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CIE 9709 2024 June Paper 41 Q4

A Level / CIE / M1

CIE 9709 2024 June Paper 41 Paper · Question 4

题目

Problem

A car of mass 1700 kg1700\,\mathrm{kg} is pulling a trailer of mass 300 kg300\,\mathrm{kg} along a straight horizontal road. The car and trailer are connected by a light inextensible cable which is parallel to the road. There are constant resistances to motion of 400 N400\,\mathrm{N} on the car and 150 N150\,\mathrm{N} on the trailer. The power of the car’s engine is 14 000 W14\,000\,\mathrm{W}.

Find the acceleration of the car and the tension in the cable when the speed is 20 m s−120\,\mathrm{m\,s^{-1}}.

(6)

题目中文翻译

一辆质量为 1700 kg1700\,\mathrm{kg} 的汽车拉着一辆质量为 300 kg300\,\mathrm{kg} 的拖车,沿一条笔直的水平道路行驶。汽车和拖车由一根平行于道路的轻且不可伸长的缆绳连接。汽车和拖车受到的恒定运动阻力分别为 400 N400\,\mathrm{N} 和 150 N150\,\mathrm{N}。汽车发动机的功率为 14 000 W14\,000\,\mathrm{W}。

求速率为 20 m s−120\,\mathrm{m\,s^{-1}} 时汽车的加速度和缆绳中的张力。

(6)

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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功率除以速率得驱动力。对整体求加速度,内部张力消去;再单独对拖车求张力。

答题过程

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The driving force at the stated speed is

F=1400020=700 N.F=\frac{14000}{20}=700\,\mathrm{N}.

For the car and trailer together,

700−400−150=(1700+300)a.700-400-150=(1700+300)a.

Thus

a=1502000=0.075 m s−2.\boxed{a=\frac{150}{2000}=0.075\,\mathrm{m\,s^{-2}}}.

For the trailer alone,

T−150=300a,T=172.5 N.T-150=300a, \qquad \boxed{T=172.5\,\mathrm{N}}.