A bar is pulled along the surface of the table at a constant speed. In this case, a force of 4 N is applied

A bar is pulled along the surface of the table at a constant speed. In this case, a force of 4 N is applied. What speed does the bar have if 4.8 J is carried out in 12 seconds?

Data: the movement of the taken bar on the table surface is uniform; F (applied force) = 4 N; t (duration of movement of the bar) = 12 s; A (perfect work) = 4.8 J.

To calculate the speed of the taken bar, we use the formula: A = F * S = F * V * t, whence V = A / (F * t).

Let’s calculate: V = 4.8 / (4 * 12) = 4.8 / 48 = 0.1 m / s.

Answer: The taken block moves along the surface of the table at a speed of 0.1 m / s.