Cross-Disciplinary Practice Questions for Year 7 AQA Engineering | 跨学科综合题型训练

📚 Cross-Disciplinary Practice Questions for Year 7 AQA Engineering | 跨学科综合题型训练

Welcome to this focused revision guide for Year 7 AQA Engineering. Engineering is inherently cross-disciplinary, blending concepts from mathematics, science, and design technology. This article provides a set of integrated practice questions designed to help you apply your knowledge across different subjects, preparing you for the style of questions you may encounter on your assessments. Each section combines key theory with worked examples, and we strongly encourage you to try solving them before reading the solutions. Work through each step carefully, paying attention to the units and the reasoning behind every answer.

欢迎来到这篇针对 Year 7 AQA 工程学的专项复习指南。工程学本质上是跨学科的，融合了数学、科学和设计技术的概念。本文提供了一组综合性练习题，旨在帮助你跨学科应用知识，为你在评估中可能遇到的题型做好准备。每一节都将关键理论与例题相结合，我们强烈鼓励你在阅读解题过程之前先自己尝试解答。请仔细完成每一步，注意单位以及每个答案背后的推理。

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1. Understanding Units and Conversions | 理解单位与换算

In engineering, accurate measurement is everything. Whether you are working with lengths, masses, or areas, you must be able to convert between different units confidently. The SI base units are the standard: metre (m) for length, kilogram (kg) for mass, and second (s) for time. However, practical work often uses millimetres, grams, kilometres, and other multiples. Knowing how to move between these systems is a core skill that links maths directly to real‑world design and manufacturing tasks.

在工程学中，精确测量至关重要。无论你处理的是长度、质量还是面积，你都必须能够自信地在不同单位之间进行换算。国际单位制的基本单位是标准：米 (m) 代表长度，千克 (kg) 代表质量，秒 (s) 代表时间。然而，实际工作中经常使用毫米、克、千米和其他倍数。懂得如何在不同的单位系统之间转换，是将数学直接联系到现实世界设计与制造任务的一项核心技能。

Let’s begin with a straightforward conversion. A steel I‑beam is specified as 4.25 m long. Express this length in millimetres.

让我们从一个直接的换算开始。一根工字钢梁的规格长度为 4.25 m。请将该长度用毫米表示。

emember the relationship: 1 m = 1000 mm. Therefore, multiply the length in metres by 1000.

请记住关系：1 m = 1000 mm。因此，将米为单位的长度乘以 1000。

Length in mm = 4.25 × 1000 = 4250 mm

Now consider an area conversion. A rectangular aluminium panel measures 120 cm by 80 cm. What is its area in square metres?

现在考虑一个面积换算。一块矩形铝板的尺寸为 120 cm × 80 cm。它的面积是多少平方米？

nsion by 100. So length = 120 cm = 1.20 m, and width = 80 cm = 0.80 m. Then find the area.

首先，把两个尺寸都转换为米。因为 1 m = 100 cm，每个尺寸除以 100。因此长 = 120 cm = 1.20 m，宽 = 80 cm = 0.80 m。然后求出面积。

Area = 1.20 m × 0.80 m = 0.96 m²

0 cm². The correct conversion factor is 9600 cm² ÷ 10 000 = 0.96 m², which matches our working above.

一个常见错误是先计算平方厘米面积 (120 × 80 = 9600 cm²)，然后试图除以 100。这是不正确的，因为 1 m² = 10 000 cm²。正确的换算因子是 9600 cm² ÷ 10 000 = 0.96 m²，与上述计算一致。

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2. Basic Mechanics: Weight, Mass and Gravity | 基础力学：重量、质量与重力

One of the first scientific ideas you apply in engineering is the relationship between mass and weight. Mass is the amount of matter in an object and is measured in kilograms (kg). Weight is a force caused by gravity pulling on that mass, and its unit is the newton (N). On Earth, the gravitational field strength is approximately 10 N/kg. This lets us use a simple equation that is critical when designing structures, selecting cables or calculating support forces.

你在工程学中应用的最早的科学概念之一就是质量与重量的关系。质量是物体所含物质的多少，以千克 (kg) 为单位。重量是由重力吸引该质量而产生的力，单位是牛顿 (N)。在地球上，重力场强度约为 10 N/kg。这使得我们能够使用一个简单的方程，它在设计结构、选择缆绳或计算支撑力时至关重要。

Weight (W) = mass (m) × gravitational field strength (g)

W = m × g

Worked example: A site worker lifts a concrete block that has a mass of 15 kg. Calculate the weight of the block.

例题：一名工地工人抬起一块质量为 15 kg 的混凝土砌块。计算该砌块的重量。

Using g = 10 N/kg, we multiply mass by 10.

使用 g = 10 N/kg，我们将质量乘以 10。

W = 15 kg × 10 N/kg = 150 N

Now try a design scenario: an overhead crane needs to lift a steel girder of mass 500 kg. Find the minimum breaking strength the lifting cable must have if we include a safety factor of 2. (The cable must be able to withstand at least twice the actual weight.)

