Complete Physics for Cambridge Secondary Workbook: Formula Derivations | 剑桥中学物理练习册公式推导

📚 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

We start from the definition of acceleration: a = (v − u) / t. Rearranging gives t = (v − u) / a. Substituting this into the displacement equation s = ut + ½at² 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

Newton’s second 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 ∝ F/m. Introducing a constant of proportionality k, we write F = kma. By defining 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

Consider a constant net force F accelerating an object from rest over a distance s. The work done is W = Fs. Using F = ma and the motion equation v² = 0² + 2as ⇒ s = v²/(2a), 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

To lift an object of mass m vertically through height h at constant speed, an upward force equal to its weight mg must be applied. The work done against 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

Power is the rate of doing work: P = W/t. When a constant force F moves an object at steady speed v, the work done in 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‑sectional area A. Its weight is mg = (ρAh)g. The pressure at the bottom is force per unit area: p = F/A = ρAhg / A = ρgh. 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

For resistors in series, the current I is the same through each. The total potential difference V across the combination equals the sum of individual 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

An ideal transformer assumes no energy loss. The rate of change of magnetic flux is the same in the primary and secondary coils. According to Faraday’s law, the induced e.m.f. is proportional to the number of turns: Vₚ ∝ Nₚ and Vₛ ∝ Nₛ for the same flux change. Since power 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。

Published by TutorHao | Physics Revision Series | aleveler.com

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