6. The brakes of a car allow it to stand on an inclined asphalt surface with an angle at the base of no more than $15^{\circ}$. Determine the minimum braking distance of this car when moving at a speed of 20 m/s on a flat horizontal road with the same surface. The acceleration due to gravity $g=10 m / s^{2}, \cos 15^{\circ} \approx 0.966, \sin 15^{\circ} \approx 0.259$. (15 points)
Answer: $74.6 m$

Solution. For the situation when the car is on an inclined plane:
$F_{n p}=m g \sin \alpha$ (2 points), $\mu m g \cos \alpha=m g \sin \alpha$ (2 points). Therefore, the coefficient of friction: $\mu=\frac{\sin 15^{\circ}}{\cos 15^{\circ}}=0.268 \quad(3 \quad$ points). For the situation when the car is moving on a horizontal surface: $F_{m p}=m a, \quad \mu m g=m a$ (2 points), $\mu g=a=2.68$ (3 points). Then the braking distance of the car: $S=\frac{v^{2}}{2 a}=\frac{20^{2}}{2 \cdot 2.68} \approx 74.6 m$ (3 points).