What is the force with which the 10 th car pulls the 11 th one?

Question

An engine pulls a train of 20 freight car, each having a mass of 5.0 *10^4 kg with a constant force. The cars move from rest to a speed of 4.0 m/s in 20.0s on a straight track. Neglection friction, what is the force with which the 10 th car pulls the 11 th one?

Answer 1

Luiz Felipe A gives the right answer, but his solution may be hard to follow.

First, the force with which the 10th car pulls the 11th one is actually the force with which the 10th car pulls the 11th through 20th cars. The amount of force that is not used to accelerate the 11th car is transmitted to the 12th car, and so on.

So we really want to know what force is being used to accelerate 10 cars (11th through 20th) to a speed of 4 m/s in 20 s. Assuming that the force and the acceleration are constant, the acceleration is 4 m/s / 20 s = 0.2 m/s^2.

To accelerate a kilogram 0.2 m/s^2 (i.e., to increase its velocity by 0.2 m/s each second), we would have to apply a force of 0.2 N. In this case, we have 10 freight cars, each having a mass of 5*10^4 kg, or a total mass of 5*10^5 kg. Multiplying this total mass by the required force of 0.2 N for each kg produces a required force of 10^5 N (as Luiz Felipe A calculated).

Hope that helps you.

Answer 2

4m/s----20s
0,2m/s--1s

a=0,2m/s²

F=ma
m=10cars=5*10^5kg

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