📚 Year 9 CIE Physics: Case Study Practice – Energy and Motion on a Slope | 案例分析实战演练：斜坡上的能量与运动
In this article we tackle a realistic physics problem involving a skier sliding down a snowy slope. The case study brings together key Year 9 concepts – gravitational potential energy, kinetic energy, work done against friction, efficiency, power and forces – and shows you how to apply them step by step. By practising such multi‑step scenarios, you build the skills needed for extended exam questions and real‑world application of energy principles.
本文将通过一个滑雪者沿雪坡滑下的真实物理问题，把九年级的关键概念——重力势能、动能、克服摩擦力做功、效率、功率和力——串联起来，一步步教你如何应用。练习这种多步骤情景分析，能培养你解答考试综合题和用能量原理解释真实世界的能力。
1. Case Overview | 案例概览
A skier of mass 75 kg starts from rest at the top of a straight ski slope. The vertical height of the slope is 25 m and its length along the snow is 80 m. A constant average frictional force of 150 N acts on the skier throughout the run. The skier reaches the bottom after 12 seconds. We will use this information to explore energy changes, speed, efficiency and power.
一位质量为 75 kg 的滑雪者从笔直雪坡顶端静止出发。坡的垂直高度为 25 m，沿雪面的长度为 80 m。整个滑行过程中，滑雪者受到 150 N 的恒定平均摩擦力。滑雪者经过 12 s 到达坡底。我们将利用这些数据探究能量转化、速率、效率与功率。
2. Energy Transformation | 能量转换
At the top of the slope, the skier has stored gravitational potential energy (GPE) because of his position in the Earth’s gravitational field. As he slides down, this GPE is transferred into kinetic energy (KE) and internal energy (thermal energy) due to friction between the skis and the snow. The total amount of energy is conserved, but some becomes no longer available as mechanical energy.
在坡顶，滑雪者因其在地球重力场中的位置而储存了重力势能。滑下时，这部分势能转化为动能以及因雪板与雪面摩擦产生的内能（热能）。总能量守恒，但一部分不再以机械能的形式存在。
The energy flow can be summarised as:
能量流动可概括为：
GPE at top → KE at bottom + work done against friction (as heat)
3. Calculating Initial GPE | 计算初始重力势能
ld strength (we take g = 10 N/kg) and h is the height difference. Substituting the values:
重力势能公式为 GPE = m g h，其中 m 为质量，g 为重力场强度（取 g = 10 N/kg），h 为高度差。代入数值：
GPE = 75 kg × 10 N/kg × 25 m = 18 750 J
So the skier starts with 18 750 joules of energy stored in the gravitational field. This is the total energy input available for the descent.
因此滑雪者在重力场中储存了 18 750 J 的能量。这是滑行过程中可用的总能量输入。
4. Work Done Against Friction | 克服摩擦力所做的功
nce moved in the direction of the force. Thus the energy transferred to thermal stores is:
150 N 的摩擦力沿 80 m 的坡面全程作用。功 = 力 × 沿力方向移动的距离。因此，转化为内能的能量为：
Wf = 150 N × 80 m = 12 000 J
This 12 000 J is the energy lost from the mechanical system because of friction; it cannot be turned back into kinetic energy of the skier.
这 12 000 J 是由于摩擦从机械系统中损失的能量，它无法再变回滑雪者的动能。
5. Useful Kinetic Energy at the Bottom | 底部的有用动能
Assuming no air resistance or other losses, the kinetic energy the skier actually has at the bottom is the initial GPE minus the work done against friction:
假设没有空气阻力及其他损失，滑雪者到达坡底时实际拥有的动能等于初始势能减去克服摩擦力做的功：
KE = GPE − Wf = 18 750 J − 12 000 J = 6 750 J
Out of the original 18 750 J, only 6 750 J remains as kinetic energy. The rest has been dissipated as heat to the snow and skies.
在初始的 18 750 J 中，仅有 6 750 J 保留为动能，其余以热的形式散失到雪和雪板中。
6. Finding Final Speed | 求最终速度
Kinetic energy is related to speed by the equation KE = ½ m v². Rearranging for v gives:
动能与速率的关系式为 KE = ½ m v²。整理得 v：
v = √2 KE / m = √2 × 6 750 J / 75 kg = √13 500 / 75 = √180 ≈ 13.4 m/s
The skier reaches the bottom at about 13.4 m/s (roughly 48 km/h). This value is realistic for a moderately steep slope with significant friction.
滑雪者以约 13.4 m/s（大约 48 km/h）的速度到达坡底。对于有显著摩擦的中等坡度雪道，这个数值是合理的。
We can double‑check using the forces approach later, but the energy method is simpler when only speed is required.
我们稍后可用力的方法验证，但在仅需求速率时，能量法更为简单。
7. Introduction to Efficiency | 效率简介
Efficiency of an energy transfer process is defined as the ratio of useful output energy to total input energy. Here the useful output is the kinetic energy of the skier:
能量转化过程的效率定义为有用输出能量与总输入能量之比。此处的有用输出为滑雪者的动能：
η = (KE / GPE) × 100% = (6 750 / 18 750) × 100% = 36%
Only 36% of the original gravitational potential energy is converted to useful kinetic energy; the remaining 64% is wasted as heat. Real‑world scenarios often have such relatively low efficiency due to friction.
