📚 Interdisciplinary Integrated Problem-Solving Training for Year 8 Engineering | 八年级工程跨学科综合题型训练
Engineering is not a subject in isolation. It blends concepts from mathematics, physics, design, and materials science to solve real-world problems. In this article, we present a series of integrated problem-solving tasks that mirror the style of CIE Year 8 checkpoint assessments. Each problem requires you to apply knowledge from different fields, interpret data, and communicate your reasoning clearly.
工程学并不是一门孤立的学科。它融合了数学、物理、设计和材料科学的概念来解决现实世界的问题。在本文中，我们提供了一系列跨学科综合题型训练，这些题目模拟了 CIE 八年级评估的风格。每个问题都需要你运用不同领域的知识、解读数据并清晰地表达你的推理过程。
1. Forces and Motion Calculations | 力与运动计算
A student pushes a trolley of mass 12 kg on a level floor. The student exerts a constant pushing force of 30 N. There is a frictional force of 6 N opposing the motion.
一名学生在水平地面上推一辆质量为 12 kg 的小车。学生施加的恒定推力为 30 N。阻碍运动的摩擦力为 6 N。
First, calculate the resultant force acting on the trolley.
首先，计算作用在小车上的合力。
Resultant force = pushing force − friction = 30 N − 6 N = 24 N
合力 = 推力 − 摩擦力 = 30 N − 6 N = 24 N
Now use Newton's second law to find the acceleration.
现在用牛顿第二定律求加速度。
F = m × a → a = F ÷ m = 24 N ÷ 12 kg = 2 m/s²
F = m × a → a = F ÷ m = 24 N ÷ 12 kg = 2 m/s²
The trolley accelerates at 2 m/s². If the student continues pushing, sketch a speed–time graph for the first 5 seconds, assuming the trolley started from rest.
小车以 2 m/s² 加速。如果学生继续推动，请画出前 5 秒的速度–时间图，假设小车从静止开始。
Speed = acceleration × time. After 1 s: 2 m/s; after 2 s: 4 m/s; … after 5 s: 10 m/s. The graph is a straight line through the origin with a slope of 2.
速度 = 加速度 × 时间。1 秒后：2 m/s；2 秒后：4 m/s；… 5 秒后：10 m/s。图像是斜率等于 2 且通过原点的直线。
2. Electrical Circuits and Power | 电路与功率
Two bulbs are connected in series across a 9 V battery. Bulb A has a resistance of 4 Ω and Bulb B has a resistance of 6 Ω. Draw the circuit diagram and calculate the total current.
两只灯泡串联在 9 V 电池上。灯泡 A 的电阻为 4 Ω，灯泡 B 的电阻为 6 Ω。请画出电路图并计算总电流。
The total resistance in series is the sum.
串联总电阻为各个电阻之和。
Rtotal = 4 Ω + 6 Ω = 10 Ω
R总 = 4 Ω + 6 Ω = 10 Ω
I = V ÷ Rtotal = 9 V ÷ 10 Ω = 0.9 A
I = V ÷ R总 = 9 V ÷ 10 Ω = 0.9 A
ermine the power consumed by Bulb A and Bulb B. Recall that power P = I² × R or P = V × I.
现在计算灯泡 A 和灯泡 B 消耗的功率。记住功率 P = I² × R 或 P = V × I。
PA = (0.9 A)² × 4 Ω = 0.81 × 4 = 3.24 W
PA = (0.9 A)² × 4 Ω = 0.81 × 4 = 3.24 W
PB = (0.9 A)² × 6 Ω = 0.81 × 6 = 4.86 W
PB = (0.9 A)² × 6 Ω = 0.81 × 6 = 4.86 W
Which bulb glows brighter? Explain using the energy transfer concept.
哪只灯泡更亮？请用能量转化的概念解释。
Bulb B dissipates more energy per second (4.86 W > 3.24 W), so it converts more electrical energy into light and heat. Therefore, Bulb B is brighter.
