Calculating Energy Changes: Methods and Worked Examples | 能量变化的计算方法与例题精讲

📚 Calculating Energy Changes: Methods and Worked Examples | 能量变化的计算方法与例题精讲

Energy is a central concept in A-Level Physics. This article explains the essential formulas for calculating changes in kinetic, potential, elastic, and thermal energy, along with worked examples that follow the CIE syllabus. You will learn how to apply conservation of energy, define efficiency, and solve multi-step problems confidently.

能量是 A-Level 物理的核心概念。本文系统讲解动能、势能、弹性势能和热能变化的计算方法，并配以符合 CIE 考纲的例题精讲。你将学会应用能量守恒定律、定义效率，并自信地解决多步骤问题。

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1. Conservation of Energy | 能量守恒定律

The principle of conservation of energy states that energy cannot be created or destroyed, only transferred from one form to another. The total energy of an isolated system remains constant.

能量守恒定律指出：能量不能被创造或消灭，只能从一种形式转化为另一种形式。孤立系统的总能量保持不变。

For calculations, the key equation is:

ΔE_total = 0 (for an isolated system)

When solving problems, always identify the initial energy forms and the final energy forms. For example, if an object falls freely, its gravitational potential energy decreases while its kinetic energy increases by the same amount (assuming no air resistance).

在解题时，务必先判断初始能量形式和末态能量形式。例如，物体自由下落时，重力势能减少，动能等量增加（假设无空气阻力）。

• Energy transfer does not change total energy.
• 能量转移不会改变总能量。
• Identify energy stores before and after the process.
• 判断过程前后的能量储存形式。

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2. Kinetic Energy and Gravitational Potential Energy | 动能与重力势能

An object of mass m moving at speed v has kinetic energy (EK) given by:

E_K = ½ m v²

质量为 m、速度为 v 的物体所具有的动能为：

E_K = ½ m v²

The gravitational potential energy (EP) of an object near the Earth’s surface is:

E_P = m g h

物体在地球表面附近的重力势能为：

E_P = m g h

Here, g is the gravitational field strength (9.81 m s⁻² on Earth) and h is the height above a chosen reference level. In CIE questions, g is often taken as 9.81 m s⁻² unless stated otherwise.

其中 g 为重力场强度（地球上取 9.81 m s⁻²），h 是相对于所选参考面的高度。在 CIE 考试中，除特别说明外，通常取 g = 9.81 m s⁻²。

Quantity 物理量	Formula 公式	Unit 单位
Kinetic energy 动能	E_K = ½ m v²	J (kg m² s⁻²)
Gravitational potential energy 重力势能	E_P = m g h	J

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3. Elastic Potential Energy | 弹性势能

For a spring or material obeying Hooke’s law, the elastic potential energy stored when extended or compressed by a displacement x from its natural length is:

E_el = ½ k x²

对于遵守胡克定律的弹簧或材料，当其从自然长度被拉伸或压缩位移 x 时，储存的弹性势能为：

E_el = ½ k x²

Here k is the spring constant (force per unit extension) in N m⁻¹. This formula assumes the elastic limit is not exceeded.

其中 k 是劲度系数（单位伸长所受的力），单位为 N m⁻¹。此公式要求不超过弹性限度。

In an energy calculation, elastic potential energy may be converted into kinetic energy, for example, when a stretched catapult is released.

在能量计算中，弹性势能可以转化为动能，例如释放被拉伸的弹弓时。

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4. Work Done and Energy Transfers | 功与能量转化

Work done by a constant force F moving an object through a displacement s in the direction of the force is:

W = F s cos θ

恒力 F 使物体沿力的方向位移 s 时所做的功为：

W = F s cos θ

where θ is the angle between the force and displacement. When θ = 0°, W = F s. Work done is equal to the energy transferred to the object.

其中 θ 是力与位移之间的夹角。当 θ = 0° 时，W = F s。做功等于物体获得的能量。

If a force opposes motion (e.g., friction), the work done by friction removes energy from the system, usually dissipated as heat.

