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 calculate a) the net accelerating force, b) the acceleration of the train and c) the force of wagon 1 on wagon 2.
​

Answer 1

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

→ Net acceleration force = 40000 - 5000

→ Net acceleration force = 35000 N.

Thus,the net acceleration force (F) exerted by the engine is 35000 newtons.

b) Calculation of the acceleration of train

The mass of 1 wagon is 2000 kg,so the mass of 5 wagons of train will be 2000 × 5 = 10000 kg.

→ Net force = Mass of train × Acceleration

→ Force = Mass × Acceleration

→ 35000 = 10000 × a

→ a = 35000 ÷ 10000 m/s²

→ a = 3.5 m/s²

So,the acceletion of the train is 3.5 m/s²...

c) Calculation of force wagon 1 on wagon 2

→ Force of wagon 1 on wagon 2 = Mass of 4 wagon × Acceleration of train

→ Force of wagon 1 on wagon 2 = 2000 × 4 × 3.5

→ Force of wagon 1 on wagon 2 = 28000 N

Thus,the force of wagon 1 on wagon 2 is of 28000 newtons.

______________________________

Answer 2

$\huge{\underline{\underline{\boxed{\sf{\purple{Answer ࿐}}}}}}$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

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

$\small \underline {\sf{ Fa = 35000 N}}$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

(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)

$\small \underline {\sf{m = 18000 kg }}$

Acceleration of the train (a) = 35000/18000

$\small \underline {\sf {a = 1.944m/s^2 }}$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

(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

$\small \underline {\sf{ F’ = 28000 N}}$

$\small \underline \red {\sf{Therefore , \ the \ force \ of \ wagon-1 \ on \ the \ wagon-2 \ is \ 28000N. }}$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━