# How Much Work is Done by Buoyant Force on Ascending Flotation Device?

## Explanation:

**Given data:**

The height of the cylinder is h = 0.588 m

The face Area is A = 4.19 m^2

The density of the cylinder is Ï = 0.346 * Ï_w

Where Ï_w is the density of freshwater which has a constant value Ï_w = 1000 kg/m^3

Let the final height of the device under the water be hf

Let the initial volume underwater be V_n

Let the initial height under water be hi

Let the final volume under water be V_f

**Calculation:**

According to the rule of floatation, the weight of the cylinder equals the upward thrust:

Ï_c g V_n = Ï_w g V_f

Ï_c A h = Ï A hf

hf / h = 0.346

The work done is mathematically represented as:

W = â«(hf to h) Ï_w g A (-h) dh

W = (g A Ï / 2) (h^2 - hf^2)

W = (g A Ï / 2) (h^2) (1 - (hf^2 / h^2))

**Substituting values:**

W = (9.8) (4.19) (10^3) / 2 (0.588)^2 (1 - 0.346)

W = 9.28 * 10^3 J