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CIE 9709 2024 November Paper 43 Q5

A Level / CIE / M1

CIE 9709 2024 Nov Paper 43 Paper · Question 5

题目

Problem

A particle of mass 12 kg12\,\mathrm{kg} is going to be pulled across a rough horizontal plane by a light inextensible string. The string is at an angle of 30∘30^\circ above the plane and has tension T NT\,\mathrm{N} (see diagram). The coefficient of friction between the particle and the plane is 0.50.5.

(a) Given that the particle is on the point of moving, find the value of TT.

(5)

(b) Given instead that the particle is accelerating at 0.2 m s−20.2\,\mathrm{m\,s^{-2}}, find the value of TT.

(3)

题目中文翻译

一个质量为 12 kg12\,\mathrm{kg} 的质点将由一根轻且不可伸长的绳拉着在粗糙水平面上运动。绳在平面上方与平面成 30∘30^\circ 角,张力为 T NT\,\mathrm{N}(见图)。质点与平面之间的摩擦系数为 0.50.5。

(a) 已知质点恰好处于即将运动的状态,求 TT 的值。

(5)

(b) 改为已知质点以 0.2 m s−20.2\,\mathrm{m\,s^{-2}} 的加速度运动,求 TT 的值。

(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

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张力向上分量减小支持力,临界静摩擦为μR;水平平衡后消去R求张力。

答题过程

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Resolving vertically,

R+Tsin⁡30∘=120.R+T\sin30^\circ=120. R=120−T2.R=120-\frac T2.

At limiting equilibrium, friction is 0.5R0.5R. Resolving horizontally,

Tcos⁡30∘= 0.5R= 0.5(120−T2).\begin{align*} T\cos30^\circ=&\,0.5R\\[4mm] =&\,0.5\bigg(120-\frac T2\bigg). \end{align*}

Thus

(32+14)T=60,\bigg(\frac{\sqrt3}{2}+\frac14\bigg)T=60,

and

T=24023+1 N≈53.8 N.\boxed{T=\frac{240}{2\sqrt3+1}\,\mathrm{N} \approx53.8\,\mathrm{N}}.

(b)

解法一

思路

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这是另一个状态,张力和支持力都要重新求;只沿水平有加速度,竖直仍平衡。

答题过程

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Again R=120−T/2R=120-T/2, but its value differs from that in (a). Newton’s second law horizontally gives

Tcos⁡30∘−0.5(120−T2)=12(0.2).T\cos30^\circ-0.5\bigg(120-\frac T2\bigg)=12(0.2).

Hence

(32+14)T=62.4,\bigg(\frac{\sqrt3}{2}+\frac14\bigg)T=62.4,

so

T=249.623+1 N≈55.9 N.\boxed{T=\frac{249.6}{2\sqrt3+1}\,\mathrm{N} \approx55.9\,\mathrm{N}}.