An Engine applies 800 N of force to run a train of 5 wagons. How much should be the frictional force
exerted to stop the train?
a. 800 N b. 4000 N c. -850 N d. 0N
Answers
Explanation:
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:
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
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