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M1 2019 June Q4

A Level / Edexcel / M1

IAL 2019 June Paper · Question 4

题目

Problem

4.

A railway truck of mass 500 kg500\,\mathrm{kg} is pushed up a straight track by a railway engine of mass 2000 kg2000\,\mathrm{kg}. The track is inclined to the horizontal at an angle θ\theta, where sin⁡θ=114\sin\theta=\frac{1}{14}, as shown in Figure 3. The engine produces a constant driving force of magnitude 3050 N3050\,\mathrm{N}. The truck experiences a constant resistance to motion of magnitude 100 N100\,\mathrm{N} and the engine experiences a constant resistance to motion of magnitude 200 N200\,\mathrm{N}. The engine and the truck are connected by a coupling which is modelled as a light rod that is parallel to the track.

Find

(i) the acceleration of the system,

(ii) the magnitude of the force exerted on the truck by the coupling.

(8)
题目中文翻译

一台质量为 2000 kg2000\,\mathrm{kg} 的火车机车将一节质量为 500 kg500\,\mathrm{kg} 的货车厢沿直轨道向上推。如图 3 所示,轨道与水平面成 θ\theta 角,其中 sin⁡θ=114\sin\theta=\frac{1}{14}。机车产生大小为 3050 N3050\,\mathrm{N} 的恒定驱动力。货车厢受到大小为 100 N100\,\mathrm{N} 的恒定运动阻力,机车受到大小为 200 N200\,\mathrm{N} 的恒定运动阻力。机车与货车厢由连接装置相连,将该装置视为平行于轨道的轻杆。

求

(i) 系统的加速度,

(ii) 连接装置对货车厢施加的力的大小。

(8)

解答

(i)

解法一

思路

展开

把机车与车厢作为整体,连接杆的力作为内力消去。沿斜面向上为正,阻力与重力分量均向下。

答题过程

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Let the acceleration up the track be aa. For the complete system,

3050−100−200−2500gsin⁡θ=2500a.3050-100-200-2500g\sin\theta=2500a.

Using sin⁡θ=1/14\sin\theta=1/14 and g=9.8g=9.8,

3050−300−1750=2500a,1000=2500a.3050-300-1750=2500a, \qquad 1000=2500a.

Hence the acceleration is 0.400 m s−2\boxed{0.400\,\mathrm{m\,s^{-2}}} up the track.

(ii)

解法一

思路

展开

单独研究车厢。连接杆向上推车厢,需同时克服车厢自身的阻力和重力分量,并使其加速。

答题过程

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Let the coupling force on the truck be CC up the track. For the truck alone,

C−100−500gsin⁡θ=500a.C-100-500g\sin\theta=500a.

Substituting the result from (i),

C−100−350=500(0.4),C=650 N.C-100-350=500(0.4), \qquad \boxed{C=650\,\mathrm{N}}.

解法二

思路

展开

也可以研究机车。车厢通过连接杆对机车施加大小相同、方向向下的反作用力。

答题过程

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For the engine alone, taking up the track as positive,

3050−200−2000gsin⁡θ−C=2000a.3050-200-2000g\sin\theta-C=2000a.

Using a=0.4a=0.4,

3050−200−1400−C=800,C=650 N.3050-200-1400-C=800, \qquad \boxed{C=650\,\mathrm{N}}.