Question

(ii) The pull of the engine is 5000 N. When the wagon's speed is increasing, how large is the
frictional force? Type in the correct answer
A. Zero
B. Between 0 and 5000N
C. 5000N
D. More than 5000N​

Answer 1

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