A train (locomotive and cars) has a total mass m = 400 t…

A train (locomotive and cars) has a total mass $m = 400\ \mathrm{t}$; the coefficient of friction (rolling resistance) is $\mu = 0.04$. What tractive force must the locomotive exert for the train to reach $v = 40\ \mathrm{km/h}$ from rest in $t = 120\ \mathrm{s}$?

Explanation of Method

Tractive force of a locomotive.

Step 1: Acceleration

\[ v = \frac{40}{3.6} = 11.11\ \mathrm{m/s}, \]

\[ a = \frac{v}{t} = \frac{11.11}{120} \approx 0.0926\ \mathrm{m/s^2}. \]

Step 2: Newton's second law

\[ F - \mu mg = ma, \]

\[ F = m\,(a + \mu g) = 4 \cdot 10^5 \cdot (0.0926 + 0.3924). \]

\[F \approx 1.94 \cdot 10^5\ \mathrm{N} = 194\ \mathrm{kN}\]

$0$ $50$ $100$ $150$ $200$ $\mathrm{kN}$ $\mu mg = 157\ \mathrm{kN}$ $ma$ $194\ \mathrm{kN}$ $F$ $a = \frac{11.1\ \mathrm{m/s}}{120\ \mathrm{s}} = 0.0926\ \mathrm{m/s^2}$

overcoming resistance
accelerating (37 kN)

Where the tractive force goes. Most of it only balances the rolling resistance; about a fifth actually speeds the train up.

Result

\[\boxed{\,F \approx 1.94 \cdot 10^5\ \mathrm{N} = 194\ \mathrm{kN}\,}\]

Remark

Check. The original result is correct: 157 kN balances the resistance and 37 kN accelerates the train.