Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2025 November Paper 43 Q3

A Level / CIE / M1

CIE 9709 2025 Nov Paper 43 Paper · Question 3

题目

Problem

A block of mass 4 kg4\,\mathrm{kg} is pulled along a rough horizontal road by a constant force of magnitude 25 N25\,\mathrm{N} acting at an angle of 36∘36^\circ above the horizontal. The block moves in a straight line passing through two points AA and BB on the road, where AB=120 mAB=120\,\mathrm{m}. The coefficient of friction between the block and the road is 0.40.4.

(a) Find the work done against friction in moving the block from AA to BB.

(3)

The speed of the block at AA is 7 m s−17\,\mathrm{m\,s^{-1}}.

(b) Use an energy method to find the speed of the block at BB.

(4)
题目中文翻译

一个质量为 4 kg4\,\mathrm{kg} 的物块,在大小为 25 N25\,\mathrm{N}、方向在水平方向上方 36∘36^\circ 的恒力作用下,沿粗糙水平道路运动。物块沿直线经过道路上的两点 AA 和 BB,其中 AB=120 mAB=120\,\mathrm{m}。物块与道路之间的摩擦系数为 0.40.4。

(a) 求物块从 AA 移动到 BB 时克服摩擦力所做的功。

(3)

物块在 AA 处的速率为 7 m s−17\,\mathrm{m\,s^{-1}}。

(b) 使用能量方法,求物块在 BB 处的速率。

(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

(a)

解法一

思路

展开

斜向上拉力减小正压力,摩擦功只取摩擦力乘120,不能与牵引功混合。

答题过程

展开

Vertical equilibrium gives

R+25sin⁡36∘=4(10),R=40−25sin⁡36∘.R+25\sin36^\circ=4(10), \qquad R=40-25\sin36^\circ.

Hence the work done against friction is

Wf= μR(120)= 0.4(40−25sin⁡36∘)(120)= 1214.657697… J≈ 1210 J (3 s.f.).\begin{align*} W_f=&\,\mu R(120)\\[4mm] =&\,0.4(40-25\sin36^\circ)(120)\\[4mm] =&\,1214.657697\ldots\text{ J}\\[4mm] \approx&\,\boxed{1210\text{ J}}\text{ (3 s.f.)}. \end{align*}

(b)

解法一

思路

展开

按指定能量法,牵引力仅水平分量做功,减未舍入摩擦功等于动能增加。

答题过程

展开

The work done by the pulling force is 25cos⁡36∘(120)25\cos36^\circ(120) joules. Using the unrounded frictional work from (a),

 12(4)vB2−12(4)(7)2= 25cos⁡36∘(120)−Wf.\begin{align*} &\,\frac12(4)v_B^2-\frac12(4)(7)^2\\[4mm] =&\,25\cos36^\circ(120)-W_f. \end{align*}

Therefore

2vB2−98=2427.050983…−1214.657697…,2v_B^2-98=2427.050983\ldots-1214.657697\ldots, vB=25.596809234…≈25.6 m s−1.v_B=25.596809234\ldots\approx\boxed{25.6\text{ m s}^{-1}}.