Solution

Statement

$2.2.2.$ An initially stationary body located on a horizontal plane began to be pulled by a rope attached to it with a constant horizontal force $F$. After a time $\Delta t$, the action of this force ceased. What friction force acted on the body during its motion if it stopped after a time $3\Delta t$ from the start of motion?

Solution

During the time $\Delta t$, the force $F$ applied to the body and the friction force change the body's momentum:
$$(F - F_{\text{fr}}) \cdot \Delta t = \Delta p = p - 0 = p \quad (1)$$

After the action of the force ceases, the body stops under the action of the friction force, which changes the acquired momentum of the body to zero (stop):
$$-F_{\text{fr}} \cdot 2\Delta t = 0 - p = -p \quad (2)$$
Here it is taken into account that the friction force acts until the stop for a time equal to:
$$3\Delta t - \Delta t = 2\Delta t$$

Equating the right and respectively the left parts of equations (1) and (2):
$$(F - F_{\text{fr}}) \cdot \Delta t = F_{\text{fr}} \cdot 2\Delta t$$
From where we find the required friction force:
$$F_{\text{fr}} = \frac{F}{3}$$

Note: the problem can also be solved by a dynamic method, for example, by setting up the forces applied to the body, writing Newton's second law, determining the acceleration in the first stage of motion, and expressing the velocity. Similarly, apply Newton's second law for the second segment of motion, considering that the final velocity in the first segment of motion is the initial velocity for the second segment. Another method is the energy method. Solve the problem using the second and third methods independently.

Alternative solution:

Time of action of force $F$: $\Delta t$, and time of action of friction force $F_{\text{fr}}$: $3\Delta t$.
The total change in momentum is $0$, then:
$$F_{\text{fr}} \cdot 3\Delta t = F \cdot \Delta t$$
$$F_{\text{fr}} = \frac{1}{3}F$$