Part of the marble sculpture, the volume of which is V = 0.4 m3, is raised to the sea surface from a depth of H

Part of the marble sculpture, the volume of which is V = 0.4 m3, is raised to the sea surface from a depth of H = 15 m. What work A was spent on lifting this sculpture? Density of marble pm = 2.7.10 ^ 3 kg / m3, density of seawater rzh 1.03.10 ^ 3 kg / m

Given:

V = 0.4 m ^ 3 – the volume of the marble sculpture, which is raised from the surface of the sea;

ro1 = 2.7 * 10 ^ 3 kg / m ^ 3 – density of marble;

ro2 = 1.03 * 10 ^ 3 kg / m ^ 3 – density of sea water;

g = 10 m / s ^ 2 – acceleration of gravity;

H = 15 meters – the depth from which the sculpture is lifted.

It is required to determine A (Joule) – the work that was spent on lifting the sculpture.

Find the weight of the sculpture in seawater:

P = F gravity – Farkhimedov;

P = ro1 * V * g – ro2 * V * g

P = V * g * (ro1 – ro2) = 0.4 * 10 * (2700 – 1030) = 4 * 1670 = 6680 Newton.

Then the work expended will be equal to:

A = P * H = 6680 * 15 = 100200 Joules = 100.2 kJ.

Answer: the work expended is equal to 100.2 kJ.