如果力阻碍运动（例如摩擦力），摩擦力做功会使系统能量减少，通常以热能形式耗散。

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5. Thermal Energy and Specific Heat Capacity | 热能变化与比热容

When a substance of mass m changes temperature by Δθ, the thermal energy change is:

Q = m c Δθ

质量为 m 的物质温度变化 Δθ 时，热能变化为：

Q = m c Δθ

Here c is the specific heat capacity in J kg⁻¹ K⁻¹. In A-Level problems, this is often used to calculate energy absorbed or released during heating.

其中 c 是比热容，单位 J kg⁻¹ K⁻¹。在 A-Level 题目中，该公式常用于计算加热或冷却时吸收或释放的能量。

For phase changes, latent heat is used: Q = m L, but for most energy-change problems in mechanics, thermal effects are either ignored or treated as ‘wasted’ energy.

对于相变，使用潜热：Q = m L。但在大多数力学能量变化问题中，热效应要么被忽略，要么被视为“耗散”能量。

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6. Efficiency | 效率

Efficiency measures the fraction of input energy that is converted into useful output energy. It is a ratio without units, often expressed as a percentage.

效率衡量输入能量中有多少转化为有用的输出能量。它是一个无量纲比值，通常用百分比表示。

Efficiency = (Useful output energy / Total input energy) × 100%

效率 =（有用输出能量 / 总输入能量）× 100%

Alternatively, using power, efficiency = (Useful output power / Total input power) × 100%.

或者用功率表示：效率 =（有用输出功率 / 总输入功率）× 100%。

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7. Power and Energy | 功率与能量

Power is the rate of energy transfer or work done. The average power is:

P = W / t = ΔE / t

功率是能量转化或做功的速率。平均功率为：

P = W / t = ΔE / t

The unit of power is the watt (W), equal to one joule per second (1 W = 1 J s⁻¹). For a constant force acting on an object moving at constant speed, P = F v.

功率的单位是瓦特（W），等于焦耳每秒（1 W = 1 J s⁻¹）。当恒力作用于匀速运动的物体时，P = F v。

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8. Example 1: A Falling Object | 例题 1：自由落体

A 2.0 kg ball is dropped from a height of 15 m. Calculate its speed just before hitting the ground. Ignore air resistance.

一个 2.0 kg 的小球从 15 m 高处自由下落。忽略空气阻力，计算它落地前瞬间的速度。

The ball initially has gravitational potential energy only. Just before impact, this energy is entirely converted into kinetic energy.

小球最初只有重力势能。落地前瞬间，这些能量全部转化为动能。

E_P = m g h = 2.0 × 9.81 × 15 = 294.3 J

This equals the kinetic energy at impact:

此动能等于落地时的动能：

E_K = ½ m v² = 294.3 J

Solving for v:

解出 v：

v = √(2 × 294.3 / 2.0) = √294.3 ≈ 17.2 m s⁻¹

Thus the ball hits the ground at approximately 17.2 m s⁻¹.

因此小球落地时速度约为 17.2 m s⁻¹。

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9. Example 2: A Block Sliding Down a Rough Slope | 例题 2：粗糙斜面滑块

A 3.0 kg block slides down a slope of length 4.0 m inclined at 30° to the horizontal. The coefficient of kinetic friction between the block and the slope is 0.20. Calculate the speed of the block at the bottom of the slope if it starts from rest.

一个 3.0 kg 的滑块从与水平面成 30°、长度为 4.0 m 的斜面顶端由静止开始下滑。滑块与斜面间的动摩擦因数为 0.20。求滑块到达斜面底端时的速度。

First, find the vertical height of the slope.

首先求斜面的竖直高度。

h = 4.0 sin 30° = 4.0 × 0.5 = 2.0 m

The initial gravitational potential energy is:

初始重力势能为：

E_P = 3.0 × 9.81 × 2.0 = 58.86 J

The normal reaction is N = m g cos 30° = 3.0 × 9.81 × 0.866 ≈ 25.5 N. The friction force is f = μ N = 0.20 × 25.5 ≈ 5.10 N.