灯泡 B 每秒耗散的能量更多（4.86 W > 3.24 W），因此它会将更多的电能转化为光能和热能。所以灯泡 B 更亮。
3. Energy Transfers and Efficiency | 能量转换与效率
A ramp is used to lift a 200 N load to a height of 1.5 m. The length of the ramp is 5 m, and an effort force of 80 N is applied parallel to the ramp.
使用斜面将一个 200 N 的重物提升到 1.5 m 的高度。斜面长度为 5 m，沿斜面方向施加的驱动力为 80 N。
Calculate the useful work output and the total work input.
计算有用功输出和总功输入。
Useful work output = load × vertical height = 200 N × 1.5 m = 300 J
有用功输出 = 荷载 × 垂直高度 = 200 N × 1.5 m = 300 J
Work input = effort force × ramp length = 80 N × 5 m = 400 J
功输入 = 驱动力 × 斜面长度 = 80 N × 5 m = 400 J
Find the efficiency of the ramp system.
求斜面系统的效率。
Efficiency = (useful output ÷ input) × 100% = (300 J ÷ 400 J) × 100% = 75%
效率 = (有用输出 ÷ 输入) × 100% = (300 J ÷ 400 J) × 100% = 75%
Suggest one way to improve the efficiency, referring to energy dissipation.
提出一种提高效率的方法，并提到能量耗散。
Reduce friction between the ramp and the load by using rollers or lubricant. This would decrease the extra work done against friction, thus raising the efficiency.
通过使用滚轮或润滑剂来减少斜面与重物之间的摩擦。这样可以减少克服摩擦做的额外功，从而提高效率。
4. Material Properties and Selection | 材料特性与选择
A designer needs to choose a material for a lightweight bicycle frame. The table shows properties of three candidate materials.
一位设计师需要为轻便自行车车架选择材料。表格给出了三种候选材料的特性。
| Material / 材料 | Density (kg/m³) / 密度 | Tensile strength (MPa) / 抗拉强度 | Cost (£/kg) / 成本 |
|---|---|---|---|
| Aluminium alloy / 铝合金 | 2700 | 310 | 2.5 |
| Titanium alloy / 钛合金 | 4500 | 900 | 15.0 |
| Carbon fibre composite / 碳纤维复合材料 | 1800 | 600 | 20.0 |
The frame volume is approximately 0.0015 m³. Calculate the mass of a frame made from each material.
车架体积约为 0.0015 m³。计算每种材料制造的车架质量。
Mass = density × volume
质量 = 密度 × 体积
0.0015 = 4.05 kg; Titanium: 4500 × 0.0015 = 6.75 kg; Carbon fibre: 1800 × 0.0015 = 2.7 kg.
铝合金：2700 × 0.0015 = 4.05 kg；钛合金：4500 × 0.0015 = 6.75 kg；碳纤维：1800 × 0.0015 = 2.7 kg。
Which material gives the best strength-to-weight ratio? Justify your answer using the strength divided by density.
哪种材料具有最佳强度–重量比？用强度除以密度证明你的答案。
bon fibre offers the highest ratio, meaning it provides high strength for very low weight.
铝合金：310 ÷ 2700 ≈ 0.115 MPa·m³/kg；钛合金：900 ÷ 4500 = 0.200；碳纤维：600 ÷ 1800 ≈ 0.333。碳纤维复合材料可提供最高的比值，这意味着它以极轻的重量提供了高强度。
If the designer has a budget of £50 per frame for material cost, which materials are affordable? Show your working.
如果设计师每个车架的材料预算为 £50，哪种材料是负担得起的？请展示计算过程。
Aluminium cost: 4.05 × 2.5 = £10.13, affordable. Titanium cost: 6.75 × 15 = £101.25, too expensive. Carbon fibre cost: 2.7 × 20 = £ slightly over budget. Only aluminium fits the cost limit, although carbon fibre is close.
