A train of mass m = 400 t speeds up uniformly from v_1 = 40…

A train of mass $m = 400\ \mathrm{t}$ speeds up uniformly from $v_1 = 40\ \mathrm{km/h}$ to $v_2 = 60\ \mathrm{km/h}$ over a level distance $s = 1\ \mathrm{km}$. What power must the locomotive develop for this (neglecting friction)?

Explanation of Method

Locomotive power to speed up a train.

Step 1: Work

\[ W = \tfrac12 m\,(v_2^2 - v_1^2), \]

\[ W = 2 \cdot 10^5 \cdot (277.8 - 123.5) \approx 3.09 \cdot 10^7\ \mathrm{J}. \]

Step 2: Time

\[ t = \frac{s}{(v_1 + v_2)/2} = \frac{1000}{13.89} = 72\ \mathrm{s}. \]

Step 3: Power

\[\bar P = \frac{W}{t} \approx 429\ \mathrm{kW}\]

The force is constant, $F = W/s \approx 30.9\ \mathrm{kN}$, so the instantaneous power $Fv$ grows from 343 kW to 514 kW; 429 kW is its average.

$P\ (\mathrm{kW})$ $t\ (\mathrm{s})$ $0$ $20$ $40$ $60$ $72$ $343$ $429$ $514$ $\text{average } 429\ \mathrm{kW}$ $P = Fv,\ F = 30.9\ \mathrm{kN}$

With a constant force (uniform acceleration) the power $Fv$ grows with the speed, from 343 kW to 514 kW; its average over the 72 s is 429 kW.

Result

\[\boxed{\,\bar P = \frac{W}{t} \approx 429\ \mathrm{kW}\,}\]

Remark

Check. The original result (428 kW) is correct as the average power; the engine must be able to deliver up to about 514 kW at the end of the stretch.