📚 AQA OxfordAQA 9665 FM03 WRE June 2023: Work, Rest and Energy in Further Mechanics | AQA OxfordAQA 9665 FM03 WRE 2023年6月：高等力学中的功、能与守恒
The June 2023 paper for AQA OxfordAQA 9665 FM03 (WRE) tests your ability to apply fundamental mechanical principles to unfamiliar situations. Although the acronym WRE in the paper code may simply identify the written response examination, for revision purposes we can remember it as Work, Rest and Energy – the three anchors of a superb Further Mechanics performance. This guide condenses the essential theory, common question styles, and exam traps you need to master.
2023年6月的 AQA OxfordAQA 9665 FM03（WRE）试卷，旨在考查你将基础力学原理应用于陌生情境的能力。尽管试卷代码中的 WRE 可能只是“书面作答考试”的标识，但在复习时我们不妨把它记作 “功、静与能”（Work, Rest and Energy）——这是取得高等力学高分的三个支柱。本指南浓缩了必修理论、常见题型和考场陷阱，助你全面备考。
1. Understand the Exam Structure | 了解试卷结构
FM03 is a one-hour written paper, usually consisting of six to eight multi-part questions. Topics are drawn from the Further Mechanics section of the OxfordAQA 9665 specification. Marks are awarded for clear method, correct substitutions, and accurate final answers. You should practise converting physical word problems into symbolic equations before inserting numbers.
FM03 是一份时长一小时的书面试卷，通常包含六至八道多小问的题目。考点覆盖 OxfordAQA 9665 考纲中的高等力学部分。得分点在于清晰的解题步骤、正确的代数代入以及准确的最终答案。你应当练习先把实际应用题转化为符号方程，再代入数值。
Always define positive direction and state units.
始终先定义正方向并注明单位。
Draw a diagram showing forces and velocities.
画出受力图与速度方向图。
Use g = 9.8 m s⁻² unless told otherwise.
如无特别说明，取 g = 9.8 m s⁻²。
2. Work Done by a Constant Force | 恒力做功
Work is the product of the force and the displacement in the direction of the force. For a constant force F acting on a body moving along a straight line, the work done is
W = F s cos θ
the angle between F and the direction of motion. When θ = 0°, W = F s; when θ = 180°, W = –F s, meaning the force resists the motion.
功是力与沿力方向上的位移的乘积。对于沿直线运动的物体上的恒力 F，做功为
W = F s cos θ
其中 s 是位移，θ 是 F 与运动方向的夹角。当 θ = 0° 时，W = F s；当 θ = 180° 时，W = –F s，表示该力阻碍运动。
If you lift a mass m vertically to height h, the work done against gravity is mgh. This is stored as gravitational potential energy. In a horizontal push with friction, work is transferred to thermal energy – a classic FM03 trap.
若将质量 m 的物体竖直提升高度 h，则克服重力做功为 mgh，这部分能量以重力势能的形式储存。在水平推动物体时，克服摩擦力做功会转化为热能——这是 FM03 的经典陷阱。
3. Work Done by a Variable Force | 变力做功
When the force varies with position, work equals the area under the F–x graph. Mathematically,
W = ∫ F dx
ooke’s law, tension T = λ x / l, and the work done in stretching from a natural length is:
W = (λ x²) / (2l)
当力随位置变化时，做功等于 F–x 图像下的面积。数学表达式为
W = ∫ F dx
在 FM03 中，你可能会遇到线性或二次型的 F(x) 关系。积分时要特别小心，并牢记功是标量。对于遵守胡克定律的弹性绳，张力 T = λ x / l，从自然长度拉伸距离 x 所做的功为：
W = (λ x²) / (2l)
Do not confuse modulus of elasticity λ with the extension x; λ has units of newtons and depends on the material and cross-section of the string.
切勿将弹性模量 λ 与伸长量 x 混淆；λ 的单位是牛顿，由材料的性质与横截面积决定。
4. Kinetic Energy and the Work–Energy Principle | 动能与动能定理
The kinetic energy of a particle of mass m moving with speed v is
Eₖ = ½ m v²
The work–energy principle states that the total work done by external forces on a particle equals its change in kinetic energy:
Wₒₜₐₗ = ΔEₖ = ½ m v² – ½ m u²
质量 m、速度 v 的质点的动能为
Eₖ = ½ m v²
动能定理表明：作用于质点的所有外力所做的总功，等于质点动能的变化量：
Wₒₜₐₗ = ΔEₖ = ½ m v² – ½ m u²
In a question involving a block sliding down a rough inclined plane, consider all works: gravitational work (positive), friction work (negative), and normal reaction work (zero). Use the energy balance to solve for the final speed.
在物体沿粗糙斜面下滑的问题中，需要计入所有功：重力做功（正功）、摩擦力做功（负功）、法向支持力做功（为零）。利用能量守恒关系求出末速度。
5. Gravitational Potential Energy and Energy Conservation | 重力势能与机械能守恒
For a particle close to the Earth’s surface, gravitational potential energy is simply mgh. In FM03, you may also encounter slices of a lamina or a chain whose centre of mass changes. Here you need to compute the change in height of the centre of mass.
