📚 Year 8 CAIE Physics: Case Study Practice | Year 8 CAIE 物理：案例分析实战演练
Welcome to this case study practical exercise, where we apply Year 8 CAIE Physics concepts to a real-world scenario. You will follow a skateboarder rolling through a skatepark, using energy, work, and power to predict motion. By the end of this exercise, you will feel confident solving similar problems on your own.
欢迎来到本案例分析实战演练，我们将运用 Year 8 CAIE 物理概念解决一个真实情境问题。你将跟随一位滑板者在滑板场中滑行，利用能量、功和功率来预测运动。通过本次练习，你将能够自信地独立解决类似问题。
1. Understanding the Skatepark Scenario | 理解滑板场情境
m, she enters a rough horizontal surface 8.0 m long, where a constant friction force of 100 N acts against her motion. After the rough patch, she attempts to climb a second ramp, 1.0 m high and also frictionless. We will use g = 10 N/kg for simplicity. The key question is whether she can reach the top of the second ramp.
本案例中，一位质量为 50 kg 的滑板者从光滑斜坡顶端由静止出发。斜坡高 2.0 m，无摩擦。到达底部后，她进入一段长 8.0 m 的粗糙水平面，这里受到恒定的 100 N 摩擦力作用。经过粗糙路面后，她试图攀爬第二个斜坡，高 1.0 m 且同样无摩擦。为简化计算，我们取 g = 10 N/kg。核心问题是：她能否到达第二个斜坡的顶部？
| Quantity | Value |
| Mass (m) | 50 kg |
| g | 10 N/kg |
| Initial ramp height (h₁) | 2.0 m |
| Rough surface length (d) | 8.0 m |
| Friction force (F) | 100 N |
| Second ramp height (h₂) | 1.0 m |
These values are summarised in the table above. Keep them handy as we work through the calculations step by step.
上表汇总了所有已知值。请记住这些数据，我们将逐步进行计算。
2. Gravitational Potential Energy at the Start | 初始重力势能
At the top of the first ramp, the skateboarder has gravitational potential energy (GPE) because of her height above the ground. GPE is given by the formula:
在第一个斜坡顶端，滑板者因离地高度而具有重力势能 (GPE)。GPE 的计算公式为：
GPE = m × g × h
= 1000 J. This stored energy is ready to be converted into kinetic energy as she descends.
代入数值：GPE = 50 kg × 10 N/kg × 2.0 m = 1000 J。这些储存的能量将在下滑过程中转化为动能。
3. Energy Conversion Down the Ramp (No Friction) | 斜坡上的能量转换（无摩擦）
Since the first ramp is frictionless, we can assume mechanical energy is conserved. This means all the GPE at the top is transformed into kinetic energy (KE) at the bottom, with no energy lost as heat.
由于第一个斜坡无摩擦，我们可以假设机械能守恒。这意味着顶端所有的 GPE 在底部全部转化为动能 (KE)，没有能量以热的形式散失。
KE at bottom = GPE at top = 1000 J
This direct conversion is a powerful tool in physics, allowing us to bypass complex forces when surfaces are smooth.
这种直接转化是物理学中的有力工具，当接触面光滑时，我们可以绕过复杂的受力分析。
4. Kinetic Energy and Speed at the Bottom | 底部动能和速度
Kinetic energy depends on mass and speed. The formula KE = ½ m v2 lets us find her velocity at the bottom. Rearranging gives:
动能取决于质量和速度。公式 KE = ½ m v2 可用于求底部的速率。整理后得到：
v = √2 × KE / m
Plug in the numbers: v = √2 × 1000 J / 50 kg = √2000/50 = √40 ≈ 6.32 m/s. So as she leaves the ramp, she is travelling at about 6.3 m/s.
