## Related questions with answers

A freight train consists of two $8.00 \times 10^{4}-\mathrm{kg}$ engines and 45 cars with average masses of $5.50 \times 10^{4}\ \mathrm{kg}.$ (a) What force must each engine exert backward on the track to accelerate the train at a rate of $5.00 \times 10^{-2}\ \mathrm{m} / \mathrm{s}^{2}$ if the force of friction is $7.50 \times 10^{5}\ \mathrm{N},$ 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?

Solution

VerifiedAcceleration is a result of net force acting on the freight train. It can be calculatzed as:

$F_{net}=ma=(2\cdot 8\cdot 10^4+45\cdot5.5\cdot 10^4)\cdot 5\cdot 10^{-2}=131750\,\,\rm{N}$

And as we know, net force is result of $\textbf{difference between pull and friction force}$.

$F_{pull}=F{net}+F_{drag}=131750+7.5\cdot 10^5=881750\,\,\rm{N}$

Per engine:

$F_{pull-e}=\frac{F_{pull}}{2}=\frac{881750}{2}$

$\boxed{F_{pull-e}=440875\,\,\rm{N}}$

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