AS CAIE Engineering: Unit Test Mock Paper Analysis | AS CAIE 工程：单元测试模拟卷解析

📚 AS CAIE Engineering: Unit Test Mock Paper Analysis | AS CAIE 工程：单元测试模拟卷解析

Unit tests in AS Engineering serve as crucial checkpoints that consolidate your understanding of core principles across mechanics, materials, thermodynamics, and electronics. This mock paper analysis walks you through ten representative problems, each targeting a key syllabus area, with step‑by‑step solutions that mirror the clarity expected in the CAIE examination. By studying these worked examples, you will sharpen your problem‑solving technique and gain confidence in applying fundamental equations under timed conditions.

AS 工程学的单元测试是巩固力学、材料、热力学和电子学核心原理的关键环节。这份模拟卷解析带领你逐一突破十道典型题目，每道题锁定一个核心考纲领域，并给出与 CAIE 考试要求相匹配的清晰分步解答。通过研读这些范例，你将打磨解题技巧，增强在规定时间内应用基础公式的信心。

---

1. Force Equilibrium and Free‑Body Diagrams | 力平衡与受力分析图

Question: A uniform beam of length 4 m and weight 200 N rests on a support at one end (A) and is held horizontal by a cable at the other end (B). The cable makes an angle of 30° with the beam. A load of 300 N is placed 1 m from A. Determine the tension in the cable and the reaction at support A.

问题：一均匀梁长 4 m，重 200 N，一端（A）由支撑支持，另一端（B）通过缆绳保持水平；缆绳与梁成 30° 角。距 A 点 1 m 处放置 300 N 的负载。求缆绳张力以及 A 点的支撑反力。

Step 1: Draw a free‑body diagram showing all forces: weight of beam 200 N acting at its centre (2 m from A), the 300 N load at 1 m from A, tension T at point B at 30° to the beam, and the reaction at A with horizontal component Rₓ and vertical component Rᵧ.

步骤 1：画出受力分析图，标出所有作用力：梁重 200 N 作用于中点（距 A 2 m），300 N 负载距 A 1 m，B 点张力 T 与梁成 30° 角，A 点反力分解为水平分力 Rₓ 和竖直分力 Rᵧ。

Step 2: Take moments about point A to eliminate the unknown reaction components. Clockwise moments: (300 N × 1 m) + (200 N × 2 m). Anti‑clockwise moment: (T sin 30° × 4 m). For rotational equilibrium, ΣMₐ = 0: T sin 30° × 4 = 300 × 1 + 200 × 2.

步骤 2：对 A 点取矩以消去未知反力。顺时针力矩：(300 N × 1 m) + (200 N × 2 m)。逆时针力矩：(T sin 30° × 4 m)。根据转动平衡 ΣMₐ = 0：T sin 30° × 4 = 300 × 1 + 200 × 2。

Step 3: Solve for T. With sin 30° = 0.5, 4T × 0.5 = 2T = 300 + 400 = 700 → T = 350 N. Then apply ΣFᵧ = 0: Rᵧ + T sin 30° − 200 − 300 = 0 → Rᵧ + 175 − 500 = 0 → Rᵧ = 325 N. ΣFₓ = 0: Rₓ − T cos 30° = 0 → Rₓ = 350 × (√3/2) ≈ 303.1 N.

步骤 3：解出 T。sin 30° = 0.5，4T × 0.5 = 2T = 300 + 400 = 700 → T = 350 N。然后利用 ΣFᵧ = 0：Rᵧ + T sin 30° − 200 − 300 = 0 → Rᵧ + 175 − 500 = 0 → Rᵧ = 325 N。ΣFₓ = 0：Rₓ − T cos 30° = 0 → Rₓ = 350 × (√3/2) ≈ 303.1 N。

---

2. Stress and Strain Calculations | 应力与应变计算

Question: A steel wire of diameter 8 mm and original length 2.0 m is subjected to a tensile force of 4.0 kN. The Young modulus for the steel is 210 GPa. Calculate (a) the stress in the wire, (b) the strain produced, and (c) the extension of the wire.

问题：一根直径 8 mm、原长 2.0 m 的钢丝承受 4.0 kN 的拉伸载荷。钢材的杨氏模量为 210 GPa。计算 (a) 钢丝内的应力，(b) 产生的应变，以及 (c) 钢丝的伸长量。

Step 1: Compute the cross‑sectional area A = πd²/4 = π × (0.008 m)² ÷ 4 = π × 6.4×10⁻⁵ ÷ 4? More directly: A = π × (8×10⁻³)² / 4 = 5.027×10⁻⁵ m².

步骤 1：计算截面积 A = πd²/4 = π × (0.008 m)² ÷ 4 = 5.027×10⁻⁵ m²。

Step 2: Determine tensile stress σ = F / A = 4000 N / 5.027×10⁻⁵ m² ≈ 79.6×10⁶ Pa = 79.6 MPa.

