Question

A freight train consists of two 8.00 \times 10^5~\text{kg}8.00×10 ​5 ​​ kg engines and 45 cars with average masses of 5.50 \times 10^5~\text{kg}5.50×10 ​5 ​​ kg. What force must each engine exert backward on the track to accelerate the train at a rate of 5.00 \times 10^{-2}~\text{m/s}^25.00×10 ​−2 ​​ m/s ​2 ​​ if the force of friction is 7.50 \times 10^5 ~\text{N}7.50×10 ​5 ​​ 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.

Answer 1

Answer:

F₁ = 4.29 x 10⁵ N

Explanation:

The total force required to move the freight train with the given acceleration is given by the following formula:

F = ma + f

where,

F = Total Force Required from both engines = ?

m = equivalent mass of system = 2(8 x 10⁵ kg) + 5.5 x 10⁵ kg = 21.5 x 10⁵ kg

a = required acceleration = 5 x 10⁻² m/s²

f = force of friction = 7.5 x 10⁵ N

Therefore,

F = (21.5 x 10⁵ kg)(0.05 m/s²) + 7.5 x 10⁵ N

F = 8.575 x 10⁵ N

Now, for identical forces in each engine can be given as:

Force exerted by each engine = F₁ = F/2

F₁ = 8.575 x 10⁵ N/2

F₁ = 4.29 x 10⁵ N