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CIE 9709 2025 June Paper 41 Q1

A Level / CIE / M1

CIE 9709 2025 June Paper 41 Paper · Question 1

题目

Problem

A block of mass 12 kg12\,\mathrm{kg} is being pulled by a rope up a rough plane. The plane is inclined at an angle of 20∘20^\circ above the horizontal. The rope pulling the block is parallel to a line of greatest slope of the plane. The coefficient of friction between the block and the plane is 0.40.4. The acceleration of the block is 2 m s−22\,\mathrm{m\,s^{-2}}.

Find the tension in the rope.

(4)
题目中文翻译

一个质量为 12 kg12\,\mathrm{kg} 的物块被一根绳子沿粗糙斜面向上拉动。斜面与水平面的夹角为 20∘20^\circ。拉动物块的绳子平行于斜面的一条最大坡度线。物块与斜面之间的摩擦系数为 0.40.4。物块的加速度为 2 m s−22\,\mathrm{m\,s^{-2}}。

求绳子中的张力。

(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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绳沿坡方向,不改变正压力;上坡时摩擦与重力分量均向下。

答题过程

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The normal contact force is

R=12(10)cos⁡20∘.R=12(10)\cos20^\circ.

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

T−0.4R−12(10)sin⁡20∘=12(2).T-0.4R-12(10)\sin20^\circ=12(2).

Hence

T= 24+48cos⁡20∘+120sin⁡20∘= 110.147663… N≈ 110 N (3 s.f.).\begin{align*} T=&\,24+48\cos20^\circ+120\sin20^\circ\\[4mm] =&\,110.147663\ldots\text{ N}\\[4mm] \approx&\,\boxed{110\text{ N}}\text{ (3 s.f.)}. \end{align*}