NCERT solution Class 9 force and laws of motion

NCERT Solutions

Class 9 Science

Force and Laws of Motion

NCERT Questions

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Q.7

An 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

(a) Force exerted by the engine, F = 40000N

Frictional force offered by the track, F_f = 5000N

Net accelerating force, F_a = F – F_f = 40000 – 5000 =35000N.

(b) Acceleration of the train = a

The engine exerts a force of 40000N on all the five wagons.

Net accelerating force on the wagons, F_a = 35000N

Mass of the wagons, m = mass of a wagon x number of wagons

Mass of wagon = 2000kg

Number of wagons = 5

Therefore m =2000 x 5 =10000kg.

Mass of the engine m’ = 8000kg

Total mass m + m’ = 18000kg

From Newton’s second law of motion:

F_a = ma

a = ( ) = ( ) = 1.944m/s²

(b) Mass of all the wagons except wagon 1 is 4 × 2000 = 8000 kg

Acceleration of the wagons = 3.5 m/s

Thus, force exerted on all the wagons except wagon 1

= 8000 × 3.5 = 28000 N

Therefore, the force exerted by wagon 1 on the remaining four wagons is 28000 N.

Hence, the force exerted by wagon 1 on wagon 2 is 28000 N.

Video solution:

Force and Laws of Motion Class 9 Science NCERT Chapter 9 NCERT Solutions

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