From the bottom of the reservoir, with the help of a light rope, a steel slab of mass m is slowly raised to the surface

From the bottom of the reservoir, with the help of a light rope, a steel slab of mass m is slowly raised to the surface of the water. Determine the elongation of the rope with rigidity k, if the density of steel is p1, the density of water is p2.

According to Newton’s second law, the sum of the forces acting on a body is equal to mass for acceleration.
In our case:
m * g + Fy + Fa = m * a
Let’s project on the Y-axis, taking into account that the body moves with equal acceleration:
-mg + Fy + Fa = 0
A buoyant force equal to the weight of the displaced fluid acts on a body immersed in a liquid, and is called the Archimedes force:
Fа = ρж * g * V, where ρж is the density of the liquid, g is the acceleration of free fall of a body raised above the Earth g = 9.8 m / s², V is the volume of the body.
Hooke’s Law: The elastic force is directly proportional to the amount of deformation.
Fcont = k * ∆l, where k is a coefficient showing the stiffness of the spring, ∆l is the change in body length.
Now Newton’s law takes the form:
ρl * g * V + k * ∆l = m * g
Body volume:
V = m / ρ
Substitute into Newton’s second law:
ρl * g * m / ρ + k * ∆l = m * g
k * ∆l = m * g-ρl * g * m / ρ
k * ∆l = m * g * (1-ρl / ρ)
∆l = (m * g * (1-ρl / ρ)) / k
According to our conditions:
∆l = (m * g * (1-ρ2 / ρ1)) / k
Answer: cable elongation ∆l = (m * g * (1-ρ2 / ρ1)) / k.