Work and Energy: AQA Physics Exam Essentials | A-Level AQA 物理：功与能量 考点精讲

📚 Work and Energy: AQA Physics Exam Essentials | A-Level AQA 物理：功与能量 考点精讲

Mastering work and energy is crucial for tackling mechanics questions in the AQA A-level Physics specification. This article breaks down every key concept—from definitions and formulas to power, efficiency, and conservation of energy—with paired English and Chinese explanations to help you revise effectively and avoid common mistakes.

掌握功与能量对于攻克 AQA A-level 物理力学题目至关重要。本文分解了从定义和公式到功率、效率以及能量守恒等每个关键概念，并配以中英文对照讲解，助你高效复习，避开常见错误。

1. Understanding Work and Energy | 理解功与能量

In physics, ‘work’ is the process of transferring energy from one form to another by means of a force. Work is done whenever a force makes something move. The unit of both work and energy is the joule (J). One joule is the work done when a force of 1 newton moves its point of application through a distance of 1 metre in the direction of the force.

在物理学中，“功”是通过力将能量从一种形式转化为另一种形式的过程。只要力使物体移动，就做了功。功和能量的单位都是焦耳 (J)。1 焦耳是 1 牛顿的力使其作用点沿力的方向移动 1 米所做的功。

It is vital to recognise that work is a scalar quantity, even though it involves force and displacement, which are both vectors. Only the component of the force parallel to the displacement contributes to the work done.

要注意功是标量，尽管它涉及力和位移这两个矢量。只有与位移平行的分力才贡献功。

2. Work Done by a Constant Force | 恒力所做的功

When a constant force F acts on an object and the object moves through a displacement s, the work done W is given by: W = F s cos θ, where θ is the angle between the force vector and the displacement vector. If the force and displacement are in the same direction, θ = 0°, cos 0° = 1, and W = F s. If the force is perpendicular to the displacement (θ = 90°), cos 90° = 0, and no work is done.

当一个恒力 F 作用在物体上，物体发生位移 s，所做的功 W 由公式 W = F s cos θ 给出，其中 θ 是力矢量与位移矢量之间的夹角。如果力和位移方向相同，θ = 0°，cos 0° = 1，W = F s。如果力垂直于位移 (θ = 90°)，cos 90° = 0，不做功。

For example, when you lift a suitcase vertically upwards, the force and displacement are parallel, so work is done on the suitcase, transferring energy to its gravitational potential store. When you carry the same suitcase horizontally at constant velocity, the upward force you exert is perpendicular to the displacement, so you do no work against gravity—only against any friction present.

例如，当你竖直向上提起行李箱时，力和位移平行，因此对行李箱做了功，将能量转移到它的重力势能储存中。当你水平匀速搬运同一个行李箱时，你施加的向上力与位移垂直，因此你对抗重力所做的功为零——只对抗可能存在的摩擦力做功。

3. Work Done by a Variable Force | 变力所做的功

When a force varies with displacement, the work done is found from the area under a force–displacement graph. For a non‑constant force, you may need to estimate the area by counting squares or using the trapezium rule. In AQA exams, this typically involves graphs where force changes linearly, so you can calculate the area of a trapezium.

当力随位移变化时，所做的功由力-位移图下的面积确定。对于非恒力，你可能需要通过数方格或使用梯形法则估算面积。在 AQA 考试中，这通常涉及力线性变化的图形，因此你可以计算梯形面积。

For instance, in stretching a spring according to Hooke’s law (F = k x), the work done in stretching the spring from extension 0 to extension x is the area under the force–extension graph, which is a triangle of area ½ F x. Substituting F = k x yields W = ½ k x², the energy stored as elastic potential energy.

例如，在根据胡克定律 (F = k x) 拉伸弹簧时，将弹簧从伸长量 0 拉伸到 x 所做的功是力-伸长量图下的面积，这是一个面积为 ½ F x 的三角形。代入 F = k x 得 W = ½ k x²，即储存为弹性势能的能量。

4. Kinetic Energy | 动能

Kinetic energy (Eₖ) is the energy an object possesses due to its motion. It is defined as: Eₖ = ½ m v², where m is the mass of the object and v is its speed. Kinetic energy is always positive or zero; it depends on speed squared, so doubling the speed quadruples the kinetic energy.

动能 (Eₖ) 是物体因运动而拥有的能量。定义为：Eₖ = ½ m v²，其中 m 是物体质量，v 是其速度。动能总是正值或零；取决于速度的平方，因此速度加倍会使动能变为原来的四倍。

In collisions, kinetic energy is not always conserved. In an elastic collision, total kinetic energy is conserved. In an inelastic collision, some kinetic energy is converted into other forms, such as thermal energy and sound, even though total energy is still conserved.

