question archive 4) A freight train consists of two 8
4) A freight train consists of two 8
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4) A freight train consists of two 8.3 • 104 kg engines and 45 cars with average masses of 5.8 • 104 kg each.
Randomized Variables
m = 8.3 • 104 kg
m2 = 5.8 • 104 kg
f = 7.75 • 105 N
(Part A) What is the magnitude of the force must each engine exert backward on the track to accelerate the train at a rate of 5.00 • 10-2 m/s2 if the force of friction is 7.75 • 105 N, assuming the engines exert identical forces? This is not a large frictional force for such a massive system. Rolling friction for trains is small, and consequently trains are very energy-efficient transportation systems.
F =
Part (b) What is the force in the coupling between the 37th and 38th cars (this is the force each exerts on the other), assuming all cars have the same mass and that friction is evenly distributed among all of the cars and engines?
F' =
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(a) F=4.569*105 N
(b) F=1.612*105 N
Step-by-step explanation
mass of an engine, M=8.3*104 kg
mass of each car, m=5.8*104 kg
thus totlal mass, M'=2*M+45*m=2*8.3*104+45*5.8*104 Kg
(a) given acceleration, a=5*10-2 m/s2
frictional force, f=7.75*105 N
let the total force from engines be, F
thus we have, F-f=M'a
ie, F=M'*a+f=[(2*8.3*104+45*5.8*104)*5*10-2]+7.75*105 =913800 N= 9.138*105 N
magnitude of force from each engine, F'=F/2=913800/2=456900 N=4.569*105 N
(b) number of cars from 38 to 45, is, 8
thus mass of 8 cars,m8=8*5.8*104
total frictional force for these 8 cars, f8=7.75*105*8/45=1.38*105 N
thus the roces on 8 cars are, coupling force and frictional force
thus, m8*a=T-f8
thus, T=m8*a+f8=(8*5.8*104*5*10-2)+1.38*105=1.612*105 N
