Determine the steady-state velocity of a body on an inclined plane that changes one direction of its velocity u to the opposite direction with high frequency. The direction of movement of the plane is shown in the figure. Coefficient of friction \mu , angle of inclination of the plane \alpha , \tan\alpha < \mu .

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    <h3> Statement </h3>
    <p>
        $2.1.44^*.$ Determine the steady-state velocity of a body on an inclined plane that changes one direction of its velocity $u$ to the opposite direction with high frequency. The direction of movement of the plane is shown in the figure. Coefficient of friction $\mu$, angle of inclination of the plane $\alpha$, $\tan\alpha < \mu$.

    <h3>Solution</h3>
    <p>
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Since the speed changes quickly, the body does not have time to move in the horizontal direction and all the time moves in the direction of $\vec{v}$

Since $v=\text{const}$, there is no acceleration in the direction along the speed $\vec{v}$ $$mg\sin\alpha = \mu mg \cos\alpha\cos\beta$$

$$\cos\beta = \frac{\tan\alpha}{\mu}$$
From geometric considerations, the moduli of the vectors $\vec{v}$ and $\vec{u}$ are related by the relation
$$v=\frac{u}{\tan\beta}$$

$$v=u \frac{\frac{\tan\alpha}{\mu}}{\sqrt{1-\frac{\tan^2\alpha}{\mu^2}}}$$

$$\boxed{v = u \frac{\tan\alpha}{\sqrt{\mu^{2} − \tan^{2} \alpha}}}$$

    <h4>Answer</h4>
    <p>
        $$v = u \frac{\\, \text{tg} \alpha}{\sqrt{\mu^{2} − \text{tg}^{2} \alpha}}$$
    </p>


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