对于接近地球表面的质点，重力势能简化为 mgh。在 FM03 中，你可能会遇到薄板或链条的重心改变。此时需要计算重心高度的变化。
If no non-gravitational force does work, the total mechanical energy is conserved:
KEₜₒₚ + PEₜₒₚ = KE_bₒₜₜₒₘ + PE_bₒₜₜₒₘ
若没有非重力做功，则总机械能守恒：
动能₁ + 势能₁ = 动能₂ + 势能₂
When a smooth bead slides on a wire, the normal reaction does no work, so the energy equation alone is sufficient. Never add the normal reaction into an energy balance.
光滑珠在细线上滑动时，法向支持力不做功，因此仅用能量方程即可。绝不要把支持力计入能量平衡。
6. Elastic Springs and Strings: Hooke’s Law and Elastic Potential Energy | 弹性弹簧与弹性绳：胡克定律与弹性势能
For a string of natural length l and modulus of elasticity λ, the tension when stretched by x is
T = (λ x) / l
The elastic potential energy stored is
Eₚ = (λ x²) / (2l)
对自然长度为 l、弹性模量为 λ 的弹性绳，当伸长量为 x 时，张力为
T = (λ x) / l
储存的弹性势能为
Eₚ = (λ x²) / (2l)
For a spring, use spring constant k = λ / l string cannot compress, so x must be non-negative; a spring can have both positive and negative x.
对于弹簧，引入劲度系数 k = λ / l，则 T = k x，Eₚ = ½ k x²。注意：弹性绳不能压缩，x ≥ 0；而弹簧可被拉伸或压缩。
Exam questions often combine elastic energy with gravitational energy. For example, a mass released from rest at the natural length of a string will reach maximum speed when its weight equals the tension. Solve this using both energy and dynamics.
考试题常将弹性势能与重力势能结合。例如，质量从弹性绳自然长度处静止释放，当重力等于张力时达到最大速度。此时需要综合使用能量方程与动力学方程求解。
7. Power and Efficiency | 功率与效率
Power is the rate of doing work. If a constant force F acts on a particle moving at speed v in the same direction, then
P = F v
When the force varies, use P = dW/dt.
功率是做功的快慢。若恒力 F 与质点速度 v 同向，则
P = F v
当力变化时，使用 P = dW/dt。
For a vehicle climbing a slope at constant speed, the engine power must overcome both resistive forces and the component of weight along the slope. Suppose resistance R and slope angle α, then
P = (R + m g sin α) v
对于沿斜坡匀速上升的车辆，发动机功率必须克服阻力和重力沿斜面的分量。设阻力为 R、斜面倾角为 α，则
P = (R + m g sin α) v
anics questions, simply treat efficiency as a decimal factor: output = efficiency × input.
效率定义为有用输出功率与输入功率之比。在力学题中，只需将效率作为小数因子：输出 = 效率 × 输入。
8. Rest and Equilibrium: Moments and Centres of Mass | 静止与平衡：力矩与质心
The word “rest” in WRE reminds us to check equilibrium conditions. For a rigid body to be at rest, both the resultant force and the resultant moment must be zero:
ΣFₓ = 0, ΣFᵧ = 0, ΣM = 0
WRE 中的“静”提醒我们检查平衡条件。对于静止的刚体，合力与合力矩都必须为零：
ΣFₓ = 0, ΣFᵧ = 0, ΣM = 0
When a rod rests on two supports, take moments about one support to find the reaction at the other. If a body is on the point of sliding, friction is at its limiting value F = μR. Remember that friction opposes relative motion – draw its direction carefully.
当杆搁在两个支点上时，对其中一个支点取矩可求得另一支点的反力。若物体即将滑动，则摩擦力达到最大值 F = μR。记住摩擦力总是阻碍相对运动——画方向时要格外小心。
Centre of mass calculation for a lamina uses the formula
x̄ = (Σ mₐ xₐ) / (Σ mₐ), ȳ = (Σ mₐ yₐ) / (Σ mₐ)
薄板的质心计算使用公式
x̄ = (Σ mₐ xₐ) / (Σ mₐ), ȳ = (Σ mₐ yₐ) / (Σ mₐ)
For a continuous uniform lamina, replace sums by integrals. A common FM03 question is to find the maximum angle of tilt before a body topples – this occurs when the plumb line through the centre of mass passes through the pivot point.
对于连续均匀薄板，将求和改为积分。FM03 的常见题型是求物体即将倾倒前的最大倾斜角——当通过质心的铅垂线恰好经过支点边缘时，物体将倾倒。
9. Simple Harmonic Motion and Energy | 简谐运动与能量
Simple harmonic motion occurs when acceleration is proportional to displacement from a fixed point and directed towards it:
a = –ω² x
The solution is x = A sin(ωt + φ). The period is T = 2π / ω.
