Question

.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 4000 N, then the correct order of magnitude for:(a) the net accelerating force;(b) the acceleration of the train; (c) distance covered in 100 seconds after starting is:- *
1 point

Answer 1

Correct Question :-

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

Answer :

a) The net acceleration force

Here the force exerted by the engine is 40000 N and the opposing force of friction offered by the track is 5000 N. So,

Net acceleration force,F

⟶ Force of engine - force of friction

⟶ 40000 - 5000

⟶ 35000 N

So,the net acceleration force exerted by the engine us 35000 N..

b) The acceleration of the train

We have to calculate that the net force exerted by the engine on the train is 35000 N. Now, the mass of 1 wagon of train is 2000 kg,so the mass of 5 wagons of train will be 2000 × 5 = 10000 kg. In other words, the mass of whole train is 10000 kg.

Now,

Net force = mass of train × acceleration

F = m × a

So,

$\longrightarrow \sf{35000 = 10000 \times a}$

$\longrightarrow \sf{a = \frac{35000}{10000} = 3.5 { \: ms}^{ - 2} \orange \bigstar }$

So,the acceleration of the train is 3.5 m/s^2...

c) Calculate of the force of wagon 1 or wagon 2

Force of wagon 1 on wagon 2 = mass of 4 wagons × acceleration of train

⟶ 2000 × 4 × 3.5

⟶ 28000 N

So,the force of wagon 1 on wagon 2 is of 28000 N...

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${\boxed{ \sf \red{Few \: Details}}}$

• a stands for Acceleration.
• m stands for Mass.
• F stands for Force.
• N stands for Newton.