铝合金成本：4.05 × 2.5 = £10.13，可负担。钛合金成本：6.75 × 15 = £101.25，太贵。碳纤维成本：2.7 × 20 = £54.00，略微超预算。只有铝合金符合成本限制，碳纤维略超。
5. Structures and Stability | 结构与稳定性
A uniform beam 4 m long is supported at its centre. A 30 N weight is hung 1.5 m to the left of the pivot, and a 20 N weight is hung 2 m to the right of the pivot.
一根 4 m 长的均匀直梁在其中心得到支撑。一个 30 N 的重物悬挂在支点左侧 1.5 m 处，一个 20 N 的重物悬挂在支点右侧 2 m 处。
Calculate the clockwise moment about the pivot.
计算绕支点的顺时针力矩。
Clockwise moment = force × distance = 20 N × 2 m = 40 N·m
顺时针力矩 = 力 × 距离 = 20 N × 2 m = 40 N·m
Calculate the anticlockwise moment.
计算逆时针力矩。
Anticlockwise moment = 30 N × 1.5 m = 45 N·m
逆时针力矩 = 30 N × 1.5 m = 45 N·m
Is the beam balanced? If not, describe the resulting motion.
横梁平衡吗？如果不平衡，请描述所产生的运动。
The anticlockwise moment exceeds the clockwise moment (45 N·m > 40 N·m), so the beam will rotate anticlockwise until it hits a stop or the weights are adjusted.
逆时针力矩大于顺时针力矩（45 N·m > 40 N·m），因此横梁将逆时针旋转，直到碰到限位装置或调整重物。
How could you achieve balance by adding a single 5 N weight? Specify its position.
你如何通过增加一个 5 N 的重物来达到平衡？请说明其位置。
t 1 m right brings clockwise moment to 40 + 5 = 45 N·m, matching the anticlockwise moment.
我们需要额外增加 5 N·m 的顺时针力矩。所以距离 = 5 N·m ÷ 5 N = 1 m，在支点右侧。在右侧 1 m 处加一个 5 N 的重物，顺时针力矩变成 40 + 5 = 45 N·m，与逆时针力矩相等。
6. Data Analysis and Graphs | 数据分析与图表
An experiment measured the extension of a spring as weights were added. The results are shown below.
一项实验测量了增加砝码时弹簧的伸长量。结果如下所示。
| Load / 载荷 (N) | 0 | 2 | 4 | 6 | 8 |
|---|---|---|---|---|---|
| Extension / 伸长量 (cm) | 0 | 3.2 | 6.4 | 9.8 | 14.5 |
Plot a graph of extension against load. Describe the trend.
画出伸长量随载荷变化的图像。描述变化趋势。
The first three points lie on a straight line through the origin, obeying Hooke's law. At 6 N, the extension is slightly above the linear trend (9.8 cm instead of 9.6 cm). At 8 N, the extension jumps to 14.5 cm, indicating the spring has exceeded its elastic limit.
前三个点位于通过原点的直线上，遵循胡克定律。在 6 N 时，伸长量略高于线性趋势（9.8 cm 而非 9.6 cm）。在 8 N 时，伸长量跃升至 14.5 cm，表明弹簧已超过其弹性极限。
Calculate the spring constant for the linear region.
计算线性区域的弹簧常数。
k = F ÷ x = 4 N ÷ 0.064 m = 62.5 N/m
k = F ÷ x = 4 N ÷ 0.064 m = 62.5 N/m
(Using the 4 N data point: 6.4 cm = 0.064 m)
（使用 4 N 数据点：6.4 cm = 0.064 m）
Estimate the extension at 5 N from the graph and explain why it is reliable.
根据图像估算 5 N 载荷下的伸长量，并解释为什么它是可靠的。
Since 5 N is within the linear range, the extension should be around 8.0 cm (halfway between 6.4 and 9.6 cm). This interpolation is reliable because the spring obeys Hooke’s law in that region.