法向反力 N = m g cos 30° = 3.0 × 9.81 × 0.866 ≈ 25.5 N。摩擦力 f = μ N = 0.20 × 25.5 ≈ 5.10 N。

Work done against friction over a distance of 4.0 m:

在 4.0 m 距离上克服摩擦力做功：

W_f = f × s = 5.10 × 4.0 = 20.4 J

The remaining energy is converted into kinetic energy:

剩余能量转化为动能：

E_K = E_P − W_f = 58.86 − 20.4 = 38.46 J

Using E_K = ½ m v²:

根据 E_K = ½ m v²：

v = √(2 × 38.46 / 3.0) = √25.64 ≈ 5.06 m s⁻¹

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10. Example 3: Efficiency of a Motor | 例题 3：电动机效率

An electric motor lifts a 500 kg mass at a steady speed of 0.40 m s⁻¹. The electrical input power is 5.0 kW. Calculate the useful output power and the efficiency of the motor.

电动机以 0.40 m s⁻¹ 的恒定速度提升一个 500 kg 的重物。电输入功率为 5.0 kW。计算有用输出功率和电动机的效率。

The useful power is the rate at which the motor does work to lift the mass against gravity.

有用功率是电动机克服重力提升重物做功的速率。

P_useful = F v = m g v = 500 × 9.81 × 0.40 = 1962 W

Convert input power: 5.0 kW = 5000 W. Efficiency is:

将输入功率转换为 5.0 kW = 5000 W。效率为：

Efficiency = (1962 / 5000) × 100% ≈ 39.2%

The motor is about 39% efficient; the remaining 61% of input energy is dissipated as heat, sound, and internal friction.

电动机的效率约为 39%；其余 61% 的输入能量以热、声和内部摩擦的形式耗散。

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11. Worked Example 4: Spring-Launched Object | 例题 4：弹簧发射物体

A spring with spring constant 800 N m⁻¹ is compressed by 0.15 m to launch a 0.50 kg block horizontally along a frictionless surface. Calculate the speed of the block immediately after release.

一根劲度系数为 800 N m⁻¹ 的弹簧被压缩 0.15 m，用来在光滑水平面上发射一个 0.50 kg 的物块。计算释放后物块的速度。

The elastic potential energy stored in the spring equals the kinetic energy gained by the block (since there is no friction).

弹簧储存的弹性势能等于物块获得的动能（因为无摩擦）。

E_el = ½ k x² = ½ × 800 × (0.15)² = 9.0 J

Set this equal to the kinetic energy:

令其等于动能：

½ m v² = 9.0 J

v = √(2 × 9.0 / 0.50) = √36 = 6.0 m s⁻¹

The block leaves the spring with a speed of 6.0 m s⁻¹.

物块离开弹簧时的速度为 6.0 m s⁻¹。

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12. Summary of Key Formulas | 核心公式总结

The following table summarises the essential energy-change formulas you need for CIE A-Level Physics.

下表总结了 CIE A-Level 物理中必备的能量变化公式。

Energy change 能量变化	Formula 公式	Notes 备注
Kinetic 动能	E_K = ½ m v²	v is speed relative to reference frame
Gravitational potential 重力势能	E_P = m g h	h is height change
Elastic potential 弹性势能	E_el = ½ k x²	Only within elastic limit
Work done 做功	W = F s cos θ	θ is angle between force and displacement
Thermal 热能	Q = m c Δθ	c is specific heat capacity
Power 功率	P = ΔE / t or P = F v	v is speed in direction of force

Always define the system and choose a reference level for potential energy before starting a calculation. Check that your final units are joules, and remember that energy is a scalar quantity.

解题前务必明确系统并选择势能参考面，检查最终单位是否为焦耳，并记住能量是标量。

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