题目

Problem

A van of mass $900\,\mathrm{kg}$ is moving up a straight road inclined at an angle $\alpha$ to the horizontal, where $\sin\alpha=\frac1{25}$ . The van is towing a trailer of mass $200\,\mathrm{kg}$ . The trailer is attached to the van by a rigid towbar which is parallel to the direction of motion of the van and the trailer, as shown in Figure 1.

The resistance to the motion of the van from non-gravitational forces is modelled as a constant force of magnitude $400\,\mathrm{N}$ . The resistance to the motion of the trailer from non-gravitational forces is modelled as a constant force of magnitude $240\,\mathrm{N}$ .

The towbar is modelled as a light rod.

The engine of the van is working at a constant rate of $15\,\mathrm{kW}$ .

(a) Find the acceleration of the van at the instant when the speed of the van is $12\,\mathrm{m\,s^{-1}}$

(5)

At the instant when the speed of the van is $14\,\mathrm{m\,s^{-1}}$ , the trailer is passing the point $A$ on the slope and the towbar breaks.

The trailer continues to move up the slope until it comes to rest at the point $B$ .

The resistance to the motion of the trailer from non-gravitational forces is still modelled as a constant force of magnitude $240\,\mathrm{N}$ .

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

(4)

题目中文翻译

质量为 $900\,\mathrm{kg}$ 的厢式货车沿一条与水平面成 $\alpha$ 角的直路上坡，其中 $\sin\alpha=\frac1{25}$ 。货车牵引质量为 $200\,\mathrm{kg}$ 的拖车。如图 1 所示，拖车通过刚性牵引杆与货车连接，牵引杆平行于货车与拖车的运动方向。

将货车所受的非重力运动阻力视为大小为 $400\,\mathrm{N}$ 的恒力。将拖车所受的非重力运动阻力视为大小为 $240\,\mathrm{N}$ 的恒力。

将牵引杆视为轻杆。

货车发动机以恒定功率 $15\,\mathrm{kW}$ 工作。

(a) 求货车速率为 $12\,\mathrm{m\,s^{-1}}$ 的瞬间的加速度。

(5)

在货车速率为 $14\,\mathrm{m\,s^{-1}}$ 的瞬间，拖车正经过斜坡上的 $A$ 点，牵引杆断裂。

拖车继续上坡，直至在 $B$ 点静止。

仍将拖车所受的非重力运动阻力视为大小为 $240\,\mathrm{N}$ 的恒力。

(b) 利用功—能原理求距离 $AB$ 。

(4)

解答

(a)

解法一

思路

展开

恒功率先换成瓦，再求当前牵引力。把两车视为整体，扣除两项非重力阻力与整体重力沿坡分量。

答题过程

展开

At speed 12 m s $^{-1}$ , the driving force is

F=\frac{15000}{12}=1250\text{ N}.

Taking up the road as positive, Newton’s second law for the combined system gives

1250-(400+240)-(900+200)g\sin\alpha=(900+200)a.

Using $\sin\alpha=1/25$ ,

1250-640-1100(9.8)\left(\frac1{25}\right)=1100a.

Thus

a=\frac{178.8}{1100},

and therefore

\boxed{a=0.163\text{ m s}^{-2}\text{ up the road}}.

(b)

解法一

思路

展开

断杆后只研究拖车，初速为 $14$ 。按功—能列初动能等于势能增加加阻力功；不再计发动机或货车质量。

答题过程

展开

Let $AB=d$ . After the towbar breaks, the trailer has no driving force and eventually stops. By the work-energy principle,

\frac12(200)(14^2)=240d+200g(d\sin\alpha).

Therefore

19600=240d+200(9.8)\left(\frac1{25}\right)d,

19600=318.4d.

Hence

\boxed{AB=61.6\text{ m}}.

M2 2024 Oct Q3

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程