题目

Problem

A truck of mass $900\,\mathrm{kg}$ is moving along a straight horizontal road with the engine of the truck working at a constant rate of $P$ watts. The resistance to the motion of the truck is modelled as a constant force of magnitude $R$ newtons. At the instant when the speed of the truck is $15\,\mathrm{m\,s^{-1}}$ , the deceleration of the truck is $0.2\,\mathrm{m\,s^{-2}}$

Later the same truck is moving down a straight road inclined at an angle $\theta$ to the horizontal, where $\sin\theta=\frac1{30}$ . The resistance to the motion of the truck is again modelled as a constant force of magnitude $R$ newtons. The engine of the truck is again working at a constant rate of $P$ watts. At the instant when the speed of the truck is $12\,\mathrm{m\,s^{-1}}$ , the acceleration of the truck is $0.4\,\mathrm{m\,s^{-2}}$

Find the value of $R$ .

(8)

题目中文翻译

质量为 $900\,\mathrm{kg}$ 的卡车沿水平直路行驶，发动机以恒定功率 $P$ 瓦工作。将卡车所受的运动阻力视为大小为 $R$ 牛顿的恒力。在卡车速率为 $15\,\mathrm{m\,s^{-1}}$ 的瞬间，其减速度大小为 $0.2\,\mathrm{m\,s^{-2}}$ 。

后来，同一卡车沿与水平面成 $\theta$ 角的直路下坡，其中 $\sin\theta=\frac1{30}$ 。仍将卡车所受的运动阻力视为大小为 $R$ 牛顿的恒力，发动机仍以恒定功率 $P$ 瓦工作。在卡车速率为 $12\,\mathrm{m\,s^{-1}}$ 的瞬间，其加速度大小为 $0.4\,\mathrm{m\,s^{-2}}$ 。

求 $R$ 的值。

(8)

解答

解法一

思路

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恒功率不代表恒牵引力：两次速率不同，牵引力分别为 $P/15$ 和 $P/12$ 。水平路段减速，因此加速度为负；下坡路段重力帮助运动。

答题过程

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On the horizontal road, take the direction of motion as positive. The acceleration is $-0.2$ m s $^{-2}$ , so

\frac P{15}-R=900(-0.2)=-180.

Therefore

P=15(R-180).

On the slope, resolve down the road. Since $\sin\theta=1/30$ ,

\frac P{12}+900g\left(\frac1{30}\right)-R=900(0.4).

With $g=9.8$ ,

\frac P{12}+294-R=360,

P=12(R+66).

Equating the two expressions for $P$ ,

15R-2700=12R+792.

Thus $R=1164$ N before rounding. To 3 significant figures,

\boxed{R=1.16\times10^3\text{ N}}.

M2 2022 June Q4

题目

解答

解法一

思路

答题过程