现在尝试一个设计场景：一台桥式起重机需要吊起质量为 500 kg 的钢梁。如果要求安全系数为 2，求吊索必须具有的最低破断强度。（吊索必须至少能承受实际重量的两倍。）

First, find the weight of the girder: W = 500 × 10 = 5000 N. Then apply the safety factor.

首先，计算钢梁重量：W = 500 × 10 = 5000 N。然后应用安全系数。

Required strength = 5000 N × 2 = 10 000 N (or 10 kN)

Engineers routinely use this kind of calculation to prevent failure. Remember always to keep mass and weight separate in your reasoning.

工程师经常使用这种计算来防止失效。切记在推理过程中始终将质量与重量区分开。

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3. Moments and Simple Machines | 力矩与简单机械

The turning effect of a force is called a moment. Moments help us understand levers, gears, and many construction tools. The moment of a force is found by multiplying the force by the perpendicular distance from the pivot to the line of action of the force. For a system to be balanced, the sum of the clockwise moments must equal the sum of the anticlockwise moments. This principle is used in everything from seesaws to bridge counterweights.

力所产生的转动效应称为力矩。力矩帮助我们理解杠杆、齿轮和许多施工工具。一个力的力矩可通过将力乘以从支点到该力作用线的垂直距离来求得。要使系统平衡，顺时针力矩的总和必须等于逆时针力矩的总和。从跷跷板到桥梁配重，这一原理在各个领域都有应用。

Moment = Force × perpendicular distance from pivot

A spanner is used to turn a bolt. A force of 30 N is applied at a distance of 0.25 m from the centre of the bolt. Calculate the moment exerted on the bolt.

一把扳手用来拧螺丝。在距螺丝中心 0.25 m 处施加了 30 N 的力。计算作用在螺丝上的力矩。

Moment = 30 N × 0.25 m = 7.5 N m

Now consider a balanced lever: a crowbar is used to lift a heavy crate. The effort force is 50 N and is applied 0.6 m from the pivot. The load is on the other side of the pivot. What distance from the pivot must the load be placed if it weighs 200 N, in order to achieve balance?

现在考虑一个平衡的杠杆：一根撬棍被用来抬起一个沉重的板条箱。动力为 50 N，施力点距支点 0.6 m。载荷在支点的另一侧。如果载荷重 200 N，要使其平衡，载荷必须放在距支点多远的位置？

N × 0.6 m = 30 N m. So the load moment must also be 30 N m. Let the unknown distance be d.

令顺时针力矩等于逆时针力矩。设动力力矩为顺时针：50 N × 0.6 m = 30 N m。因此载荷力矩也必须为 30 N m。设未知距离为 d。

200 N × d = 30 N m

d = 30 N m ÷ 200 N = 0.15 m (or 15 cm)

When you design a tool or a lifting device, these calculations help you decide the lengths and positions that make the job possible with minimal effort.

当你设计一个工具或提升装置时，这些计算有助于你确定长度和位置，从而用最小的力气完成工作。

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4. Work, Energy and Power | 功、能量与功率

d in the direction of the force. This is measured in joules (J). Energy can be stored as gravitational potential energy (Ep = mgh) when an object is raised, and power is the rate at which work is done or energy is transferred, in watts (W).

在工程学中，我们经常谈论当一个力移动一个物体时所做的功。功的科学定义是：功 = 力 × 沿力的方向移动的距离。它的单位是焦耳 (J)。当物体被抬高时，能量可以储存为重力势能 (Ep = mgh)，而功率则是指做功或能量转移的速率，单位是瓦特 (W)。

Worked example: A worker uses a trolley to move a load. She pushes with a constant force of 40 N over a distance of 12 m. Calculate the work done.

例题：一名工人使用手推车搬运货物。她以恒定的 40 N 的力推行了 12 m 的距离。计算所做的功。

Work = Force × distance = 40 N × 12 m = 480 J

Now imagine a small electric winch lifts a component of mass 8 kg up through a height of 5 m. First, find the work done against gravity, and then determine the power output if the lift takes 4 seconds.

现在设想一台小型电动绞盘将一个质量为 8 kg 的部件提升 5 m 的高度。首先，求克服重力所做的功，然后如果提升用时 4 秒，求功率输出。

g = 8 kg × 10 N/kg = 80 N. Work done = 80 N × 5 m = 400 J. Power = work done ÷ time taken.

部件的重量：W = m × g = 8 kg × 10 N/kg = 80 N。做的功 = 80 N × 5 m = 400 J。功率 = 做的功 ÷ 所用时间。

Power = 400 J ÷ 4 s = 100 W

In many engineering tasks, you also need to consider efficiency: not all the input energy is converted to useful work.

在许多工程任务中，你还需要考虑效率：并非所有输入的能量都能转换为有用功。

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5. Materials and Density | 材料与密度

Choosing the right material for a job depends on several properties: strength, stiffness, density, and corrosion resistance. Density links mass and volume and is vital in weight‑critical applications, such as aircraft or racing bicycles. The formula for density is:

为一项工作选择合适的材料取决于几个特性：强度、刚度、密度和耐腐蚀性。密度将质量与体积联系起来，在飞机或竞速自行车等对重量敏感的应用中至关重要。密度的公式为：

Density (ρ) = mass (m) ÷ volume (V)

Here are some typical densities for common engineering materials.