由于 5 N 处于线性范围内，伸长量应约为 8.0 cm（在 6.4 cm 与 9.6 cm 之间）。这种插值是可靠的，因为弹簧在该区域遵循胡克定律。
7. Design and Sketching | 设计与草图
Design a simple device to lift a bucket of water from a well using a rotating handle. Your design must include a labelled sketch showing the wheel and axle mechanism, and state how the mechanical advantage is achieved.
设计一种利用旋转手柄从水井提水的简单装置。你的设计必须包含带标注的示意图，显示轮轴机构，并说明如何实现机械效益。
The sketch should show a large wheel (handle) attached to a smaller axle around which a rope is wound. Turning the handle rotates the axle, lifting the bucket. The mechanical advantage is the ratio of the wheel radius to the axle radius. A larger handle radius makes the effort force required smaller.
草图应显示一个较大的轮（手柄），它连接着一根较细的轴，绳索缠绕在轴上。转动手柄带动轴旋转，从而提升水桶。机械效益为轮半径与轴半径之比。手柄半径越大，所需的驱动力越小。
If the wheel radius is 0.4 m and the axle radius is 0.1 m, calculate the theoretical mechanical advantage.
如果轮半径为 0.4 m，轴半径为 0.1 m，计算理论机械效益。
Mechanical advantage = 0.4 m ÷ 0.1 m = 4
机械效益 = 0.4 m ÷ 0.1 m = 4
Now, if friction causes 20% energy loss, estimate the actual effort required to lift a 120 N bucket.
现在，如果摩擦导致 20% 能量损失，估算提升 120 N 的水桶所需的实际驱动力。
Ideal effort = load ÷ MA = 120 N ÷ 4 = 30 N. With 20% loss, efficiency is 80%, so actual effort = ideal effort ÷ efficiency = 30 N ÷ 0.8 = 37.5 N.
理想驱动力 = 载荷 ÷ 机械效益 = 120 N ÷ 4 = 30 N。因有 20% 能量损失，效率为 80%，因此实际驱动力 = 理想驱动力 ÷ 效率 = 30 N ÷ 0.8 = 37.5 N。
8. Thermal Energy and Insulation | 热能转换与隔热
A metal block is heated to 80 °C and placed in an insulating container with 200 g of water at 20 °C. The specific heat capacity of water is 4200 J/(kg·°C). After a while, the final temperature of the water and block is 25 °C.
将一块金属加热至 80 °C 并放入装有 200 g 20 °C 水的隔热容器中。水的比热容为 4200 J/(kg·°C)。一段时间后，水和金属块的最终温度为 25 °C。
Calculate the thermal energy gained by the water.
计算水获得的热能。
ΔTwater = 25 °C − 20 °C = 5 °C
水的温度变化 = 25 °C − 20 °C = 5 °C
Q = m × c × ΔT = 0.2 kg × 4200 J/(kg·°C) × 5 °C = 4200 J
Q = m × c × ΔT = 0.2 kg × 4200 J/(kg·°C) × 5 °C = 4200 J
Assuming no heat loss to the surroundings, how much thermal energy did the metal block lose?
假设没有热散失到周围环境中，金属块损失了多少热能？
The block lost the same amount of energy that the water gained: 4200 J.
金属块损失的热能等于水获得的热能：4200 J。
If the block has a mass of 0.5 kg, calculate its specific heat capacity.
如果金属块的质量为 0.5 kg，计算其比热容。
c = Q ÷ (m × ΔT) = 4200 J ÷ (0.5 kg × 55 °C) = 4200 ÷ 27.5 ≈ 152.7 J/(kg·°C)
c = Q ÷ (m × ΔT) = 4200 J ÷ (0.5 kg × 55 °C) = 4200 ÷ 27.5 ≈ 152.7 J/(kg·°C)
(ΔTblock = 80 °C − 25 °C = 55 °C)
（金属块温度变化 = 80 °C − 25 °C = 55 °C）
9. Environmental Impact of Materials | 材料的环境影响
A company produces 10,000 plastic bottles, each weighing 40 g. The plastic has a carbon footprint of 3.2 kg CO₂ per kg of plastic. Alternatively, they could use recycled plastic with a footprint of 1.1 kg CO₂ per kg, but recycled plastic costs £0.20 more per bottle.
