The velocity of a body of mass m in a viscous liquid decreases with the distance l traveled according to the law v = v_0 - \beta l, where v_0 is the initial velocity, and \beta is a constant coefficient. How does the viscous friction force acting on a body from the fluid side depend on the velocity of the body?

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    <h3 id="back-link"><a href="/#2.1">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $2.1.36.$ The velocity of a body of mass $m$ in a viscous liquid decreases with the distance $l$ traveled according to the law $v = v_0 - \beta l$, where $v_0$ is the initial velocity, and $\beta$ is a constant coefficient. How does the viscous friction force acting on a body from the fluid side depend on the velocity of the body?
    </p>

    <h3>Solution</h3>
    <p>
        The equation of Newton's second law for the direction of motion:

$$ma = F_с$$

$$\frac{F_с}{m}=\frac{dv_x}{dt}$$
Derivative of velocity with respect to time
$$\frac{dv_x}{dt}=\frac{d}{dt}(v_0-βx)$$

$$\frac{dv_x}{dt}=\frac{dv_0}{dt}-β\frac{dx}{dt}=-βt$$
the minus sign shows that the acceleration vector is directed in the direction opposite to the velocity vector.
Combining the equations, we obtain the value of the resistance force as a function of velocity
$$\boxed{F=βmv}$$

    </p>

    <h4>Answer</h4>
    <p>
        $$F = βmv$$
    </p>


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