题目

Problem

Two children, Alan and Bhavana, are standing on the horizontal floor of a lift, as shown in Figure 3.

The lift has mass $250\,\mathrm{kg}$ . The lift is raised vertically upwards with constant acceleration by a vertical cable which is attached to the top of the lift. The cable is modelled as being light and inextensible. While the lift is accelerating upwards, the tension in the cable is $3616\,\mathrm{N}$ .

As the lift accelerates upwards, the floor of the lift exerts a force of magnitude $565\,\mathrm{N}$ on Alan and a force of magnitude $226\,\mathrm{N}$ on Bhavana.

Air resistance is modelled as being negligible and Alan and Bhavana are modelled as particles.

(a) By considering the forces acting on the lift only, find the acceleration of the lift.

(3)

(b) Find the mass of Alan.

(3)

题目中文翻译

如图 3 所示，两个孩子 Alan 和 Bhavana 站在电梯的水平地板上。

电梯质量为 $250\,\mathrm{kg}$ ，由连接在电梯顶部的竖直缆绳以恒定加速度竖直向上提升。将缆绳视为轻且不可伸长的绳。电梯向上加速时，缆绳中的张力为 $3616\,\mathrm{N}$ 。

电梯向上加速时，地板对 Alan 施加大小为 $565\,\mathrm{N}$ 的力，对 Bhavana 施加大小为 $226\,\mathrm{N}$ 的力。

将空气阻力视为可忽略，将 Alan 和 Bhavana 视为质点。

(a) 仅考虑作用在电梯上的力，求电梯的加速度。

(3)

(b) 求 Alan 的质量。

(3)

解答

(a)

解法一

思路

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只研究电梯时，孩子对地板的作用力也属于电梯受到的外力。根据牛顿第三定律，这两个力均向下，大小分别等于电梯对孩子的支持力。

答题过程

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By Newton’s third law, Alan and Bhavana exert downward forces of 565 N and 226 N on the lift.

Considering the lift alone and taking upwards as positive,

\begin{align*} 3616-250g-565-226=&\,250a,\\ 375=&\,250a,\\ a=&\,\boxed{1.5\ \mathrm{m\,s^{-2}}\text{ upwards}}. \end{align*}

(b)

解法一

思路

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对 Alan 单独列牛顿第二定律。他与电梯具有相同的向上加速度，支持力与重力的差提供合力。

答题过程

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Let Alan’s mass be $m_A$ kg. Taking upwards as positive for Alan,

\begin{align*} 565-m_Ag=&\,1.5m_A,\\ 565=&\,(9.8+1.5)m_A,\\ m_A=&\,\boxed{50\ \mathrm{kg}}. \end{align*}

M1 2022 Oct Q4

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程