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:
$$v = \sqrt{4gh - 2A/m}$$