1. The weight of a car is 12 metric tons, its power is 15 horsepower, what is the maximum achievable speed (1) on a horizontal road where $\rho=\frac{1}{20}$;

(2) on a road that rises in the same conditions in the ratio $=1_{\overline{30}}$;

(3) if we load 4 more metric tons of cargo on it and take the previous two roads as a basis?[^0]


[^0]:    ${ }^{1}$ In this problem, $\rho$ is the coefficient of friction, and $e$ is the sine of the angle of inclination.
The machine has a power of 15 horsepower, so it can perform $15 \cdot 75=1125 \mathrm{mkg}$ of work per minute. This work is equal to the frictional work on a flat track, and the sum of the frictional and lifting work on an incline. Here, as in problems 1133 and 1134, the normal force can be equated to the weight of the body $(\cos \alpha=1)$.

1. $\quad \frac{1}{20} \cdot 1200 \cdot v=15,75, \quad$ from which $\quad v=18,75 \frac{\mathrm{m}}{\mathrm{sec}}=67,5 \frac{\mathrm{km}}{\text { hour }}$.
2. $\quad \frac{1}{20} \cdot 1200 \cdot v^{\prime}+\frac{1}{30} \cdot 1200 \cdot v^{\prime}=15,75, \quad$ from which $\quad v^{\prime}=11,25 \frac{\mathrm{m}}{\mathrm{sec}}=40,5 \frac{\mathrm{km}}{\text { hour }}$.
3. Calculating in the same way, but with the pulling force being $1600 \mathrm{~kg}$ instead of $1200 \mathrm{~kg}$ weight

$$
\begin{aligned}
& v_{1}=14,06 \frac{\mathrm{m}}{\mathrm{sec}}=50,6 \frac{\mathrm{km}}{\text { hour }} \\
& v_{1}^{\prime}=8,44 \frac{\mathrm{m}}{\mathrm{sec}}=30,4 \frac{\mathrm{km}}{\text { hour }}
\end{aligned}
$$

(József Székely, Budapest.)

The problem was also solved by: L. Bánó, I. Csada, R. Földes, A. Frank, V. Gráf, H. Grünhut, A. Haar, Gy. Jánosy, I. Koritsánszky, O. Schwarz.