Answer
$m = 2.1\times 10^6~kg$
Work Step by Step
We can find the energy produced by the locomotive in 6.0 minutes:
$E = P~t$
$E = (1.5\times 10^6~W)(360~s)$
$E = 5.4\times 10^8~J$
We can find the mass of the train:
$\Delta K = 5.4\times 10^8~J$
$K_2-K_1 = 5.4\times 10^8~J$
$\frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2 = 5.4\times 10^8~J$
$m = \frac{(2)(5.4\times 10^8~J)}{v_2^2-v_1^2}$
$m = \frac{(2)(5.4\times 10^8~J)}{(25~m/s)^2-(10~m/s)^2}$
$m = 2.1\times 10^6~kg$
