Correct Answer - Option 1 : 640 kW

**Concept:**

**Tractive Effort (F**_{t}) for Propulsion of Train:

The tractive effort is a force developed by the traction unit at the rim of driving wheels for moving the unit itself and its train.

The tractive effort required to train propulsion on a level track is given as,

**F**_{t} = F_{a} + F_{r}

If gradient is involved,

F_{t} = F_{a} + F_{r} + F_{g} (For ascending gradient)

F_{t} = F_{a} + F_{r} - F_{g} (For descending gradient)

Where,

F_{a} is the force required to giving linear acceleration

F_{g} is the force required to overcome the effect of gravity

F_{r} is the force required to overcome the resistance of train motion

**1. Value of F**_{a}: If M is the stationary mass of the train and a is linear acceleration.

F_{a} = Ma

Apart from the stationary mass train has a rotating part like the wheel, axles, motor-generator set, etc and it is about (8% to 15 %) of stationary mass.

**2. Value of F**_{g}:

F_{g} = W sin θ = Mg sin θ

**3. Value of F**_{r}: It consists of mechanical resistance and wind resistance.

**Power output (P**_{0}) from Driving Axles:

If F_{t} is tractive effort and v is the train velocity, then

P_{o} = F_{t} × v Watt

**Calculation:**

Given,

F_{t} = 57600 N

\(v=40 km/hr=40× \frac{5}{18}=11.11\ m/sec\)

From the above concept,

P_{0} = F_{t} × v = 57600 × 11.11 ≈ 640 × 10^{3} W = **640 kW**