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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 and
(b) the acceleration of the train.
Answers
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 force (Fa) = F – Ff
Fa = 40000 N – 5000 N
Fa = 35000 N
(b) Let us consider the acceleration of the train to 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
The 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) F21 = m4w × a
F21 = Force applied on Wagon 2 by Wagon 1
m4w = mass of 4 wagons
a = acceleration of the train
The force of wagon 1 on wagon 2 = Force of wagon 2 on wagon 1
⇒ (4 × 2000 × 1.944)
= 15552 N
∴ The force of wagon-1 on the wagon-2 is 15552 N.
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:
Given parameters
a)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 force (Fa) = F – Ff
Fa = 40000 N – 5000 N
Fa = 35000 N
(b) Let us consider the acceleration of the train to 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
The 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) F21 = m4w × a
F21 = Force applied on Wagon 2 by Wagon 1
m4w = mass of 4 wagons
a = acceleration of the train
The force of wagon 1 on wagon 2 = Force of wagon 2 on wagon 1
⇒ (4 × 2000 × 1.944)
= 15552 N
∴ The force of wagon-1 on the wagon-2 is 15552 N.
Explanation:
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