Question

A 8000 kg engine pulls a 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the tracks offers friction force of 5000 N, Then calculate ;
(a) The Net Accelerating force and
(b) The Acceleration of the Train.

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Answer 1

Answer:

Given : 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.

Answer 2

Answer:

Net accelerating force (Fa) = F – Ff

Fa = 40000 N – 5000 N

Fa = 35000 N

(b) Let us consider the acceleration of the train be a m/s2

Acceleration is the ratio of net acceleration force by mass, i.e

a = Fa/m

Where m is the mass of the train, it can be calculated as follows

Total mass of the train (m) = Mass of the engine + (Mass of the wagons × Number of wagons)

m = 8000 + (5 × 2000)

m = 18000 kg

Acceleration of the train (a) = 35000/18000

a = 1.944m/s2

(c) The external force is only applied to waggon 1 directly. On the last four waggons, the net force is equal to the force added to waggon 2 by waggon 1.

Let us consider the acceleration of the wagons is aw

35000 = 5m × aw

aw = 35000/5m

aw = 35000/(5 × 2000)

aw = 3.5 m/s2

Then the mass of last four wagons can be considered as mw

mw = 2000 × 4

mw = 8000 kg

Now let us calculate the net force on the last four wagons

F’ = mw× aw

F’ = 8000 × 3.5

F’ = 28000 N

∴ The force of wagon-1 on the wagon-2 is 28000N.

Explanation:

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