A boat is pulled along the water with a constant force $\bar{F}$. Its velocity is constant and equal to $\bar{v}$. What is the distance $S$ the boat will travel in the water after the force ceases, if the force resisting the boat's motion is proportional to its velocity relative to the water?
Decision:
Under the action of the force $\bar{F}$ the boat was moving uniformly, therefore, according to Newton's I law, there was still a drag force acting on it equal to $\bar{F_{c}} = -\bar{F}$. According to the problem $F_{c} = -\alpha \bar{v}$, where $\alpha$ is the coefficient of friction. As you can see, $\alpha = \frac{F}{v}$. After the force $F$ ceases, the boat moves sluggishly because the drag force acts on it as long as there is velocity relative to the water. For an infinitesimal amount of time $\delta t$ we can write down Newton's second law:
$m \delta v = - \alpha v \delta t$ (1)
Using the definition of velocity $v = \frac{ \delta x}{ \delta t}$, we rewrite Eq. (1) in the form
$m \delta v = - \alpha \delta x$. (2)
Equality (2), unlike (1), is valid not only for small, but also for any change in $\delta v$ and $\delta t$. The total change in velocity from the beginning of braking (when the boat speed was equal to $v$) to the stop is $\delta v = -v$, and the total distance traveled is $\delta x = S$. Substituting these values into (2), we obtain the relationship $mv = \alpha S = \frac{FS}{v}$, whence
$S = \frac{mv^{2}}{F}$.
Decision:
Under the action of the force $\bar{F}$ the boat was moving uniformly, therefore, according to Newton's I law, there was still a drag force acting on it equal to $\bar{F_{c}} = -\bar{F}$. According to the problem $F_{c} = -\alpha \bar{v}$, where $\alpha$ is the coefficient of friction. As you can see, $\alpha = \frac{F}{v}$. After the force $F$ ceases, the boat moves sluggishly because the drag force acts on it as long as there is velocity relative to the water. For an infinitesimal amount of time $\delta t$ we can write down Newton's second law:
$m \delta v = - \alpha v \delta t$ (1)
Using the definition of velocity $v = \frac{ \delta x}{ \delta t}$, we rewrite Eq. (1) in the form
$m \delta v = - \alpha \delta x$. (2)
Equality (2), unlike (1), is valid not only for small, but also for any change in $\delta v$ and $\delta t$. The total change in velocity from the beginning of braking (when the boat speed was equal to $v$) to the stop is $\delta v = -v$, and the total distance traveled is $\delta x = S$. Substituting these values into (2), we obtain the relationship $mv = \alpha S = \frac{FS}{v}$, whence
$S = \frac{mv^{2}}{F}$.