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.
Answers
Total Force applied to wagon is 40000-5000 i.e 35000 N
we know Force = ma ( Newton second law of motion)
i.e. 35000 = (8000+(5×2000) × a
i.e 35000/18000 = a
a= 1.94m/s²
Force of wagon is same as the force of engine as they are attached horizontally (if we take tension doesnot act in it)
If tension is acting then comment below i will update the ans
Answer:
Explanation:
Given : Mass of engine M = 8000M=8000 kg
Mass of each wagon m =2000 m=2000 kg
Frictional force acting in backward direction f = 5000 f=5000 N
(a) : The net force acting on the train F' = F - f = 40000- 5000 = 35000F′=F−f =40000−5000=35000 N
(b) : Let the acceleration of the train be aa
\therefore∴ F' = (5m+M) aF′=(5m+M) a
35000 = (5\times 2000+ 8000) a35000= (5×2000+8000) a \implies a =1.944⟹ a=1.944 ms^{-2}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'a′
35000 = (5m) a'35000 =(5m)a′ \implies 35000 = 10000 \times a'⟹ 35000=10000×a′
Acceleration of the wagons a' = 3.5a′=3.5 ms^{-2}ms−2
Mass of last 4 wagons m' = 4 \times 2000m′=4×2000 kg
\therefore∴ Net force on last 4 wagons F_1 = 8000 \times 3.5F1=8000×3.5 = 28000=28000 N
Thus force on wagon 2 by wagon 1 is 28000 N.
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