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CIE 9709 2025 November Paper 41 Q8

A Level / CIE / M1

CIE 9709 2025 Nov Paper 41 Paper · Question 8

题目

Problem

A block AA of mass 2 kg2\,\mathrm{kg} and a particle BB of mass 0.5 kg0.5\,\mathrm{kg} are connected by a light inextensible string inclined at 15∘15^\circ to the horizontal. They are pulled across a horizontal surface with acceleration 1.2 m s−21.2\,\mathrm{m\,s^{-2}} by a force of magnitude 6 N6\,\mathrm{N}, applied to AA, acting at 20∘20^\circ above the horizontal as shown in the diagram. The string and the force applied to AA are in the same vertical plane.

The contact between BB and the surface is smooth and the contact between AA and the surface is rough.

(a) Find the tension in the string.

(2)

(b) Find the coefficient of friction between AA and the surface.

(6)
题目中文翻译

一个质量为 2 kg2\,\mathrm{kg} 的物块 AA 和一个质量为 0.5 kg0.5\,\mathrm{kg} 的质点 BB,由一根与水平面成 15∘15^\circ 角的轻质不可伸长细绳连接。一个大小为 6 N6\,\mathrm{N} 的力作用于 AA,方向在水平方向上方 20∘20^\circ,如图所示,拉着它们以 1.2 m s−21.2\,\mathrm{m\,s^{-2}} 的加速度沿水平表面运动。细绳与作用于 AA 的力位于同一竖直平面内。

BB 与表面之间的接触光滑,AA 与表面之间的接触粗糙。

(a) 求细绳中的张力。

(2)

(b) 求 AA 与表面之间的摩擦系数。

(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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B的表面光滑,只有张力的水平分量提供向左的加速度。

答题过程

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Taking leftwards as positive, Newton’s second law for BB gives

Tcos⁡15∘=0.5(1.2)=0.6.T\cos15^\circ=0.5(1.2)=0.6.

Therefore

T=0.6cos⁡15∘=0.621165708…≈0.621 N.T=\frac{0.6}{\cos15^\circ}=0.621165708\ldots \approx\boxed{0.621\text{ N}}.

(b)

解法一

思路

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A所受张力向右下,竖直分量增加正压力;水平运动向左,摩擦向右。

答题过程

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For AA, the tension acts to the right and downwards. Vertical equilibrium gives

R+6sin⁡20∘=2(10)+Tsin⁡15∘.R+6\sin20^\circ=2(10)+T\sin15^\circ.

Using the unrounded tension from (a),

R=20+Tsin⁡15∘−6sin⁡20∘=18.108648656… N.R=20+T\sin15^\circ-6\sin20^\circ =18.108648656\ldots\text{ N}.

Taking leftwards as positive, the horizontal equation for AA is

6cos⁡20∘−Tcos⁡15∘−F=2(1.2).6\cos20^\circ-T\cos15^\circ-F=2(1.2).

Thus

F=6cos⁡20∘−0.6−2.4=2.638155725… N.F=6\cos20^\circ-0.6-2.4=2.638155725\ldots\text{ N}.

Since F=μRF=\mu R,

μ= 2.638155725…18.108648656…= 0.145684848…≈0.146.\begin{align*} \mu=&\,\frac{2.638155725\ldots}{18.108648656\ldots}\\[4mm] =&\,0.145684848\ldots\approx\boxed{0.146}. \end{align*}