题目

Problem

A vehicle of mass $450\,\mathrm{kg}$ is moving on a straight road that is inclined at angle $\theta$ to the horizontal, where $\sin\theta=\frac1{15}$

At the instant when the vehicle is moving down the road at $12\,\mathrm{m\,s^{-1}}$

the engine of the vehicle is working at a rate of $P$ watts
the acceleration of the vehicle is $0.5\,\mathrm{m\,s^{-2}}$
the resistance to the motion of the vehicle is modelled as a constant force of magnitude $R$ newtons

At the instant when the vehicle is moving up the road at $12\,\mathrm{m\,s^{-1}}$

the engine of the vehicle is working at a rate of $2P$ watts
the deceleration of the vehicle is $0.5\,\mathrm{m\,s^{-2}}$
the resistance to the motion of the vehicle from non-gravitational forces is modelled as a constant force of magnitude $R$ newtons

Find the value of $P$ .

(8)

题目中文翻译

质量为 $450\,\mathrm{kg}$ 的汽车沿一条与水平面成 $\theta$ 角的直路行驶，其中 $\sin\theta=\frac1{15}$ 。

在汽车以 $12\,\mathrm{m\,s^{-1}}$ 的速率沿公路下坡的瞬间：

在汽车以 $12\,\mathrm{m\,s^{-1}}$ 的速率沿公路上坡的瞬间：

求 $P$ 的值。

(8)

解答

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分别以各自运动方向为正列方程。下坡时加速，上坡时减速，因此第二个方程的加速度为负。两次速率相同但功率加倍，牵引力也加倍；阻力都与当时运动方向相反。

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The component of the weight along the road is

450g\sin\theta=30g=294\text{ N}.

Moving down the road, the driving force is $P/12$ N. Taking down the road as positive,

\frac P{12}+294-R=450(0.5),

\frac P{12}-R=-69.

Moving up the road, the driving force is $2P/12=P/6$ N. Taking up the road as positive, the acceleration is $-0.5$ m s $^{-2}$ , so

\frac P6-294-R=450(-0.5),

and therefore

\frac P6-R=69.

Subtracting the first equation from the second gives

\frac P{12}=138,\qquad P=1656.

Thus, to 3 significant figures,

\boxed{P=1.66\times10^3\text{ W}}.