How Is Tension Calculated Between Train Cars During Acceleration?

[spoonthrower]

A train consists of 50 cars, each of which has a mass of 6.6*10^3 kg. The train has an acceleration of 6.0*10^-2 m/s2. Ignore friction and determine the tension in the coupling at the following places.

A)between the 30th and 31st cars

B)between the 49th and 50th cars

I know f=ma that's about it.

 

[PhanthomJay]

A train consists of 50 cars, each of which has a mass of 6.6*10^3 kg. The train has an acceleration of 6.0*10^-2 m/s2. Ignore friction and determine the tension in the coupling at the following places.

A)between the 30th and 31st cars

B)between the 49th and 50th cars

I know f=ma that's about it.

The cars are all coupled together, so whether you look at one car, 10 cars, 30 cars, or 50 cars, each and all are accelerating at the same rate.
So between the 30th and 31st, isolate that end part of the train using a free body diagram and sure, use F_{net} = ma
where F_{net} is the only force acting in the horizontal direction, the tension in the coupling betweeen the 30th and 31st cars; and for m, that's just the mass of each car times 20 cars (31 thru 50). Do the same for the caboose coupling tension calculation, m will be a lot less, and therefore so will the coupling tension be less than part A result.