## College Physics (4th Edition)

The last car to the west pulls on the rest of the train with a force of $2.0\times 10^5~N$ (directed to the west).
Let $M$ be the mass of each train car. We can find the acceleration of the train: $a = \frac{F}{10~M} = \frac{2.0\times 10^6~N}{10~M} = \frac{2.0\times 10^5~N}{M}$ We can find the force $F_p$ with which the ninth car pulls on the last car: $F_p = Ma = M~(\frac{2.0\times 10^5~N}{M}) = 2.0\times 10^5~N$ By Newton's third law, the last car pulls on the ninth car with a force of the same magnitude. The last car to the west pulls on the rest of the train with a force of $2.0\times 10^5~N$ (directed to the west).