只有 36% 的原有重力势能转化为有用的动能，其余 64% 以热的形式被浪费。在真实世界中，因摩擦导致的效率往往比较低。
High efficiency means very little energy is wasted as heat – for example, a well‑lubricated almost frictionless system.
高效率意味着以热形式浪费的能量极少——例如润滑良好、近乎无摩擦的系统。
Low efficiency tells us that a large proportion of energy is dissipated and not used for the intended purpose.
低效率表明一大部分能量被耗散，未能用于预定目的。
8. Power Consideration | 功率考量
Power is the rate of doing work or transferring energy. Knowing the skier takes 12 s to descend, we can calculate average power values.
功率是做功或转移能量的速率。已知滑雪者下坡用时 12 s，我们可以计算平均功率。
The average power delivered by gravity is the rate at which gravitational potential energy is transferred:
重力提供的平均功率即重力势能的转移速率：
Pgravity = GPE / t = 18 750 J / 12 s = 1 562.5 W
The average power wasted against friction is:
克服摩擦力所浪费的平均功率为：
Pfriction = Wf / t = 12 000 J / 12 s = 1 000 W
Therefore, the useful mechanical power output is about 562.5 W, reflecting the remaining energy that becomes kinetic energy.
因此，有用的机械功率输出约为 562.5 W，反映了最终转化为动能的那部分能量。
This breakdown helps link the concepts of energy, time and power in a single context.
这一分解有助于在同一个情境中串联能量、时间和功率的概念。
9. Forces in Action | 作用力分析
Besides energy, we can examine the forces acting on the skier down the slope. The component of weight acting parallel to the slope is responsible for the skier’s acceleration, while friction opposes the motion.
除了能量，我们还可以考察沿坡面作用在滑雪者上的力。重力沿坡面的分量使滑雪者加速，而摩擦力阻碍运动。
The slope is a right‑angled triangle with height 25 m and slope length 80 m, so sin θ = opposite / hypotenuse = 25/80 = 0.3125.
该坡面形成直角三角形，高 25 m、坡长 80 m，因此 sin θ = 对边 / 斜边 = 25/80 = 0.3125。
The downhill component of the skier’s weight is:
滑雪者重力沿坡面向下的分量为：
Fparallel = m g sin θ = 75 × 10 × 0.3125 = 234.375 N
The frictional force opposes motion, so the net force accelerating the skier is:
摩擦力阻碍运动，因此使滑雪者加速的合力为：
Fnet = Fparallel − f = 234.375 N − 150 N = 84.375 N
Using Newton’s second law, acceleration a = Fnet / m = 84.375 N / 75 kg = 1.125 m/s².
由牛顿第二定律，加速度 a = Fnet / m = 84.375 N / 75 kg = 1.125 m/s²。
can verify the final speed from kinematic equation: v² = u² + 2 a s, with u = 0. Thus v = √2 × 1.125 m/s² × 80 m = √180 ≈ 13.4 m/s, matching the energy result perfectly. This cross‑check confirms consistency.
我们可用运动学方程验证末速度：v² = u² + 2 a s，u = 0。因此 v = √2 × 1.125 m/s² × 80 m = √180 ≈ 13.4 m/s，完全符合能量计算结果。这样交叉检验了结论的一致性。
10. Tips for Case Study Questions | 案例分析题型技巧
When you face a multi‑concept physics case study in an exam, keep the following strategies in mind:
考试中遇到综合性物理案例分析，牢记以下策略：
Identify the energy stores at the beginning and end of the process, and note any pathways for energy dissipation (like friction, air resistance).
明确过程始末的能量储存，并标出任何能量耗散途径（如摩擦、空气阻力）。
Write down given quantities with units, and convert if necessary. Draw a simple labelled diagram showing heights, distances and forces.
写下已知量及单位，必要时转换单位。画一个简明的标有高度、距离和力的示意图。
energy = total final energy, remembering to include work done against dissipative forces.
运用能量守恒原理：初始总能量 = 最终总能量，务必计入克服耗散力所做的功。
Choose the most efficient route – energy methods often avoid calculating acceleration and time when only speed or distance is needed.
选择最高效的路径——当仅需速率或距离时，能量法常可避免计算加速度和时间。
Check your answers with an alternative method if time allows (e.g. forces vs. energy).
若时间充裕，用另一种方法验证答案（如力 vs. 能量）。
Always state the efficiency or power to show you can relate energy to practical quantities.
始终写出效率或功率，以展现你能将能量与实际量联系起来。
Practising with numerical values, as in this slope case, builds familiarity with units and orders of magnitude that examiners expect.
像这个斜坡案例一样用实际数值进行练习，能让你熟悉考试所要求掌握的单位和数量级。
Published by TutorHao | Physics Revision Series | aleveler.com
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