# Thread: train mass and acceleration problem

1. ## train mass and acceleration problem

A train consists of 50 cars, each of which has a mass of 7.0 x 10^3 kg. The train has an acceleration of 6.0 x 10^-2 m/s^2. Ignore friction and determine the tension in the coupling at the following places.

tension in between the 30th and 31st cars
in Newtons

tension in between the 49th and 50th cars
in Newtons

2. Originally Posted by rcmango
A train consists of 50 cars, each of which has a mass of 7.0 x 10^3 kg. The train has an acceleration of 6.0 x 10^-2 m/s^2. Ignore friction and determine the tension in the coupling at the following places.

a) tension in between the 30th and 31st cars
in Newtons

b) tension in between the 49th and 50th cars
in Newtons
Hello,

the tension in a coupling describes the force which is necessary to pull all following waggons.

to a) The coupling between the 30th and the 31rst waggon has to pull 20 waggons. $m_{20}=20 \cdot 7.0 \cdot 10^3\ kg$

You are supposed to know how to calculate the force if you know the mass and the acceleration: $F = m \cdot a$. Thus:

$F= 20 \cdot 7.0 \cdot 10^3\ kg \cdot 0.06\ \frac m{s^2} = 8400 \ N = 8.4\ kN$

to b) Now the coupling has to pull only one waggon: $m_1 = 7.0 \cdot 10^3\ kg$ that means:

$F=20 \cdot 7.0 \cdot 10^3\ kg \cdot 0.06\ \frac m{s^2}=420
N$

A few personal remarks:
i) A train waggon with a mass of only 7 t must be empty.
ii) The electrical lokomotive E103 which was build to pull (relatively light) passenger waggons has an initial pulling force of 312 kN. So either I made an ugly mistake or the problem isn't very realistic.

3. thanks, its hard for me to break it down, especially with the scientific notation. thanks alot.

..upon reviewing this, your right, the problem wasn't very realistic. but the second part shouldn't have the (20 trailers to pull, but the answer is right because you didn't calculate it in there. i see it. Thanks alot!)