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