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CIE 9709 2024 November Paper 43 Q3

A Level / CIE / M1

CIE 9709 2024 Nov Paper 43 Paper · Question 3

题目

Problem

A car of mass 1600 kg1600\,\mathrm{kg} travels up a slope inclined at an angle of sin⁡−10.08\sin^{-1}0.08 to the horizontal. There is a constant resistance of magnitude 240 N240\,\mathrm{N} acting on the car.

(a) It is given that the car travels at a constant speed of 32 m s−132\,\mathrm{m\,s^{-1}}.

Find the power of the engine of the car.

(3)

(b) Find the acceleration of the car when its speed is 24 m s−124\,\mathrm{m\,s^{-1}} and the engine is working at 95%95\% of the power found in (a).

(3)

题目中文翻译

一辆质量为 1600 kg1600\,\mathrm{kg} 的汽车沿与水平面成 sin⁡−10.08\sin^{-1}0.08 角的斜坡向上行驶。汽车受到大小为 240 N240\,\mathrm{N} 的恒定阻力。

(a) 已知汽车以 32 m s−132\,\mathrm{m\,s^{-1}} 的恒定速率行驶。

求汽车发动机的功率。

(3)

(b) 当汽车的速率为 24 m s−124\,\mathrm{m\,s^{-1}},发动机以 (a) 中所求功率的 95%95\% 工作时,求汽车的加速度。

(3)

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

匀速上坡时推力平衡阻力及重力沿坡分量,再用功率等于力乘速率。

答题过程

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Let the inclination be α\alpha, so sin⁡α=0.08\sin\alpha=0.08. At constant speed, the driving force is

F=240+1600(10)(0.08)=1520 N.F=240+1600(10)(0.08)=1520\,\mathrm{N}.

Therefore the engine power is

P=Fv=1520(32)=48640 W.P=Fv=1520(32)=\boxed{48640\,\mathrm{W}}.

(b)

解法一

思路

展开

功率按95%调整,给定速率下求新的推力,再扣除两个向下的力。

答题过程

展开

The power is 0.95(48640)=46208 W0.95(48640)=46208\,\mathrm{W}, so the driving force at speed 24 m s−124\,\mathrm{m\,s^{-1}} is

F=4620824=57763 N.F=\frac{46208}{24}=\frac{5776}{3}\,\mathrm{N}.

Resolving up the slope,

1600a=57763−240−1280=12163.1600a=\frac{5776}{3}-240-1280=\frac{1216}{3}.

Hence

a=1975 m s−2≈0.253 m s−2.\boxed{a=\frac{19}{75}\,\mathrm{m\,s^{-2}}} \approx0.253\,\mathrm{m\,s^{-2}}.