题目

Problem

A car of mass $600\,\mathrm{kg}$ tows a trailer of mass $200\,\mathrm{kg}$ up a hill along a straight road that is inclined at angle $\theta$ to the horizontal, where $\sin\theta=\frac1{20}$ . The trailer is attached to the car by a light inextensible towbar. The resistance to the motion of the car from non-gravitational forces is modelled as a constant force of magnitude $150\,\mathrm{N}$ . The resistance to the motion of the trailer from non-gravitational forces is modelled as a constant force of magnitude $300\,\mathrm{N}$ .

When the engine of the car is working at a constant rate of $P\,\mathrm{kW}$ the car and the trailer have a constant speed of $15\,\mathrm{m\,s^{-1}}$

(a) Find the value of $P$ .

(5)

Later, at the instant when the car and the trailer are travelling up the hill with a speed of $20\,\mathrm{m\,s^{-1}}$ , the towbar breaks. When the towbar breaks the trailer is at the point $X$ . The trailer continues to travel up the hill before coming to instantaneous rest at the point $Y$ . The resistance to the motion of the trailer from non-gravitational forces is again modelled as a constant force of magnitude $300\,\mathrm{N}$ .

(b) Use the work-energy principle to find the distance $XY$ .

(4)

题目中文翻译

质量为 $600\,\mathrm{kg}$ 的汽车牵引质量为 $200\,\mathrm{kg}$ 的拖车，沿一条与水平面成 $\theta$ 角的直路上坡，其中 $\sin\theta=\frac1{20}$ 。拖车通过不可伸长的轻牵引杆与汽车连接。将汽车所受的非重力运动阻力视为大小为 $150\,\mathrm{N}$ 的恒力，将拖车所受的非重力运动阻力视为大小为 $300\,\mathrm{N}$ 的恒力。

当汽车发动机以恒定功率 $P\,\mathrm{kW}$ 工作时，汽车与拖车以恒定速率 $15\,\mathrm{m\,s^{-1}}$ 行驶。

(a) 求 $P$ 的值。

(5)

后来，在汽车与拖车以速率 $20\,\mathrm{m\,s^{-1}}$ 上坡的瞬间，牵引杆断裂。断裂时拖车位于 $X$ 点。拖车继续上坡，直至在 $Y$ 点瞬时静止。仍将拖车所受的非重力运动阻力视为大小为 $300\,\mathrm{N}$ 的恒力。

(b) 利用功—能原理求距离 $XY$ 。

(4)

解答

(a)

解法一

思路

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把汽车与拖车视为整体可消去牵引杆内力。匀速时牵引力平衡两项阻力及两车的下坡重力分量，最后把瓦换成千瓦。

答题过程

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Let the driving force be $F$ N. Resolving up the road for the car and trailer together, with zero acceleration,

F-150-300-(600+200)g\left(\frac1{20}\right)=0.

Hence

F=450+40g=842\text{ N}.

The power is $Fv$ , so

1000P=842(15)=12630.

Therefore

\boxed{P=12.6}\quad\text{(power }12.6\text{ kW, to 3 s.f.).}

(b)

解法一

思路

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断杆后只研究拖车。它最初以 $20$ 运动，动能用于克服 $300$ 牛顿阻力及增加重力势能；本问必须使用功—能原理。

答题过程

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Let $XY=d$ metres. After the towbar breaks there is no towing force on the trailer. Its loss of kinetic energy equals its gain in gravitational potential energy plus the work done against resistance:

\frac12(200)(20)^2=200g\left(\frac d{20}\right)+300d.

Thus

40000=(10g+300)d=398d.

Therefore

\boxed{XY=101\text{ m}}\quad\text{(3 s.f.).}

M2 2022 Jan Q2

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程