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M1 2021 June Q4

A Level / Edexcel / M1

IAL 2021 June Paper · Question 4

题目

Problem

4.

Figure 1 shows a large bucket used by a crane on a building site to move materials between the ground and the top of the building. The mass of the bucket is 15 kg15\,\mathrm{kg}.

The bucket is attached to a vertical cable with the bottom of the bucket horizontal.

The cable is modelled as light and inextensible.

When the bucket is on the ground, a bag of cement of mass 25 kg25\,\mathrm{kg} is placed in the bucket.

The bucket with the bag of cement moves vertically upwards with constant acceleration 0.2 m s−20.2\,\mathrm{m\,s^{-2}}. Air resistance is modelled as being negligible.

(a) Find the tension in the cable.

(3)

At the top of the building, the bag of cement is removed. A box of tools of mass 12 kg12\,\mathrm{kg} is now placed in the bucket.

Later on the bucket with the box of tools is moving vertically downwards with constant deceleration 0.1 m s−20.1\,\mathrm{m\,s^{-2}}. Air resistance is again modelled as being negligible.

(b) Find the magnitude of the normal reaction between the bucket and the box of tools.

(3)
题目中文翻译

图 1 显示建筑工地起重机使用的一个大吊桶,用来在地面和楼顶之间搬运物料。吊桶的质量为 15 kg15\,\mathrm{kg}。

吊桶连接在一根竖直缆绳上,桶底保持水平。将缆绳视为轻且不可伸长的绳。

吊桶位于地面时,将一袋质量为 25 kg25\,\mathrm{kg} 的水泥放入桶内。

吊桶连同水泥袋以恒定加速度 0.2 m s−20.2\,\mathrm{m\,s^{-2}} 竖直向上运动。将空气阻力视为可忽略。

(a) 求缆绳中的张力。

(3)

到达楼顶后,取出水泥袋,将一个质量为 12 kg12\,\mathrm{kg} 的工具箱放入桶内。

之后,吊桶连同工具箱竖直向下运动,恒定减速度为 0.1 m s−20.1\,\mathrm{m\,s^{-2}}。仍将空气阻力视为可忽略。

(b) 求吊桶与工具箱之间法向反作用力的大小。

(3)

解答

(a)

解法一

思路

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把桶和水泥作为整体,内部接触力互相抵消。向上加速,所以拉力必须大于整体重力。

答题过程

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Treat the bucket and cement as one system of mass 15+25=4015+25=40 kg. Taking upwards as positive, Newton’s second law gives

T−40g= 40(0.2),T= 40(9.8+0.2)=400 N.\begin{align*} T-40g=&\,40(0.2),\\ T=&\,40(9.8+0.2)=\boxed{400\ \mathrm N}. \end{align*}

(b)

解法一

思路

展开

向下运动但正在减速,因此加速度向上。要求的是桶与工具箱之间的支持力,应只对工具箱列方程。

答题过程

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The tools are moving downwards but slowing down, so their acceleration is 0.1 m s−20.1\ \mathrm{m\,s^{-2}} upwards.

Let RR be the upward normal reaction on the 12 kg box of tools. Taking upwards as positive,

R−12g= 12(0.1),R= 12(9.8+0.1)=118.8 N.\begin{align*} R-12g=&\,12(0.1),\\ R=&\,12(9.8+0.1)=118.8\ \mathrm N. \end{align*}

The required magnitude is 119 N\boxed{119\ \mathrm N} (3 s.f.).