Thordin

answers: 1

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Answer:

Answer:

Mass of engine M=8000 kg

Mass of each wagon m=2000 kg

Frictional force acting in backward direction f=5000 N

(a) : The net force acting on the train F

′

=F−f=40000−5000=35000 N

(b) : Let the acceleration of the train be a

∴ F

′

=(5m+M)a

35000=(5×2000+8000)a ⟹a=1.944 ms

−2

(c) : External force is applied directly to wagon 1 only. The net force on the last four wagons is equal to the force applied by wagon 2 by wagon 1.

Let the acceleration of the wagons be a

′

35000=(5m)a

′

⟹35000=10000×a

′

Acceleration of the wagons a

′

=3.5 ms

−2

Mass of last 4 wagons m

′

=4×2000 kg

∴ Net force on last 4 wagons F

1

=8000×3.5 =28000 N

Thus force on wagon 2 by wagon 1 is 28000 N.

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