A trains consists of 20 cars, each with a mass of 6.8 X 103 kg. The train accele

Question

A trains consists of 20 cars, each with a mass of 6.8 X 103 kg. The train accelerates at a rate of 8.4 X 10-2 m/sec2. (You may ignore any minor effects of friction in this problem.)
1) What is magnitude of the force that engine must pull forward on the 1st car?
|F20| =

N
2) The train has picked up some additional frieght and is now 50 cars long. If it accelerates as before, what is magnitude of the force that the 30th car must pull forward on the 31st car?
|F30-31| =

N
3) What is magnitude of the force that the 31st car must pull backward on the 30th car?
|F31-30| =

N
4) What is magnitude of the force that the 31st car must pull forward on the 32nd car?
|F31-32| =

N
5) What is magnitude of the force that the 32nd car must pull backward on the 31st car?
|F32-31| =

N
6) The train is now disassembled into individual cars in preparation for unloading. One of the individual cars in pulled forwards with a force equal to the answer to part (b) and pulled backwards with a force equal to the answer to part (e). What is the magnitude of the acceleration of this individual car?
|a| =

Explanation / Answer

It depends on which car moves the train. If all cars move the train then there are no tension in any couplings. If only the front (1st) car moves the train, then the last car will receive the least force that is F=m.a This force is applied to coupling no 49 , we say C49=m.a, hence C48 = 2.m.a and C47=3.m.a or in general Cn= (50-n).m.a This formula answer your question as follows a. between 30th and 31st = C30 =(50-30).6400.0.07=8960N b. between 49th and 50th =C49 =(50-49).6400.0.07=448N

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