Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2023 November Paper 42 Q1

A Level / CIE / M1

CIE 9709 2023 Nov Paper 42 Paper · Question 1

题目

Problem

A block of mass 15 kg15\,\mathrm{kg} slides down a line of greatest slope of an inclined plane. The top of the plane is at a vertical height of 1.6 m1.6\,\mathrm{m} above the level of the bottom of the plane. The speed of the block at the top of the plane is 2 m s−12\,\mathrm{m\,s^{-1}} and the speed of the block at the bottom of the plane is 4 m s−14\,\mathrm{m\,s^{-1}}.

Find the work done against the resistance to motion of the block.

(4)

题目中文翻译

一个质量为 15 kg15\,\mathrm{kg} 的物块沿斜面的最大坡度线向下滑动。斜面顶端比底端所在水平面高 1.6 m1.6\,\mathrm{m}。物块在斜面顶端的速率为 2 m s−12\,\mathrm{m\,s^{-1}},在斜面底端的速率为 4 m s−14\,\mathrm{m\,s^{-1}}。

求克服物块运动阻力所做的功。

(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

展开

初动能加势能损失,扣除末动能,余下就是克服阻力的功,不必知道斜面长或倾角。

答题过程

展开

Let WW be the work done against resistance. The work-energy equation gives

12(15)(22)+15(10)(1.6)= 12(15)(42)+W.\begin{align*} \frac12(15)(2^2)+15(10)(1.6) =&\,\frac12(15)(4^2)+W. \end{align*}

Hence

W=30+240−120=150 J.W=30+240-120=\boxed{150\,\mathrm{J}}.