Answer
The work done by the buoyant force is $~~2350~J$
Work Step by Step
We can find an expression for the mass of the device:
$m = \rho_d~V_d$
This mass of water is displaced by the device when the device floats on the water in equilibrium. We can find the volume of water that is displaced:
$V_w = \frac{m}{\rho_w}$
$V_w = \frac{\rho_d~V_d}{\rho_w}$
We can find the ratio $\frac{V_w}{V_d}$:
$V_w = \frac{\rho_d~V_d}{\rho_w}$
$\frac{V_w}{V_d} = \frac{\rho_d}{\rho_w}$
$\frac{V_w}{V_d} = 0.400$
The fraction of the volume of the device that would be above the water surface is $~~1-0.400~~$ which is $~~0.600$
We can find the change in vertical position of the device when the buoyant force pushes it upward to the equilibrium point:
$\Delta h = (0.600)(0.500~m) = 0.300~m$
We can find the increase in gravitational potential energy:
$\Delta U = mg~\Delta h$
$\Delta U = \rho~V~g~\Delta h$
$\Delta U = (400~kg/m^3)(4.00~m^2)(0.500~m)(9.8~m/s^2)(0.300~m)$
$\Delta U = 2350~J$
The work done by the buoyant force is $~~2350~J$
