Question

A 8000kg engine pulls a train of 5 wagons, each of 2000kg, along a horizontal track. If the engine exerts a force of 40000N and the track offers a friction force of 5000N, then calculate:
(a) The net accelerating force:
(b) The acceleration of the train; and
(c) The force of wagon 1 on wagon. In your school if this question is fone tell me the answer your school has done.​

Answer 1

Answer:

Here is your answer friends

Explanation:

Given parameters

Mass of the engine (M) = 8000 kg

Number of wagons = 5

Mass of the wagons (m) = 2000 kg

Force exerted by the engine (F ) = 40000 N

Frictional force offered by the track (Ff) = 5000 N

(a) 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.

Answer 2

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 F1=8000×3.5 =28000 N

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