A train of mass m = 400 t moves on rails; the friction force…

A train of mass $m = 400\ \mathrm{t}$ moves on rails; the friction force is $6/1000$ of its weight. What speed does the train reach if a force $F = 60\ \mathrm{kN}$ acts on it for $\Delta t = 50\ \mathrm{s}$, starting from rest?

Explanation of Method

Speed of a train after 50 s of pulling.

Step 1: Net force

\[ F_f = 0.006\,mg = 0.006 \cdot 4 \cdot 10^5 \cdot 9.81 \approx 23\,544\ \mathrm{N}, \]

\[ F - F_f = 36\,456\ \mathrm{N}. \]

Step 2: Speed

\[ a = \frac{F - F_f}{m} = \frac{36\,456}{4 \cdot 10^5} \approx 0.0911\ \mathrm{m/s^2}, \]

\[ v = a\,\Delta t. \]

\[v \approx 4.56\ \mathrm{m/s} \approx 16.4\ \mathrm{km/h}\]

$v\ (\mathrm{m/s})$ $t\ (\mathrm{s})$ $0$ $10$ $20$ $30$ $40$ $50$ $2$ $4$ $4.56\ \mathrm{m/s}$ $a = \frac{F - F_f}{m} = 0.091\ \mathrm{m/s^2}$

Speed against time. Of the 60 kN pull, 23.5 kN goes to friction; the remaining 36.5 kN accelerates the 400 t train uniformly.

Result

\[\boxed{\,v \approx 4.56\ \mathrm{m/s} \approx 16.4\ \mathrm{km/h}\,}\]

Remark

Check. The original result is correct. Equivalently, the net impulse $(F - F_f)\Delta t$ equals the momentum $mv$.