Answer:
The workdone is [tex]W = 9.28 * 10^{3} J[/tex]
Explanation:
From the question we are told that
The height of the cylinder is [tex]h = 0.588\ m[/tex]
The face Area is [tex]A = 4.19 \ m^2[/tex]
The density of the cylinder is [tex]\rho = 0.346 * \rho_w[/tex]
Where [tex]\rho_w[/tex] is the density of freshwater which has a constant value
[tex]\rho_w = 1000 kg/m^3[/tex]
Now
Let the final height of the device under the water be [tex]= h_f[/tex]
Let the initial volume underwater be [tex]= V_n[/tex]
Let the initial height under water be [tex]= h_i[/tex]
Let the final volume under water be [tex]= V_f[/tex]
According to the rule of floatation
The weight of the cylinder = Upward thrust
This is mathematically represented as
[tex]\rho_c g V_n = \rho_w gV_f[/tex]
[tex]\rho_c A h = \rho A h_f[/tex]
So [tex]\frac{0.346 \rho_w}{\rho_w} = \frac{h_f}{h}[/tex]
=> [tex]\frac{h_f}{h_c} = 0.346[/tex]
Now the work done is mathematically represented as
[tex]W = \int\limits^{h_f}_{h} {\rho_w g A (-h)} \, dh[/tex]
[tex]= \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.[/tex]
[tex]= \frac{g A \rho}{2} [h^2 - h_f^2][/tex]
[tex]= \frac{g A \rho}{2} (h^2) [1 - \frac{h_f^2}{h^2} ][/tex]
Substituting values
[tex]W = \frac{(9.8 ) (4.19) (10^3)}{2} (0.588)^2 (1 - 0.346)[/tex]
[tex]W = 9.28 * 10^{3} J[/tex]
