Question

Question 7: A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate:(a) the net accelerating force;
(b) the acceleration of the train; and
(c) the force of wagon 1 on wagon 2.

Answer 1

≡



9th science

Force and Laws of Motion

NCERT Exercise Solution - part 2

Question 7: A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate:

(a) the net accelerating force;
(b) the acceleration of the train; and
(c) the force of wagon 1 on wagon 2.

Answer: 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, 35000 N = (8000 kg + 10000 kg) X a
Or, 35000N = 18000 kg X 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⇒F=1552N

Thus, (a) The net accelerating force = 35000N

(b) The acceleration of train = 1.944 ms–2

(c) The force of wagon 1 on 2 = 1552 N

Answer 2

 


Force exerted by the engine= 40000 N 

force of friction ecerted by the track = 5000N

as the force of friction always acts on opposite side of applied of force 

a) net accelerating force= 

=40000- 5000= 35000N 

b) mass of engine + 5 wagon 

 = 8000 + (2000*5)= 18000 kg 

    F=ma

 =35000=18000*a 

a=35000/18000= 1.94 m/s square 

c) force of 1 wagon on wagon 2 

f=m*a 

f= 8000 x 1.94 

f= 15520 N