在碰撞中，动能并不总是守恒。在弹性碰撞中，总动能守恒。在非弹性碰撞中，部分动能转化为其他形式，如热能和声能，尽管总能量仍然守恒。

5. Gravitational Potential Energy | 重力势能

Gravitational potential energy (Eₚ or GPE) is the energy stored in an object due to its position in a gravitational field. Near the Earth’s surface, the change in GPE when an object of mass m is raised through a vertical height Δh is: ΔEₚ = m g Δh, where g is the gravitational field strength (9.81 N kg⁻¹ on Earth).

重力势能 (Eₚ 或 GPE) 是物体因在引力场中的位置而储存的能量。在地球表面附近，当质量为 m 的物体被提升竖直高度 Δh 时，重力势能的变化为：ΔEₚ = m g Δh，其中 g 是重力场强度（地球上为 9.81 N kg⁻¹）。

The zero of GPE can be chosen arbitrarily; only changes in GPE matter. Typically, ground level is taken as the zero reference. When a mass falls, its GPE decreases as it is converted into kinetic energy (if no other forces act).

重力势能的零点可以任意选取；只有重力势能的变化才是重要的。通常选择地平面为零参考点。当物体下落时，其重力势能减少，转化为动能（如果其他力作用不计）。

6. Elastic Potential Energy | 弹性势能

Elastic potential energy is the energy stored in a deformed elastic object, such as a stretched spring or a compressed rubber band. For a spring that obeys Hooke’s law (F = k x, where k is the spring constant), the elastic potential energy stored when stretched or compressed by an amount x from its natural length is: Eₑ = ½ k x².

弹性势能是储存在形变的弹性物体中的能量，例如被拉伸的弹簧或压缩的橡皮筋。对于符合胡克定律 (F = k x，其中 k 为弹簧常数) 的弹簧，当其从自然长度被拉伸或压缩距离 x 时储存的弹性势能为：Eₑ = ½ k x²。

This energy is recoverable: if the spring is released, the stored Eₑ can do work, converting into kinetic energy or other forms. The formula assumes the spring’s elastic limit is not exceeded, so it returns to its original shape.

这种能量是可回收的：若释放弹簧，储存的 Eₑ 可以做功，转化为动能或其他形式。公式假设未超出弹簧的弹性极限，因此它能恢复原状。

7. The Principle of Conservation of Energy | 能量守恒原理

The principle of conservation of energy states that energy cannot be created or destroyed; it can only be transferred from one form to another or from one place to another. In a closed system, the total energy remains constant.

能量守恒原理指出，能量既不能被创造也不能被消灭；它只能从一种形式转移到另一种形式，或从一个地方转移到另一个地方。在一个封闭系统中，总能量保持不变。

In mechanics problems, this principle is often applied by equating the initial total energy to the final total energy, accounting for any work done against resistive forces. For example, for a falling object without air resistance: loss in GPE = gain in KE, so m g h = ½ m v², which allows the speed to be found.

在力学问题中，该原理通常通过令初始总能量等于最终总能量来应用，并将对抗阻力所做的功考虑在内。例如，对于无空气阻力的下落物体：减少的重力势能 = 增加的动能，因此 m g h = ½ m v²，从而可以求出速度。

When non‑conservative forces such as friction or air resistance do work, some mechanical energy is dissipated as thermal energy, so total mechanical energy is not conserved, but total energy (including thermal) is still conserved.

当摩擦力或空气阻力等非保守力做功时，部分机械能被耗散为热能，因此总机械能不守恒，但总能量（包括热能）仍然守恒。

8. The Work–Energy Theorem | 功–能定理

The work–energy theorem states that the net work done on an object is equal to the change in its kinetic energy: W_net = ΔEₖ = ½ m v² − ½ m u², where u is the initial speed and v is the final speed. This theorem is extremely useful for solving problems where forces vary or multiple forces act, because it avoids detailed analysis of acceleration and time.

功–能定理指出，物体上所做的净功等于其动能的变化：W_net = ΔEₖ = ½ m v² − ½ m u²，其中 u 为初速度，v 为末速度。该定理对于解决受力变化或多力作用的问题非常有用，因为它避免了加速度和时间的详细分析。

For example, if a car of mass 800 kg, initially at rest, experiences a net forward force of 2000 N over 50 m, the work done by the net force is 2000 N × 50 m = 100 000 J. By the work–energy theorem, this equals ½ m v², so 100 000 = ½ × 800 × v², giving v = 15.8 m s⁻¹.

例如，一辆质量为 800 kg 的汽车，最初静止，受到 2000 N 的合外力作用行驶 50 m，合外力做功为 2000 N × 50 m = 100 000 J。根据功–能定理，这等于 ½ m v²，因此 100 000 = ½ × 800 × v²，解得 v = 15.8 m s⁻¹。

9. Power | 功率

Power (P) is the rate of doing work or the rate of transferring energy. It is defined as: P = W / t, where W is the work done in time t. The SI unit of power is the watt (W), where 1 W = 1 J s⁻¹. Another useful expression for mechanical power is P = F v, where F is a constant force acting in the direction of velocity v. This formula is particularly useful when an engine maintains a steady speed against resistive forces.