简谐运动发生在当加速度与离开平衡位置的位移成正比，并且方向始终指向平衡位置时：
a = –ω² x
运动学解为 x = A sin(ωt + φ)。周期为 T = 2π / ω。
Energy conservation in SHM gives a powerful shortcut:
½ m v² + ½ m ω² x² = ½ m ω² A²
简谐运动中的能量守恒提供了高效解法：
½ m v² + ½ m ω² x² = ½ m ω² A²
At the maximum amplitude, all energy is potential; at the equilibrium position, all energy is kinetic. If a question gives speed at a certain displacement, use this equation to find ω or A.
在最大振幅处，全部能量为势能；在平衡位置处，全部能量为动能。如果题目给出某一位移处的速度，可利用该方程求 ω 或 A。
10. Vertical Circular Motion and Work–Energy | 竖直圆周运动与功-能关系
A bead moving on a smooth vertical loop has a varying speed because gravitational work changes the kinetic energy. At the bottom, the normal reaction is maximum; at the top, it is minimum. To find the minimum speed at the top to maintain contact, set the centripetal force equal to the weight:
m v² / r = m g ⇒ vmin = √g r
光滑竖直圆环上的小珠速度会变化，因为重力做功改变了动能。在最低点，法向支持力最大；在最高点，支持力最小。为使物体在最高点仍保持接触，所需的最小速度满足向心力等于重力：
m v² / r = m g ⇒ vmin = √g r
Combine this with an energy equation between the bottom and top to find the minimum projection speed at the bottom. The formula is
vbₒₜₜₒₘ ² = vtop ² + 4 g r
将最高点条件与底部到最高点之间的能量方程结合，即可求得底部的最小初速度。公式为
vbₒₜₜₒₘ ² = vtop ² + 4 g r
Notice that the normal reaction does no work in any circular motion, so energy is still conserved in the tangential direction. Use radial equations for forces and tangential equations for speed.
注意：在任何圆周运动中，法向支持力都不做功，因此沿切线方向能量仍然守恒。用力学方程处理径向，用能量方程处理切向速度。
11. Impulse and Momentum in Spring–Mass Systems | 弹簧-质量系统中的冲量与动量
Although the acronym WRE focuses on work and energy, FM03 questions often blend in impulse and momentum. When two particles collide and then engage a spring, use conservation of momentum during the collision, and energy conservation after the collision.
尽管 WRE 聚焦功与能量，FM03 的题目也常融合冲量动量。当两个粒子碰撞并压缩弹簧时，碰撞过程中运用动量守恒，碰撞后运用能量守恒。
For a head-on collision between masses m₁ and m₂ with velocities u₁, u₂ and final velocities v₁, v₂:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
If the collision is elastic, the relative speed of separation equals the relative speed of approach. If it is perfectly inelastic, the masses coalesce and share a common final velocity.
对于质量 m₁、m₂，初速度 u₁、u₂，末速度 v₁、v₂ 的正碰：
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
若碰撞完全弹性，则分离相对速度等于接近相对速度。若完全非弹性，则两物体粘合在一起，具有共同末速度。
In a mass–spring system oscillating on a smooth table, the maximum compression can be found by setting the total kinetic energy after collision equal to the elastic potential energy at maximum compression. Be careful that during the interaction both masses are moving; use reduced mass in two-body energy analysis.
在光滑水平桌上的弹簧-质量系统中，最大压缩量可通过将碰撞后的总动能等于最大压缩时的弹性势能来求解。注意相互作用过程中两个质量都在运动；分析两体能量时宜使用约化质量。
12. Exam Strategy and Common Pitfalls | 应试策略与常见陷阱
Always list the known quantities, then classify the question into single-particle dynamics, work–energy, or impulse–momentum. Do not mix positive and negative work signs. Write numerical answers to three significant figures unless told otherwise.
始终先列出已知量，再判断题目的类型：单质点动力学、功能关系还是冲量动量。切勿混淆正功与负功的符号。未特殊说明时，最终数值保留三位有效数字。
| Common mistake | Correction |
| Assuming friction always does negative work | Friction can do positive work if it is the only external force driving the body |
| Forgetting that the normal reaction is perpendicular to the displacement | Thus the work done by the normal reaction is zero |
| Using v² = u² + 2as when acceleration is not constant | Use work–energy or integration for variable acceleration |
常见错误
纠正方法
认为摩擦力总做负功
若摩擦力是唯一驱动力，则其做功为正
忘记法向支持力垂直于位移方向
因此支持力做功为零
加速度非恒定仍使用 v² = u² + 2as
应使用功能关系或积分处理变加速度
As you practise the June 2023 paper, focus on the mark scheme’s method statements. The most common lost marks in FM03 come from missing constant of integration, incorrect signs in energy equations, and neglecting to state that work done by a force is zero when the force is perpendicular to displacement.
在练习 2023 年 6 月试卷时，请重点关注评分标准中的方法表述。FM03 最常见的失分原因包括：漏掉积分常数、能量方程符号错误，以及未说明当力垂直于位移时做功为零。
Published by TutorHao | Further Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图