Question

An 8 metric tonne engine pulls a train v of 5 wagon each of 2 metric tonnes along a horizontal track. If the engine exerts a force of 3.5 x10^4 n and the track is smooth, then the force exerted by wagon 1 on wagon 2 is

Answer 1

8 meteric tonne=8000 kg

Given, force of engine = 40000 N

Force of friction = 5000 N
Mass of engine = 8000 kg
Total weight of wagons = 5 x 2000 kg = 10000 kg

(a) The net accelerating force 
= Force exerted by engine – Force of fricition
= 40000 N – 5000 N = 35000 N

(b) The acceleration of the train
We know that, F = mass x acceleration
Or,
35000 N = (mass of engine + mass of 5 wagons) X a
Or, 
35000N=(8000kg+10000kg)×a35000N
=(8000kg+10000kg)×a
Or, 
35000N=18000kg×a35000N
=18000kg×a
⇒a=35000N18000 kg=1.944ms−2
⇒a=35000N18000 kg=1.944ms-2

(c) The force of wagon 1 on wagon 2 = mass of four wagons x acceleration
⇒F=4×2000 kg×1.944ms−2
⇒F=4×2000 kg×1.944ms-2
⇒F=8000 kg×1.944ms−2
⇒F=8000 kg×1.944ms-2
⇒F=1552N