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]
  Â
Now Â
   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 Â
     [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]
