📚 Derivation of Kinetic Energy Formula (Ek = ½mv²) | 动能公式推导 (Ek = ½mv²)
eed v, the kinetic energy is given by Ek = ½mv². This formula can be derived from the work– aning of kinetic energy but also provides a firm foundation for tackling related problems in the PH02 International AS Physics examination.
动能是物体由于运动而具有的能量。对于质量为 m、速度为 v 的物体，动能公式为 Ek = ½mv²。这一公式可以通过功–能原理与匀加速运动方程联合推导出来。透彻理解推导过程，不仅能加深对动能物理含义的把握，也能为解答 PH02 国际 AS 物理考试中的相关题目打下坚实基础。
1. Work–Energy Principle | 功–能原理
When a net force does work on an object, it transfers energy to that object. If no other energy transfers occur, the work done on the object equals the change in its kinetic energy. This is the work–energy principle, a cornerstone of classical mechanics: Wnet r a single constant force accelerating an object from an initial speed u to a final speed v over a displacement s.
当有净力对物体做功时，它会把能量传递给该物体。如果没有其他能量转移，对物体所做的功等于其动能的变化量。这就是功–能原理，它是经典力学的基石：Wnet = ΔEk。为了推导公式，我们考虑一个恒力使物体产生加速度，使其速度从初速 u 变化到末速 v，位移为 s。
2. Defining Work Done by a Constant Force | 恒力做功的定义
imple relation is valid only when the force and displacement are parallel. In our derivation, we assume the object moves along a straight line under the action of a constant net force.
对于沿位移 s 方向作用的恒力 F，它所做的功 W 定义为力与位移的乘积：W = F s。这一简单关系式只在力与位移同向时才成立。在我们的推导中，假设物体在恒定的净力作用下沿直线运动。
3. Applying Newton's Second Law | 应用牛顿第二定律
Newton' is into the work equation gives W = (m a) s. At this stage, we have expressed work in terms of mass, acceleration, and displacement.
牛顿第二定律指出，作用在恒定质量 m 物体上的净力 F 与其加速度 a 成正比：F = m a。将它代入功的表达式，得到 W = (m a) s。到这一步，我们已经用质量、加速度和位移把功表示出来了。
4. Key Kinematic Equation | 关键运动学方程
a, and displacement s: v² = u² + 2 a s. This equation is derived from the definitions of velocity and acceleration and is valid when acceleration is constant. Rearranging it allows us to eliminate the product a s.
对于匀加速运动，其中一个标准 SUVAT（运动学）方程将初速度 u、末速度 v、加速度 a 和位移 s 联系在一起：v² = u² + 2 a s。该方程由速度和加速度的定义导出，且仅在加速度恒定时成立。通过移项整理，我们可以消去乘积项 a s。
5. Expressing 'a s' from Kinematics | 从运动学中表达 a s
entral to the derivation because it introduces the dependence on the square of the speeds.
从运动学方程中分离出项 a s：2 a s = v² − u²，因此 a s = (v² − u²) / 2。这个表达式在推导中至关重要，因为它引入了对速度平方的依赖关系。
6. Substituting into the Work Expression | 代入功的表达式
y the net force is now expressed purely in terms of mass and the change in the square of speed.
现在我们将功的表达式 W = m a s 与结果 a s = (v² − u²)/2 结合起来。得到：W = m × (v² − u²)/2，或者更整齐地写成 W = ½ m (v² − u²)。这样，净力所做的功就完全用质量和速度平方的变化量表示了。
7. Linking Work to Kinetic Energy | 将功与动能联系起来
According to the work–ergy principle, the net work done on an object equals its change in kinetic energy: W = Ekfinal − Ekinitial. If we define the kinetic energy of an object moving at speed v as ½ m v², then the change in kinetic energy is exactly ½ m v² − ½ m u², which matches our expression for W.
依据功–能原理，对物体所做的净功等于其动能的变化量：W = Ek末 − Ek初。如果我们将以速度 v 运动的物体的动能定义为 ½ m v²，那么动能的变化量恰好就是 ½ m v² − ½ m u²，这与我们得到的 W 表达式完全一致。
8. The Derived Formula | 推导出的公式
Ek = ½ m v²
e brings it from zero kinetic energy to ½ m v². This confirms that the kinetic energy of any object of mass m moving at speed v is indeed ½mv². The derivation shows no dependence on the type of force—only that it is constant and does net work.
因此，如果物体从静止开始（u = 0），净力对它所做的功就会使其动能从零增加到 ½ m v²。这就证实了任何质量为 m、速度为 v 的物体的动能确实是 ½ m v²。推导过程表明，这一结果与力的种类无关——唯一的要求是力保持恒定并做净功。
9. Assumptions Made | 所作的假设
The derivation rests on several assumptions: the mass m is constant, the net force is constant in both magnitude and direction, the motion is in a straight line, and no energy is lost to other forms (e.g., heat, sound). These idealizations are typical for AS-level mechanics and yield the exact kinetic energy expression used throughout physics.
这个推导建立在若干假设之上：质量 m 恒定，净力的大小和方向不变，运动沿直线进行，并且能量没有转化为其他形式（如热能、声能）。这些理想化条件是 AS 阶段力学中常见的简化处理，推导出的动能表达式在整个物理学中都被准确使用。
10. Units and Dimensions | 单位与量纲
² = J. Dimensional analysis gives [M][L]²[T]⁻², which matches the dimensions of work and all forms of energy, confirming internal consistency.
动能是标量，在国际单位制中用焦耳（J）来度量。分解单位：质量用 kg，速度用 m/s，因此 kg·(m/s)² = kg·m²/s² = J。量纲分析给出 [M][L]²[T]⁻²，这与功和各种形式能量的量纲相同，验证了体系的内在一致性。
11. Illustrative Example | 示例说明
0 000 J. If the acceleration occurred over a distance of 150 m, the average net force can be found from W = F s, giving F = 120 000 / 150 = 800 N. This illustrates the direct link among force, work, and kinetic energy change.
一辆质量为 800 kg 的汽车从 10 m/s 匀加速到 20 m/s。其动能的增加量为 ΔEk = ½ × 800 × (20² − 10²) = 400 × (400 − 100) = 400 × 300 = 120 000 J。若这一加速过程发生在 150 m 的距离内，则可由 W = F s 求出平均净力，F = 120 000 / 150 = 800 N。这直观展示了力、功和动能变化之间的直接联系。
12. Summary and Exam Tips | 总结与应试提示
ked in a PH02-style question, always state the assumptions clearly, lay out the equations step by step, and be explicit about the link between net work and change in kinetic energy. Memorising the derivation helps with understanding related concepts such as gravitational potential energy and elastic potential energy, which follow analogous reasoning.
推导的核心在于将 W = F s、F = m a 和 v² = u² + 2 a s 结合起来。当在 PH02 类试题中遇到此类问题时，务必清晰地陈述假设条件，逐步列出方程，并明确净功与动能变化之间的联系。记住这个推导有助于理解重力势能和弹性势能等相关概念，因为它们遵循类似的推理过程。
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