An 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track.

Solution:
(a) Force exerted by the engine $=40,000 \mathrm{~N}$

Force of friction exerted by the tracks $=5000 \mathrm{~N}$

As the force of friction always acts opposite to the direction of applied force,

$\therefore$ Net accelerating force of engine

$=40000-5000=35000 \mathrm{~N}$

(b) Mass of 5 wagons $=2000 \times 5=10000 \mathrm{~kg}$

We know, $\mathrm{F}=\mathrm{ma}$

or $35000=10000 \times$ a

$\Rightarrow \quad a=\frac{35000}{10000}=3.5 \mathrm{~m} / \mathrm{s}^{2}$

(c) Force of wagon 1 on wagon $2=$

mass of 4 wagons behind wagon $1 \times$

acceleration

$\mathrm{F}=4 \times 2000 \times 3.5=28000 \mathrm{~N}$