A subway car weighing m = 25 t departs from the station and moves uphill with an acceleration of a = 0.6 m / s2

A subway car weighing m = 25 t departs from the station and moves uphill with an acceleration of a = 0.6 m / s2. The slope of the mountain is a = 0.03. Determine the strength of the current I, which will flow through the winding of the electric motor when the car travels a distance s = 0.5 km. Movement resistance coefficient k = 0.01, line voltage U = 2.5 kV, efficiency h = 0.8.

Let’s find the resulting force that imparts acceleration to the car:
F = m * a + m * g * k + m * g * α =
F = m * (a + g * k + g * α),
where α = 0.03 and
g = 10m / s ^ 2;
F = 25000kg * (0.6 + 0.1 + 0.3) m / s2 = 25000N
Mechanical work performed on path s
A1 = F * s = 25000N * 500m = 12.5 * 10 ^ 6 J;
Electric motor operation:
A = U * I * t;
Taking into account efficiency:
A1 = h * A = h * U * I * t;
From here you can find the current:
I = A1 / (h * U * t);
From the equation for uniformly accelerated motion, we find the time t:
s = (a / 2) * t ^ 2;
t = sqrt (2 * s / a) = sqrt (1000 / 0.6) = 40.82c;
I = 12.5 * 10 ^ 6 / (0.8 * 2500 * 40.82) = 153.1A.