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CIE 9709 2025 March Paper 42 Q3

A Level / CIE / M1

CIE 9709 2025 March Paper 42 Paper · Question 3

题目

Problem

An aeroplane is flying at a constant speed.

(a) The aeroplane is flying horizontally. The aeroplane’s engines are producing a constant power of 5500 kW5500\,\mathrm{kW}, and the aeroplane experiences a constant horizontal resistance force of 25 kN25\,\mathrm{kN}.

Find the speed of the aeroplane.

(2)

(b) The aeroplane then ascends 300 m300\,\mathrm{m} in 50 s50\,\mathrm{s}, while maintaining the same speed. The resistance force is no longer constant, and the work done against the resistance force in ascending 300 m300\,\mathrm{m} is 270 000 kJ270\,000\,\mathrm{kJ}. The mass of the aeroplane is 60 000 kg60\,000\,\mathrm{kg}.

Find the average power of the aeroplane’s engines.

(4)
题目中文翻译

一架飞机以恒定速率飞行。

(a) 飞机水平飞行。飞机发动机产生的功率恒为 5500 kW5500\,\mathrm{kW},飞机受到大小恒为 25 kN25\,\mathrm{kN} 的水平阻力。

求飞机的速率。

(2)

(b) 随后飞机在保持相同速率的同时,在 50 s50\,\mathrm{s} 内上升 300 m300\,\mathrm{m}。阻力不再恒定,在上升这 300 m300\,\mathrm{m} 的过程中,克服阻力所做的功为 270 000 kJ270\,000\,\mathrm{kJ}。飞机的质量为 60 000 kg60\,000\,\mathrm{kg}。

求飞机发动机的平均功率。

(4)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

Where a numerical value for the acceleration due to gravity (gg) is needed, use 10 m s−210\,\mathrm{m\,s^{-2}}.

需要使用重力加速度 gg 的数值时,取 10 m s−210\,\mathrm{m\,s^{-2}}。

解答

(a)

解法一

思路

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水平匀速时推力等于阻力。统一单位后用功率等于力乘速率。

答题过程

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Since the aeroplane travels horizontally at constant speed, its thrust is 25000 N25000\,\mathrm{N}. Using P=FvP=Fv,

5500000=25000v,v=220 m s−1.5500000=25000v, \qquad \boxed{v=220\,\mathrm{m\,s^{-1}}}.

(b)

解法一

思路

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速率不变,所以动能不变。发动机做功等于重力势能增加与克服阻力做功之和,再除以总时间求平均功率。

答题过程

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The increase in gravitational potential energy is

60000(10)(300)=180000000 J.60000(10)(300)=180000000\,\mathrm{J}.

The speed is unchanged, so the change in kinetic energy is zero. The work done by the engines is therefore

W= 180000000+270000(1000)= 450000000 J.\begin{align*} W=&\,180000000+270000(1000)\\[4mm] =&\,450000000\,\mathrm{J}. \end{align*}

Hence the average power is

P‾= 45000000050= 9000000 W=9000 kW.\begin{align*} \overline P=&\,\frac{450000000}{50}\\[4mm] =&\,9000000\,\mathrm{W}=\boxed{9000\,\mathrm{kW}}. \end{align*}