Answer: 49. Solution. The kinetic energy of the train $\frac{m v^{2}}{2}$ will be equal to the potential energy of the part of the train that has entered the hill $\frac{1}{2} \frac{L}{4} \sin \alpha \frac{m}{4} g$.

Then we get $V^{2}=\frac{L}{4} \frac{1}{8} *(3.6)^{2}=\frac{6000 *(3.6)^{2}}{32}=9 \sqrt{30}$.

Since the following double inequality is true $49<9 \sqrt{30}<49.5$ (verified by squaring), the answer is 49.
4.1. A train of length $L=600$ m, moving by inertia, enters a hill with an angle of inclination $\alpha=30^{\circ}$ and stops when exactly a quarter of the train is on the hill. What was the initial speed of the train $V$ (in km/h)? Provide the nearest whole number to the calculated speed. Neglect friction and assume the acceleration due to gravity is $g=10 \mathrm{m} /$ sec $^{2}$.