A train travels at a maximum speed of $v=12 \frac{\mathrm{m}}{\mathrm{sec}}$ on a track where $\rho=0.004$.

(1.) If the steam is shut off, how far will it travel on a horizontal track due to its kinetic energy, and how far on a track with an incline of $e=0.025$?

(2.) How long will these movements take, assuming that the deceleration is uniform?[^0]


[^0]:    ${ }^{1}$ In this problem, $\rho$ is the coefficient of friction, and $e$ is the sine of the incline angle.
In one case, the kinetic energy is converted into frictional work, and in the other case, it is converted into frictional and lifting work. Given the available accuracy, the pressing force can again be equated with the weight $(\cos \alpha=1)$. If $s$ and $s^{\prime}$ represent the sought distances, then

$$
\begin{gathered}
\frac{1}{2} m v^{2}=\rho m \cdot g \cdot s \\
\frac{1}{2} m v^{2}=\rho m \cdot g \cdot s^{\prime}+e \cdot m g \cdot s^{\prime} \\
s=\frac{v^{2}}{2 \cdot \rho \cdot g}=180 \frac{\mathrm{m}}{\mathrm{sec}} \\
s^{\prime}=\frac{v^{2}}{2 g \cdot(\rho+e)}=250 \frac{\mathrm{m}}{\mathrm{sec}}
\end{gathered}
$$

Since the motion is uniformly decelerating, based on the well-known formulas,

$$
\begin{aligned}
t & =\frac{2 s}{v}=300 \mathrm{sec} \\
t^{\prime} & =\frac{2 s^{\prime}}{v}=42 \mathrm{sec}
\end{aligned}
$$

The problem was solved by: Bánó L., Blum J., Csada I., Ehrenstein P., Földes R., Gráf V., Grünhut H., Haar A., Heimlich P., Jánosy Gy., Koritsánszky I., Merse P., Pözel T., Schwarz O., Schuster Gy., Székely J.