代入数字：v = √2 × 1000 J / 50 kg = √2000/50 = √40 ≈ 6.32 m/s。因此在离开斜坡时，她的速度约为 6.3 m/s。
5. Work Done Against Friction on the Rough Surface | 粗糙水平面上克服摩擦力做功
When the skateboarder moves across the rough horizontal section, a constant friction force of 100 N opposes her motion. Work is done against this force, removing energy from the system. The work done against friction is:
当滑板者经过粗糙水平面时，恒定的 100 N 摩擦力阻碍其运动。克服该力做功会从系统中消耗能量。克服摩擦做功的计算为：
W = F × d
f energy is dissipated as heat and sound, and is no longer available as mechanical energy.
因此 W = 100 N × 8.0 m = 800 J。这 800 J 的能量以热和声的形式耗散，不再作为机械能使用。
6. Remaining Kinetic Energy After the Rough Patch | 经过粗糙路面后的剩余动能
By the time she reaches the end of the rough surface, the friction has eaten away 800 J from her initial 1000 J of kinetic energy. The remaining useful kinetic energy is simply:
当她到达粗糙路面末端时，摩擦力已经消耗掉初始动能 1000 J 中的 800 J。剩余有用动能为：
KEremaining = 1000 J - 800 J = 200 J
This remaining energy determines how high she can climb the second ramp. She still has some speed, but much less than before.
这些剩余能量决定了她能攀爬第二个斜坡的高度。她仍有一定速度，但远不如之前。
7. Climbing the Second Ramp: How High? | 攀爬第二个斜坡：能有多高？
As she rolls up the second frictionless ramp, her remaining kinetic energy is converted back into gravitational potential energy. We can find the maximum height she can reach by setting KEremaining = m g hmax.
当她冲上第二个光滑斜坡时，剩余动能会重新转化为重力势能。通过令 KE剩余 = m g hmax 可以求出她能达到的最大高度。
hmax = KEremaining / (m × g) = 200 J / (50 kg × 10 N/kg) = 200 / 500 = 0.40 m
The calculated height is just 0.40 m, far below the 1.0 m needed. Therefore, with the given friction and distance, she cannot reach the top of the second ramp. She will roll partway up and then roll back down.
计算得到的高度仅为 0.40 m，远低于所需的 1.0 m。因此，在给定的摩擦和距离下，她无法到达第二个斜坡的顶端。她会部分上冲然后滚落回来。
8. Adjusting Variables: Making It to the Top | 调整变量：如何才能冲到坡顶
at ramp. This means the work done against friction must be at most 1000 J - 500 J = 500 J.
要成功到达 1.0 m 高的斜坡顶端，她至少需要 m g h₂ = 50×10×1.0 = 500 J 的动能在坡底。这意味着克服摩擦做功必须不超过 1000 J - 500 J = 500 J。
0 m. So shortening the rough patch to 5.0 m or less would let her conquer the second ramp.
如果摩擦力仍为 100 N，则允许的粗糙面长度为 d = W / F = 500 J / 100 N = 5.0 m。因此将粗糙路段缩短至 5.0 m 或更短，她就能冲上第二个斜坡。
9. What If Friction Were Zero? | 如果摩擦力为零会怎样？
Imagine a completely smooth skatepark with no friction anywhere. Then no mechanical energy would be lost. Starting from 2.0 m, she would convert all her initial GPE into KE, and then back into GPE on any ramp, reaching exactly 2.0 m again. She would easily clear the 1.0 m ramp and continue higher.
想象一个完全光滑的滑板场，处处无摩擦。那么机械能不会有任何损失。从 2.0 m 高处出发，她会将初始 GPE 全部转化为 KE，然后在任何斜坡上再转回 GPE，恰好能再次达到 2.0 m 高度。她会轻松越过 1.0 m 坡并继续升高。
This comparison highlights how friction dissipates energy and limits real motion. Energy conservation in idealised cases gives the upper limit of what is possible.
这一对比突显了摩擦如何耗散能量并限制真实运动。理想情况下的能量守恒给出了可能实现的最高极限。
10. Introducing Power: Time to Slide Down | 引入功率：下滑时间
Power measures the rate at which energy is transferred or work is done. If we time her descent down the first ramp and find it takes 2.0
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