t is W = Fs, and the displacement s = vt. Substituting gives P = F(vt)/t = Fv. Thus, P = Fv for an object moving at constant speed against a force or being driven by a force.
功率是做功的快慢：P = W/t。当恒力 F 使物体以恒定速度 v 运动时，在时间 t 内做功 W = Fs，而位移 s = vt。代入得 P = F(vt)/t = Fv。因此，对于恒速运动或受力驱动的情形，有 P = Fv。
6. Deriving Pressure in a Liquid p = ρgh | 推导液体压强 p = ρgh
Consider a liquid column of density ρ, height h, and cross‑ . This shows that liquid pressure depends only on density, gravitational field strength, and depth, not on the area.
考虑截面积为 A、高为 h、密度为 ρ 的液柱。其重力为 mg = (ρAh)g。底部的压强为压力除以面积：p = F/A = ρAhg / A = ρgh。这表明液体压强只取决于密度、重力场强度和深度，而与截面积无关。
7. Deriving Resistors in Series R = R₁ + R₂ | 推导串联电阻 R = R₁ + R₂
l p.d.s: V = V₁ + V₂. Using Ohm’s law, V = IR, V₁ = IR₁, V₂ = IR₂. Therefore, IR = IR₁ + IR₂ ⇒ R = R₁ + R₂. The effective resistance is the sum.
对于串联电阻，通过每个电阻的电流 I 相同。组合两端的总电压 V 等于各电阻电压之和：V = V₁ + V₂。运用欧姆定律，V = IR，V₁ = IR₁，V₂ = IR₂。因此 IR = IR₁ + IR₂ ⇒ R = R₁ + R₂。等效电阻为各电阻之和。
8. Deriving the Transformer Voltage Ratio Vₚ/Vₛ = Nₚ/Nₛ | 推导变压器电压比 Vₚ/Vₛ = Nₚ/Nₛ
According to Faraday’s law, the induced e.m.f. is proportional to the number of turns: Vₚ ∝ Nₚ and Vₛ ∝ ower input equals power output, VₚIₚ = VₛIₛ, but the voltage ratio alone follows directly: Vₚ/Vₛ = Nₚ/Nₛ.
理想变压器假设无能量损失。初级和次级线圈的磁通量变化率相同。根据法拉第定律，在相同磁通变化下，感应电动势与线圈匝数成正比：Vₚ ∝ Nₚ，Vₛ ∝ Nₛ。虽然输入输出功率相等，VₚIₚ = VₛIₛ，但电压比可直接得出：Vₚ/Vₛ = Nₚ/Nₛ。
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