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M2 2024 Jan Q5

A Level / Edexcel / M2

IAL 2024 Jan Paper · Question 5

题目

Problem

5.

A van of mass 600 kg600\,\mathrm{kg} is moving up a straight road inclined at an angle α\alpha to the horizontal, where sin⁡α=114\sin\alpha=\frac1{14}. The van is towing a trailer of mass 200 kg200\,\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 4.

The resistance to the motion of the van from non-gravitational forces is modelled as a constant force of magnitude 250 N250\,\mathrm{N}.

The resistance to the motion of the trailer from non-gravitational forces is modelled as a constant force of magnitude 150 N150\,\mathrm{N}.

The towbar is modelled as a light rod.

At the instant when the speed of the van is 16 m s−116\,\mathrm{m\,s^{-1}}, the engine of the van is working at a rate of 10 kW10\,\mathrm{kW}.

(a) Find the deceleration of the van at this instant.

(5)

(b) Find the tension in the towbar at this instant.

(4)
题目中文翻译

质量为 600 kg600\,\mathrm{kg} 的厢式货车沿一条与水平面成 α\alpha 角的直路上坡，其中 sin⁡α=114\sin\alpha=\frac1{14}。货车牵引质量为 200 kg200\,\mathrm{kg} 的拖车。如图 4 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 250 N250\,\mathrm{N} 的恒力。

将拖车所受的非重力运动阻力视为大小为 150 N150\,\mathrm{N} 的恒力。

将牵引杆视为轻杆。

在货车速率为 16 m s−116\,\mathrm{m\,s^{-1}} 的瞬间，货车发动机以 10 kW10\,\mathrm{kW} 的功率工作。

(a) 求该瞬间货车的减速度大小。

(5)

(b) 求该瞬间牵引杆中的张力。

(4)

解答

(a)

解法一

思路

展开

将两车视为整体消去杆内力。取上坡为正，计算得到负加速度；题目问减速度大小，最后给正数。

答题过程

展开

The engine power is 10000 W, so at speed 16 m s−1^{-1} the driving force is

F=1000016=625 N.F=\frac{10000}{16}=625\text{ N}.

For the van and trailer together, take up the road as positive:

625−(250+150)−(600+200)gsin⁡α=(600+200)a.625-(250+150)-(600+200)g\sin\alpha=(600+200)a.

Using sin⁡α=1/14\sin\alpha=1/14,

625−400−800(9.8)(114)=800a.625-400-800(9.8)\left(\frac1{14}\right)=800a.

Hence

a=625−400−560800=−335800 m s−2.a=\frac{625-400-560}{800}=-\frac{335}{800}\text{ m s}^{-2}.

Therefore the deceleration is

0.419 m s−2.\boxed{0.419\text{ m s}^{-2}}.

(b)

解法一

思路

展开

单独研究拖车求张力。仍取上坡为正，须沿用上一问的负加速度，不能将减速度大小直接当正加速度代入。

答题过程

展开

Let the tension in the towbar be TT N. Resolving up the road for the trailer,

T−150−200gsin⁡α=200a.T-150-200g\sin\alpha=200a.

Using the unrounded signed acceleration from (a),

T−150−200(9.8)(114)=200(−335800).T-150-200(9.8)\left(\frac1{14}\right)=200\left(-\frac{335}{800}\right).

Thus

T=150+140−83.75=206.25 N.T=150+140-83.75=206.25\text{ N}.

Therefore

T=206 N.\boxed{T=206\text{ N}}.