步骤 2：计算拉应力 σ = F / A = 4000 N / 5.027×10⁻⁵ m² ≈ 79.6×10⁶ Pa = 79.6 MPa。

Step 3: Strain ε = σ / E. Since E = 210 GPa = 210×10⁹ Pa, ε = 79.6×10⁶ / 210×10⁹ = 3.79×10⁻⁴ (no units). The extension ΔL = ε × original length = 3.79×10⁻⁴ × 2.0 m = 7.58×10⁻⁴ m ≈ 0.758 mm.

步骤 3：应变 ε = σ / E。E = 210 GPa = 210×10⁹ Pa，所以 ε = 79.6×10⁶ / 210×10⁹ = 3.79×10⁻⁴（无量纲）。伸长量 ΔL = ε × 原长 = 3.79×10⁻⁴ × 2.0 m = 7.58×10⁻⁴ m ≈ 0.758 mm。

---

3. Kinematics of Constant Acceleration | 匀加速运动学

Question: A car accelerates from rest at a constant rate of 2.5 m/s² for 8 seconds. Find its final velocity and the distance it covers during this period.

问题：一辆汽车从静止开始以 2.5 m/s² 的恒定加速度行驶 8 秒。求末速度以及在这段时间内行驶的距离。

Step 1: For uniformly accelerated motion from rest (u = 0), use v = u + a t. v = 0 + 2.5 × 8 = 20 m/s.

步骤 1：对于从静止开始的匀加速运动（初速 u = 0），运用 v = u + a t。v = 0 + 2.5 × 8 = 20 m/s。

Step 2: Displacement s = u t + ½ a t² = 0 + ½ × 2.5 × (8)² = ½ × 2.5 × 64 = 80 m.

步骤 2：位移 s = u t + ½ a t² = 0 + ½ × 2.5 × (8)² = ½ × 2.5 × 64 = 80 m。

Step 3: Alternative check using average velocity: v_avg = (u + v)/2 = 10 m/s, s = v_avg × t = 10 × 8 = 80 m, confirming the result.

步骤 3：也可用平均速度验证：v_avg = (u + v)/2 = 10 m/s，s = v_avg × t = 10 × 8 = 80 m，结果一致。

---

4. Newton’s Second Law and Friction | 牛顿第二定律与摩擦力

Question: A box of mass 80 kg rests on a horizontal surface. A horizontal force of 250 N is applied. If the coefficient of kinetic friction is 0.25, determine the acceleration of the box. (Take g = 9.81 m/s²)

问题：一质量为 80 kg 的箱子置于水平面上，施加 250 N 的水平力。若动摩擦系数为 0.25，求箱子的加速度。（取 g = 9.81 m/s²）

Step 1: Find the normal reaction N. Since no vertical acceleration, N = weight = m g = 80 × 9.81 = 784.8 N.

步骤 1：求法向反力 N。因竖直方向无加速度，N = 重力 = m g = 80 × 9.81 = 784.8 N。

Step 2: Calculate kinetic friction force f = μ N = 0.25 × 784.8 = 196.2 N.

步骤 2：计算动摩擦力 f = μ N = 0.25 × 784.8 = 196.2 N。

Step 3: Apply Newton’s second law horizontally: F_net = applied force − friction = 250 − 196.2 = 53.8 N. Acceleration a = F_net / m = 53.8 / 80 = 0.6725 m/s² ≈ 0.67 m/s².

步骤 3：水平方向应用牛顿第二定律：合力 F_net = 作用力 − 摩擦力 = 250 − 196.2 = 53.8 N。加速度 a = F_net / m = 53.8 / 80 = 0.6725 m/s² ≈ 0.67 m/s²。

---

5. Work, Energy and Power | 功、能与功率

Question: A crane lifts a mass of 600 kg vertically through a height of 15 m at a constant speed. If the crane’s motor has an efficiency of 70%, determine the useful work done, the total electrical energy input, and the average power if the lift takes 20 seconds.

问题：起重机将 600 kg 的重物以恒定速度垂直提升 15 m。若电动机效率为 70%，求有用功、输入的总电能，以及提升用时 20 s 情况下的平均功率。

Step 1: Useful work done against gravity W = m g h = 600 × 9.81 × 15 = 88,290 J ≈ 88.3 kJ.

步骤 1：克服重力所做的有用功 W = m g h = 600 × 9.81 × 15 = 88,290 J ≈ 88.3 kJ。

Step 2: Efficiency η = useful work / total input energy → total input = W / η = 88,290 / 0.70 = 126,129 J ≈ 126.1 kJ.

步骤 2：效率 η = 有用功 / 总输入能量 → 总输入 = W / η = 88,290 / 0.70 = 126,129 J ≈ 126.1 kJ。

Step 3: Average power output P_avg = useful work / time = 88,290 / 20 = 4414.5 W ≈ 4.41 kW. (Alternatively, input power =

Published by TutorHao | AS 工程 Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）