Solution

Statement

$2.3.13.$ A load of mass $m$ is slowly pulled up to a height $h$ along an inclined plane using a block and a cable. The work done during this process is $A$. Then the cable is released, and the load slides down. What velocity will it gain by the time it reaches the starting point?

Solution

The difference between the work of the pulling force and the work of friction is equal to the change in potential energy:
$$A - A_{\text{f}} = mgh$$
During the slide down from the top, friction performs the same amount of work. Let us write the law of conservation of energy:
$$A_{\text{f}} + \frac{mv^2}{2} = mgh$$
Substitute the work of friction to express the kinetic energy:
$$\frac{mv^2}{2} = 2mgh - A$$
From this, we find the velocity the load will gain upon reaching the starting point:
$$v = \sqrt{4gh - 2A/m}$$