Answer :

(a) Given,

Force of engine = 40,000 N

Force of friction = 5,000 NSo, let us consider the complete train as single body on which a force of 40000 N is applied by engine of mass 8000 kg on which a opposing force(friction force) is applied in direction opposite to direction of motion of body.

Net accelerating force, F = Force of engine – Force of friction= 40000 – 5000

=**35000 N**

(b)

given:

Net force exerted by the engine = 35000 N

The mass of 5 wagons of train = 2000 × 5= 10000 kg

The mass of engine = 8000 kg

The mass of train = 10000 +8000 =18000 kg

We know that,

**Net force = mass of train × acceleration**

**F = m × a**

*a*

a = m/s^{2}** = 1.9444 m/s ^{2}**

Therefore, the acceleration of the train is 1.9444 m/s^{2}

(c)

There are total 5 wagons behind the engine (which have been marked 1, 2, 3, 4 and 5 in the figure) and there are only 4 wagons behind wagon 1.

Force of wagon 1 = Mass of 4 wagons × Acceleration of train

(Behind Wagon 1)

= 2000 × 4 × 1.9444 (We know that weight of one Wagon is 2000 kg)

= 15555.2** N**

Thus, the force of Wagon 1 on Wagon 2 is 15555.2 N

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