Given m = 60 kg and v = 40 km/h, the kinetic energy is K = 1/2 * m * v^2 = 3.703 kJ Acceleration is not going to be 100% efficient, so let's round that up t

Given m = 60 kg and v = 40 km/h, the kinetic energy is K = 1/2 * m * v^2 = 3.703 kJ Acceleration is not going to be 100% efficient, so let's round that up to 10 kJ. We also need to decelerate, so let's assume the same amount of energy being put in to do that. 20 kJ every 2 minutes for 12 hours comes out to 2 kWh total. For electric trains, you can map that to a cost using your local electricity prices, but keep in mind that public transit operators purchase electricity wholesale and probably get a cheaper price than residential users. (Also note that my estimations are intentionally pessimistic to map out the worst-case scenario. Energy efficiency is probably better than my assumptions, and trains may have some sort of regenerative braking.)