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CIE 9709 2024 June Paper 41 Q3

A Level / CIE / M1

CIE 9709 2024 June Paper 41 Paper · Question 3

题目

Problem

A train of mass 180 000 kg180\,000\,\mathrm{kg} ascends a straight hill of length 1.5 km1.5\,\mathrm{km}, inclined at an angle of 1.5∘1.5^\circ to the horizontal. As it ascends the hill, the total work done to overcome the resistance to motion is 12 000 kJ12\,000\,\mathrm{kJ} and the speed of the train decreases from 45 m s−145\,\mathrm{m\,s^{-1}} to 40 m s−140\,\mathrm{m\,s^{-1}}.

Find the work done by the engine of the train as it ascends the hill, giving your answer in kJ\mathrm{kJ}.

(4)

题目中文翻译

一列质量为 180 000 kg180\,000\,\mathrm{kg} 的火车沿一段长 1.5 km1.5\,\mathrm{km}、与水平面成 1.5∘1.5^\circ 角的笔直山坡上行。上坡过程中,克服运动阻力所做的总功为 12 000 kJ12\,000\,\mathrm{kJ},火车的速率从 45 m s−145\,\mathrm{m\,s^{-1}} 降至 40 m s−140\,\mathrm{m\,s^{-1}}。

求火车上坡过程中发动机所做的功,答案以 kJ\mathrm{kJ} 为单位。

(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

解法一

思路

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用能量方程,发动机做功等于势能增加、克服阻力做功及动能变化之和。动能在减少,所以该项应为负;最后转成千焦。

答题过程

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The vertical rise is 1500sin⁡1.5∘ m1500\sin1.5^\circ\,\mathrm{m}. The work done by the engine is

W= 180000(10)(1500sin⁡1.5∘) +12000000 +12(180000)(402−452)= 70677760.431…+12000000 −38250000= 44427760.431… J.\begin{align*} W=&\,180000(10)(1500\sin1.5^\circ)\\[4mm] &\,\hspace{2pt}+12000000\\[4mm] &\,\hspace{4pt}+\frac12(180000)(40^2-45^2)\\[4mm] =&\,70677760.431\ldots+12000000\\[4mm] &\,\hspace{2pt}-38250000\\[4mm] =&\,44427760.431\ldots\,\mathrm{J}. \end{align*}

Hence, in the requested units,

W≈44400 kJ(3 s.f.).W\approx\boxed{44400\,\mathrm{kJ}}\quad\text{(3 s.f.)}.