$2.1.36.$ 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?
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 $$ \frac{dv_x}{dt}=\frac{d}{dt}(v_0-\beta x) $$ $$ \frac{dv_x}{dt}=\frac{dv_0}{dt}-\beta\frac{dx}{dt}=-\beta t $$ 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 $$ \boxed{F=\beta mv} $$
$$F = \beta mv$$