题目

Problem

A car of mass $900\,\mathrm{kg}$ is moving up a hill inclined at $\sin^{-1}0.12$ to the horizontal. The initial speed of the car is $11\,\mathrm{m\,s^{-1}}$ . After $12\,\mathrm{s}$ , the car has travelled $150\,\mathrm{m}$ up the hill and has speed $16\,\mathrm{m\,s^{-1}}$ . The engine of the car is working at a constant rate of $24\,\mathrm{kW}$ .

(a) Find the work done against the resistive forces during the $12\,\mathrm{s}$ .

(5)

The car then travels along a straight horizontal road. There is a resistance to the motion of the car of $(1520+4v)\,\mathrm{N}$ when the speed of the car is $v\,\mathrm{m\,s^{-1}}$ . The car travels at a constant speed with the engine working at a constant rate of $32\,\mathrm{kW}$ .

(b) Find this speed.

(3)

题目中文翻译

一辆质量为 $900\,\mathrm{kg}$ 的汽车沿与水平面夹角为 $\sin^{-1}0.12$ 的山坡上行。汽车的初始速率为 $11\,\mathrm{m\,s^{-1}}$ 。经过 $12\,\mathrm{s}$ ，汽车沿山坡上行了 $150\,\mathrm{m}$ ，速率为 $16\,\mathrm{m\,s^{-1}}$ 。汽车发动机以 $24\,\mathrm{kW}$ 的恒定功率工作。

(a) 求这 $12\,\mathrm{s}$ 内克服阻力所做的功。

(5)

随后汽车沿水平直路行驶。当汽车的速率为 $v\,\mathrm{m\,s^{-1}}$ 时，所受的运动阻力为 $(1520+4v)\,\mathrm{N}$ 。汽车以恒定速率行驶，发动机以 $32\,\mathrm{kW}$ 的恒定功率工作。

(b) 求该速率。

(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 ( $g$ ) is needed, use $10\,\mathrm{m\,s^{-2}}$ .

需要使用重力加速度 $g$ 的数值时，取 $10\,\mathrm{m\,s^{-2}}$ 。

解答

(a)

解法一

思路

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恒功率乘时间得到发动机总功；扣掉动能和重力势能的增加量，剩余即克服阻力的功。不能假定加速度恒定。

答题过程

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The engine does work

W_E=24000(12)=288000\,\mathrm{J}.

The vertical rise is $150(0.12)=18\,\mathrm{m}$ , so the gain in gravitational potential energy is

\Delta E_P=900(10)(18)=162000\,\mathrm{J}.

The gain in kinetic energy is

\begin{align*} \Delta E_K=&\,\frac12(900)(16^2-11^2)\\[4mm] =&\,60750\,\mathrm{J}. \end{align*}

Therefore the work done against resistance is

\begin{align*} W_R=&\,W_E-\Delta E_P-\Delta E_K \\[4mm] =&\,288000-162000-60750 \\[4mm] =&\,\boxed{65250\,\mathrm{J}}. \end{align*}

(b)

解法一

思路

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水平路上速率恒定，牵引力等于阻力；用功率等于力乘速率列二次方程，并舍去负速率。

答题过程

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At constant speed the driving force equals the resistance, so

\frac{32000}{v}=1520+4v, \qquad v>0.

Multiplying by $v$ and rearranging,

\begin{align*} 4v^2+1520v-32000=&\,0, \\[4mm] v^2+380v-8000=&\,0, \\[4mm] (v-20)(v+400)=&\,0. \end{align*}

Rejecting $v=-400$ , the speed is $\boxed{20\,\mathrm{m\,s^{-1}}}$ .

CIE 9709 2022 June Paper 42 Q6

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程