功率 (P) 是做功的速率或能量传递的速率。定义为：P = W / t，其中 W 是在时间 t 内所做的功。功率的国际单位是瓦特 (W)，1 W = 1 J s⁻¹。另一个有用的机械功率表达式是 P = F v，其中 F 是作用于速度方向的恒力。该公式在发动机对抗阻力保持稳定速度时特别有用。

In many AQA questions, you will be asked to find the power required to lift a load or drive a vehicle at constant speed. For instance, lifting a 10 kg mass through 2 m in 4 s against gravity (g = 9.81 N kg⁻¹) requires work W = m g h = 10 × 9.81 × 2 = 196.2 J, so power P = 196.2 / 4 = 49.05 W.

在许多 AQA 题中，你会被要求求出提升重物或驱动车辆匀速行驶所需的功率。例如，用 4 s 将 10 kg 的物体逆重力提升 2 m（g = 9.81 N kg⁻¹），需要做功 W = m g h = 10 × 9.81 × 2 = 196.2 J，因此功率 P = 196.2 / 4 = 49.05 W。

10. Efficiency | 效率

Efficiency (η) is the ratio of useful output energy (or power) to total input energy (or power), often expressed as a percentage: η = (useful output / total input) × 100%. In any real process, efficiency is less than 100% because of energy dissipation mainly through heating.

效率 (η) 是有用输出能量（或功率）与总输入能量（或功率）的比值，通常以百分比表示：η = （有用输出 / 总输入）× 100%。在任何实际过程中，效率都小于 100%，主要是因为能量通过加热等方式耗散。

For example, an electric motor lifts a mass, using 200 J of electrical energy to gain 150 J of gravitational potential energy. Its efficiency is (150 / 200) × 100% = 75%. The remaining 50 J is dissipated as thermal energy in the motor and its surroundings.

例如，一台电动机提升物体，使用了 200 J 的电能却获得了 150 J 的重力势能。其效率为 (150 / 200) × 100% = 75%。剩下的 50 J 以热能形式耗散在电机及其周围环境中。

11. Energy in Systems: Slopes and Friction | 系统能量：斜面与摩擦

When an object slides down a smooth inclined plane (no friction), the loss in GPE is completely converted into kinetic energy. If friction is present, some mechanical energy is converted into thermal energy, reducing the kinetic energy gained. The work done against friction is typically the frictional force multiplied by the distance along the slope.

当一个物体沿光滑斜面（无摩擦）下滑时，减少的重力势能完全转化为动能。如果存在摩擦力，部分机械能被转化为热能，从而减少了获得的动能。克服摩擦力所做的功通常等于摩擦力乘以沿斜面的距离。

For a rough slope, the energy equation becomes: m g h = ½ m v² + F_friction × d, where d is the distance moved along the plane. This allows you to find v, F_friction, or d as required. Understanding these energy transfers is a core AQA skill.

对于粗糙斜面，能量方程变为：m g h = ½ m v² + F_friction × d，其中 d 是沿斜面的运动距离。由此可根据需要求出 v、F_friction 或 d。理解这些能量转换是 AQA 的核心技能。

12. Exam Tips and Common Pitfalls | 考试技巧与常见错误

(1) Always use SI units: mass in kg, displacement in m, force in N, time in s, energy in J. (2) Remember that work done by a force perpendicular to motion is zero. (3) When using ΔEₚ = m g Δh, Δh is the vertical height change, not the path length. (4) For springs, check whether the question asks for the extension from the natural length or the total length; Eₑ uses extension (or compression) x. (5) In calculations involving efficiency, don’t confuse input and output. (6) The work–energy theorem saves time in multiforce problems, but ensure you are using the net work, not just the work done by one force.

(1) 始终使用国际单位：质量以 kg、位移以 m、力以 N、时间以 s、能量以 J 为单位。(2) 记住垂直于运动方向的力做功为零。(3) 使用 ΔEₚ = m g Δh 时，Δh 是竖直高度的变化，而非路径长度。(4) 对于弹簧，确认题目要求的是距自然长度的伸长量还是总长度；Eₑ 使用伸长量（或压缩量）x。(5) 在涉及效率的计算中，不要混淆输入和输出。(6) 功–能定理在多力问题中可节省时间，但要确保使用净功，而不仅仅是某一个力所做的功。

Also, be attentive to ‘smooth’ (no friction) or ‘rough’ (friction present) in exam questions. Watch for key words like ‘constant speed’—if an object moves at constant speed, net force is zero and net work is zero, so kinetic energy doesn’t change. Finally, always draw a clear diagram and mark all forces, displacements, and angles.

此外，留心考题中的“光滑”（无摩擦）或“粗糙”（有摩擦）表述。注意像“恒定速度”这样的字眼——若物体匀速运动，合力为零，净功为零，因此动能不变。最后，务必画一张清晰的图，标出所有力、位移和角度。

Published by TutorHao | Physics Revision Series | aleveler.com

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