A 1000 kg engine puls a train of 4 wagons each of 2500 kg along a horizontal railway tra Example 13 the engineers a force of 50000 N on wagons and track offers force of friction 2500 N di when an cacướp O The ne accelerating force O The acceleration of the train The force of wagon 1 on wagon 2
Answers
Explanation:
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(a)
we know that Force = mass x acceleration
given:
Force = 50000 N
Frictional Force = 5000 N
thus, Net Force = Force - frictional Force = 50000 N - 10000N = 40000 N
(b)
Now,
total mass = mass of engine + mass of train = 1000 kg + (4 x 2500 kg) = 11000 kg
thus, acceleration will be
a = net force / total mass = F/m = 40000/11000
or
a = 3.63 m/s2
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Answer:
Given, force of engine = 40000 N
Force of friction = 5000 N
Mass of engine = 8000 kg
Total weight of wagons = 5 x 2000 kg = 10000 kg
(a) The net accelerating force
= Force exerted by engine – Force of fricition
= 40000 N – 5000 N = 35000 N
(b) The acceleration of the train
We know that, F = mass x acceleration
Or, 35000 N = (mass of engine + mass of 5 wagons) X a
Or, 35000 N = (8000 kg + 10000 kg) X a
Or, 35000N = 18000 kg X a
`=> a=(35000N)/(18000\ kg)=1.944ms^(-2)`
(c) The force of wagon 1 on wagon 2
Since, net accelerating force = 35000 N
Mass of all 5 wagons = 10000 kg
We know that, F = m x a
Therefore, 35000N = 10000 kg x a
`=>a = (35000N)/(10000\ kg)=3.5ms^(-2)`
Therefore, acceleration of wagons `=3.5m//s^2`
Thus, force of wagon 1 on 2 = mass of four wagons x acceleration
`=> F= 4xx2000\ kg xx 3.5ms^(-2)`
`=>F= 8000\ kg xx 3.5ms^(-2)`
`=>F=28000N`
Thus, (a) The net accelerating force = 35000N
(b) The acceleration of train = 1.944 ms–2
(c) The force of wagon 1 on 2 = 28000N