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CIE 9709 2022 March Paper 42 Q4

A Level / CIE / M1

CIE 9709 2022 March Paper 42 Paper · Question 4

题目

Problem

The total mass of a cyclist and her bicycle is 70 kg70\,\mathrm{kg}. The cyclist is riding with constant power of 180 W180\,\mathrm{W} up a straight hill inclined at an angle α\alpha to the horizontal, where sin⁡α=0.05\sin\alpha=0.05. At an instant when the cyclist’s speed is 6 m s−16\,\mathrm{m\,s^{-1}}, her acceleration is −0.2 m s−2-0.2\,\mathrm{m\,s^{-2}}. There is a constant resistance to motion of magnitude F NF\,\mathrm{N}.

(a) Find the value of FF.

(4)

(b) Find the steady speed that the cyclist could maintain up the hill when working at this power.

(2)

题目中文翻译

一名骑车者与她的自行车的总质量为 70 kg70\,\mathrm{kg}。她以 180 W180\,\mathrm{W} 的恒定功率沿直山坡上行,山坡与水平面的夹角为 α\alpha,其中 sin⁡α=0.05\sin\alpha=0.05。在她的速率为 6 m s−16\,\mathrm{m\,s^{-1}} 的某瞬间,加速度为 −0.2 m s−2-0.2\,\mathrm{m\,s^{-2}}。运动中受到大小为 F NF\,\mathrm{N} 的恒定阻力。

(a) 求 FF 的值。

(4)

(b) 求她以此功率骑行时能够维持的上坡稳定速率。

(2)

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

展开

由恒功率与瞬时速率求驱动力。加速度是沿上坡正方向的负值,列式时要保留其符号。

答题过程

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At speed 6 m s−16\,\mathrm{m\,s^{-1}}, the driving force is

D=Pv=1806=30 N.D=\frac Pv=\frac{180}{6}=30\,\mathrm{N}.

Taking up the slope as positive, Newton’s second law gives

30−F−70(10)(0.05)=70(−0.2).30-F-70(10)(0.05)=70(-0.2).

Thus 30−F−35=−1430-F-35=-14, giving F=9\boxed{F=9}.

(b)

解法一

思路

展开

稳态速率对应加速度为零,此时驱动力等于阻力与重力下坡分量之和。

答题过程

展开

At steady speed vv, the resultant force is zero:

180v−9−70(10)(0.05)=0.\frac{180}{v}-9-70(10)(0.05)=0.

Thus 180/v=44180/v=44, and

v=4511≈4.09 m s−1(3 s.f.).v=\frac{45}{11}\approx\boxed{4.09\,\mathrm{m\,s^{-1}}}\quad\text{(3 s.f.)}.