题目

Problem

A car has mass $1600\,\mathrm{kg}$ .

(a) The car is moving along a straight horizontal road at a constant speed of $24\,\mathrm{m\,s^{-1}}$ and is subject to a constant resistance of magnitude $480\,\mathrm{N}$ .

Find, in $\mathrm{kW}$ , the rate at which the engine of the car is working.

(2)

The car now moves down a hill inclined at an angle of $\theta$ to the horizontal, where $\sin\theta=0.09$ . The engine of the car is working at a constant rate of $12\,\mathrm{kW}$ . The speed of the car is $24\,\mathrm{m\,s^{-1}}$ at the top of the hill. Ten seconds later the car has travelled $280\,\mathrm{m}$ down the hill and has speed $32\,\mathrm{m\,s^{-1}}$ .

(b) Given that the resistance is not constant, use an energy method to find the total work done against the resistance during the ten seconds.

(5)

题目中文翻译

一辆汽车的质量为 $1600\,\mathrm{kg}$ 。

(a) 汽车沿笔直的水平道路以 $24\,\mathrm{m\,s^{-1}}$ 的恒定速率行驶，受到大小恒为 $480\,\mathrm{N}$ 的阻力。

求汽车发动机做功的功率，以 $\mathrm{kW}$ 为单位。

(2)

汽车现在沿一个与水平面成 $\theta$ 角的山坡向下行驶，其中 $\sin\theta=0.09$ 。汽车发动机以 $12\,\mathrm{kW}$ 的恒定功率做功。汽车在坡顶的速率为 $24\,\mathrm{m\,s^{-1}}$ 。十秒后，汽车沿山坡向下行驶了 $280\,\mathrm{m}$ ，速率为 $32\,\mathrm{m\,s^{-1}}$ 。

(b) 已知阻力不恒定，使用能量法，求这十秒内克服阻力所做的总功。

(5)

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

展开

水平匀速，驱动力等于480牛，功率需转换为千瓦。

答题过程

展开

At constant speed, the driving force equals the resistance. Hence

P=480(24)=11520\,\mathrm{W} =\boxed{11.52\,\mathrm{kW}}.

(b)

解法一

思路

展开

阻力不恒定，按要求用全程能量。发动机做功是功率乘时间；下降损失的势能加上发动机功，一部分转为动能，其余克服阻力。

答题过程

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The engine does work

W_E=12000(10)=120000\,\mathrm{J}.

The loss of potential energy is

1600(10)(280)(0.09)=403200\,\mathrm{J}.

If $W$ is the work against resistance, the work-energy equation is

\begin{align*} 120000+403200-W=&\,800(32^2-24^2). \end{align*}

Therefore

W=523200-358400=\boxed{164800\,\mathrm{J}}.

CIE 9709 2023 November Paper 43 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