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CIE 9709 2023 June Paper 42 Q4

A Level / CIE / M1

CIE 9709 2023 June Paper 42 Paper · Question 4

题目

Problem

An athlete of mass 84 kg84\,\mathrm{kg} is running along a straight road.

(a) Initially the road is horizontal and he runs at a constant speed of 3 m s−13\,\mathrm{m\,s^{-1}}. The athlete produces a constant power of 60 W60\,\mathrm{W}.

Find the resistive force which acts on the athlete.

(1)

(b) The athlete then runs up a 150 m150\,\mathrm{m} section of the road which is inclined at 0.8∘0.8^\circ to the horizontal. The speed of the athlete at the start of this section of road is 3 m s−13\,\mathrm{m\,s^{-1}} and he now produces a constant driving force of 24 N24\,\mathrm{N}. The total resistive force which acts on the athlete along this section of road has constant magnitude 13 N13\,\mathrm{N}.

Use an energy method to find the speed of the athlete at the end of the 150 m150\,\mathrm{m} section of road.

(6)

题目中文翻译

一名质量为 84 kg84\,\mathrm{kg} 的运动员沿笔直道路跑步。

(a) 起初道路水平，他以 3 m s−13\,\mathrm{m\,s^{-1}} 的恒定速率跑步。运动员输出的功率恒为 60 W60\,\mathrm{W}。

求作用于运动员的阻力。

(1)

(b) 随后运动员沿一段长 150 m150\,\mathrm{m}、与水平面成 0.8∘0.8^\circ 角的道路向上跑。他在这段道路起点的速率为 3 m s−13\,\mathrm{m\,s^{-1}}，现在产生的驱动力恒为 24 N24\,\mathrm{N}。在这段道路上，作用于运动员的总阻力大小恒为 13 N13\,\mathrm{N}。

使用能量法，求运动员到达这段 150 m150\,\mathrm{m} 道路终点时的速率。

(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

(a)

解法一

思路

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水平匀速时推力等于阻力，用功率除以速率。

答题过程

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At constant speed on the horizontal road, the resistance equals the driving force. Hence

R=603=20 N.\boxed{R=\frac{60}{3}=20\,\mathrm{N}}.

(b)

解法一

思路

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按照指定能量法，驱动力功扣去阻力功及势能增加，等于动能变化。

答题过程

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Let the final speed be vv. The work-energy equation is

24(150)−13(150)= 126000sin⁡0.8∘ +42(v2−32).\begin{align*} 24(150)-13(150)=&\,126000\sin0.8^\circ\\[4mm] &\,\hspace{2pt}+42(v^2-3^2). \end{align*}

Thus

42v2=1650+378−126000sin⁡0.8∘.42v^2=1650+378-126000\sin0.8^\circ.

Taking the positive root,

v=2028−126000sin⁡0.8∘42≈2.53 m s−1.v=\sqrt{\frac{2028-126000\sin0.8^\circ}{42}} \approx\boxed{2.53\,\mathrm{m\,s^{-1}}}.