27. A freight train consists of two 8.00x 101-kg engines and 45 cars with averag

Question

27. A freight train consists of two 8.00x 101-kg engines and 45 cars with average masses of 5.50x 10 kg . (a) What force must each engine exert backward on the track to accelerate the train at a rate of 5.00x 102 m/s2 if the force of friction is 7.50x 10 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. (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?

Explanation / Answer

we can use newton's second law here

The force exerted by train on the tracks is given as

F = ma + Friction force

Here m is the total mass

m = (2*8e4) + (45*5.5e4)

m = 2635000 Kg

Therefore,

F = (2635000 Kg)(5e-2 m/s2) + (7.5e5 N)

F = 881750 N

However, we know that there are two engines applying same force in same directions,

so,

F = 881750 / 2

F = 440875 N

_____________________________________________________________________

After 37th car, we have 8 more cars, so the total mass will be 8 multiplied by mass of each car.

Therefore, the coupling force is

Force of friction is equally distributed,

Ffriction = 7.5e5 / 47 = 15957.44681 N

Therefore, the coupling force is

Fc = 8*((5.5e4*0.05) + 15957.44681)

Fc = 149659.57 N

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