一家公司生产 10,000 个塑料瓶，每个重 40 g。该塑料的碳足迹为每千克塑料 3.2 kg CO₂。另一替代方案是使用再生塑料，碳足迹为每千克 1.1 kg CO₂，但每个瓶子的成本增加 £0.20。
Calculate the total mass of plastic and the CO₂ emissions for both options.
计算两种方案下的塑料总质量和 CO₂ 排放量。
in plastic emissions: 400 × 3.2 = 1280 kg CO₂. Recycled emissions: 400 × 1.1 = 440 kg CO₂.
总质量 = 10,000 × 0.04 kg = 400 kg。原生塑料排放：400 × 3.2 = 1280 kg CO₂。再生塑料排放：400 × 1.1 = 440 kg CO₂。
Describe the trade-off between environmental benefit and economic cost.
描述环境效益与经济成本之间的权衡。
y 840 kg CO₂, which is beneficial for the environment. However, the extra cost is 10,000 × £0.20 = £ company must decide if the reduction in carbon footprint is worth the additional expense.
使用再生塑料可减少 840 kg CO₂ 排放，对环境有利。然而，额外成本为 10,000 × £0.20 = £2000。公司必须决定减少碳足迹是否值得增加这笔费用。
Suggest one other sustainable practice for the bottle manufacturer.
为瓶子制造商提出另一项可持续实践建议。
The company could design a refillable bottle to reduce the total number of bottles produced, thus lowering overall environmental impact.
公司可以设计可重复灌装的瓶子，以减少生产的瓶子总数，从而降低整体环境影响。
10. Automation and Programming Logic | 自动化与编程逻辑
A temperature control system in a greenhouse uses a sensor and a microcontroller. The system should turn on a heater when the temperature falls below 15 °C and turn on a fan when it exceeds 28 °C.
温室中的温度控制系统使用传感器和微控制器。该系统应在温度低于 15 °C 时打开加热器，在温度超过 28 °C 时打开风扇。
Write a pseudocode algorithm to describe this control logic.
编写伪代码算法来描述此控制逻辑。
'START, READ temperature, IF temperature < 15 THEN turn heater ON, ELSE turn heater OFF. IF temperature > 28 THEN turn fan ON, ELSE turn fan OFF. REPEAT.'
'开始，读取温度，若温度 < 15 则打开加热器，否则关闭加热器。若温度 > 28 则打开风扇，否则关闭风扇。重复。'
If the sensor fails and always reads 10 °C, what would happen? Explain the safety risk and suggest a fail-safe measure.
如果传感器故障并始终显示 10 °C，会发生什么？解释安全风险并提出故障安全措施。
The heater would stay on continuously, overheating the greenhouse and possibly damaging plants. A fail-safe measure could be a second independent sensor that cuts off power if the temperature exceeds a critical limit, say 35 °C.
加热器将一直开启，使温室过热并可能损坏植物。故障安全措施可以是安装第二个独立传感器，如果温度超过临界值（例如 35 °C）就切断电源。
11. Integrated Challenge: Bridge Design | 综合挑战：桥梁设计
idge must support a 500 g mass at its centre without breaking. Sketch your design and label the reinforcements.
仅使用 20 根木条（每根 300 mm × 10 mm × 5 mm）和胶水，设计一座跨越 1.2 m 间隙的简支梁桥。桥梁必须在中央支撑 500 g 的质量而不断裂。画出设计草图并标注加固部分。
The bridge
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