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CIE 9709 2024 November Paper 43 Q1

A Level / CIE / M1

CIE 9709 2024 Nov Paper 43 Paper · Question 1

题目

Problem

An athlete has mass m kgm\,\mathrm{kg}. The athlete runs along a horizontal road against a constant resistance force of magnitude 24 N24\,\mathrm{N}. The total work done by the athlete in increasing his speed from 5 m s−15\,\mathrm{m\,s^{-1}} to 6 m s−16\,\mathrm{m\,s^{-1}} while running a distance of 5050 metres is 1541 J1541\,\mathrm{J}.

Find the value of mm.

(4)

题目中文翻译

一名运动员的质量为 m kgm\,\mathrm{kg}。运动员沿一条水平道路跑步,受到大小为 24 N24\,\mathrm{N} 的恒定阻力。运动员在跑过 5050 米的过程中,将速率从 5 m s−15\,\mathrm{m\,s^{-1}} 增加到 6 m s−16\,\mathrm{m\,s^{-1}},所做的总功为 1541 J1541\,\mathrm{J}。

求 mm 的值。

(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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总做功分成增加的动能与克服阻力的功;没有给恒定驱动力,不能默认匀加速。

答题过程

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The work done against resistance is

24(50)=1200 J.24(50)=1200\,\mathrm{J}.

By the work-energy equation,

1541=12m(62−52)+1200.1541=\frac12m(6^2-5^2)+1200.

Thus 5.5m=3415.5m=341, giving

m=62.\boxed{m=62}.