以下是一些常用工程材料的典型密度。

Material / 材料	Density (g/cm³) / 密度
Aluminium / 铝	2.7
Steel / 钢	7.8 – 8.0
Copper / 铜	8.9
Nylon / 尼龙	1.1 – 1.2
Oak wood / 橡木	0.6 – 0.9

Question: A metal bracket has a mass of 160 g and a volume of 20 cm³. Using the table, identify which material it is most likely to be.

问题：一个金属支架的质量为 160 g，体积为 20 cm³。请利用上表，判断它最可能是什么材料。

Density = 160 g ÷ 20 cm³ = 8.0 g/cm³

Looking at the table, a density of 8.0 g/cm³ is close to that of steel. This indicates the bracket is most likely made from a grade of steel.

查看表格，8.0 g/cm³ 的密度接近于钢的密度。这表明该支架最可能由某种钢材制成。

Think about a bicycle frame: why might a designer choose aluminium instead of steel, even though steel is stronger? Using the density values, calculate the mass of a 200 cm³ volume of each material and compare. This inter‑disciplinary reasoning combines maths and material science.

想想自行车车架：为什么设计师可能会选择铝而不是钢，尽管钢的强度更高？利用密度值，计算出两种材料各 200 cm³ 体积的质量并进行比较。这种跨学科的推理综合了数学和材料科学。

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6. Introduction to Electrical Circuits | 电路入门

Electricity is another scientific area that engineers routinely combine with mechanical and structural work. In Year 7 you will have met voltage (V), current (I) and resistance (R). Ohm’s law relates these three quantities. It is essential for understanding wiring, motors and electronic controls in any engineered product.

电学是工程师经常与机械和结构工作结合使用的另一个科学领域。在 Year 7 你已经遇到了电压 (V)、电流 (I) 和电阻 (R)。欧姆定律将这三者联系起来。它对理解任何工程产品中的布线、电机和电子控制都至关重要。

V = I × R

A small toy robot uses a 9 V battery. The motor has a resistance of 18 Ω. Calculate the current flowing through the motor.

一个小型玩具机器人使用 9 V 电池。电机的电阻为 18 Ω。计算流过电机的电流。

I = V ÷ R = 9 V ÷ 18 Ω = 0.5 A

Now we can find the power consumed by the motor. Electrical power P = I × V.

现在我们可以求出电机消耗的电功率。电功率 P = I × V。

P = 0.5 A × 9 V = 4.5 W

Imagine you are designing a lighting system for a model bridge. Two LEDs are connected in series to a 12 V supply. Each LED requires 2 V and a current of 20 mA. Determine the value of the series resistor needed to protect the LEDs.

设想你正在为一座模型桥设计照明系统。两个 LED 串联到一个 12 V 电源上。每个 LED 需要 2 V 电压和 20 mA 的电流。确定保护 LED 所需的串联电阻值。

Total voltage drop across LEDs = 2 V + 2 V = 4 V. Voltage across resistor = 12 V – 4 V = 8 V. Current = 0.02 A. Using R = V ÷ I.

LED 上的总电压降 = 2 V + 2 V = 4 V。电阻两端的电压 = 12 V – 4 V = 8 V。电流 = 0.02 A。利用 R = V ÷ I。

R = 8 V ÷ 0.02 A = 400 Ω

This is exactly the kind of cross‑disciplinary question that requires you to use circuit theory and apply it to a straightforward engineering design choice.

这正是那种要求你运用电路理论并将其应用于一项简单工程设计选择的跨学科题型。

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7. Interpreting Graphs and Data | 图表与数据解读

Data handling is a skill shared between mathematics, science and engineering. You will often be given graphs of force against extension, temperature against time, or load against deflection. Analysing these graphs helps you pull out key values, identify trends, and make predictions. This section practises reading tables and a typical force–extension graph used to test materials.

数据处理是数学、科学和工程学共享的一项技能。你经常会遇到力与伸长量、温度与时间或载荷与挠度之间的关系图。分析这些图表有助于你提取关键数值、识别趋势并进行预测。本节将练习阅读表格以及一种用于测试材料的典型力‑伸长量图。

A student tests a spring by hanging different weights and measuring its length. The results are shown below.

一名学生通过悬挂不同的重物来测试弹簧，并测量其长度。结果如下表所示。

Weight (N) / 重量 (N)	Length (mm) / 长度 (mm)
0	100
5	120
10	140
15	160
20	185
25	215

Question: Plot a graph of weight (x‑axis) against extension (y‑axis). First calculate the extension for each weight. Which weight provides the first evidence that the spring has passed its elastic limit?

问题：绘制重量（x 轴）‑伸长量（y 轴）图。首先计算每个重量下的伸长量。哪个重量最先证明弹簧已超过其弹性极限？

mm. Extensions: 0 mm, 20 mm, 40 mm, 60 mm, 85 mm, 115 mm. Notice that from 0 to 15 N, each

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