Work done when submerging an object into fluid How does one calculate the work needed to submerge, e.g. wooden, object into water?
For example, you want to submerge wooden cube, density 800 $\mathrm {kg/m^3}$ and side's length a=0.2 m.
The way I was doing it is I calculated force of gravity $G$ and buoyancy $U$, then I got total force $F=U-G$. Then, to get length of path along which one must apply that force, I calculated percentage of cube's volume that is inside the fluid when in equilibrium $(G=U)$. For that I got 0.8, so length of path is $s=0.2a$, and work needed would be $W=F\times s$.

But I can't get the result right.  What did I miss?
 A: An alternative approach, avoiding integration, is to calculate the increase in gravitational potential energy when the block descends from the floating position (left diagram) to the just-submerged position (right diagram) and the water moves up.  

As the block is submerged, the water level rises from $h_2$ to $h_3$ measured from the base of the container, while the bottom of the block moves down from $h_1$ to the base. For convenience I assume the block just touches the base of the container when it is just submerged (right diagram). The top portion of the block (grey) moves to replace the water below the block (dark blue), while this water moves up to cause the increase in water level. All other portions of block or water remain in the same position, so they can be ignored. The grey and dark blue volumes are equal. 
The condition for the block to float is
$\frac{h_2-h_1}{h_3}=\frac{\rho}{\rho_w}$
where $\rho, \rho_w$ are the densities of the block and water respectively.
The CG of the water which moves up is initially $\frac12h_1$ and finally $\frac12(h_3+h_2)$ above the base. The volume of this water is the same in both positions, so
$(h_3-h_2)(A-a)=h_1a$
where $A, a$ are the cross-sectional areas of the container and block respectively. 
The CG of the grey portion of the block moves down by distance $h_3$. The volume of this portion is $h_1a$. 
The above equations should be sufficient for you to calculate the overall increase in GPE when the block is submerged. This equals the work required to submerge the block.
If the block is being submerged in a large body of water instead of a container, then you should apply the limit $\frac{a}{A} \to 0$.
A: I can think of at least four things you would need to clear up/think about before you can answer this question:


*

*Does "submerging" start when the block first touches the water, and end when it is just submerged?

*Can we assume the block is oriented so that the surface of the cube is parallel to the surface of the water?

*Do we expect the water level to rise as the cube is submerged (see Sammy Gerbil's answer)

*How does the buoyancy force change as more of the cube is submerged?

*What is the role of gravity in all this - as the cube moves down, gravity will do some of the work. Does that count in the "work done"?


When you have thought about all these things, you may be able to get the right answer yourself.
A: Don't be upset, I want to help you too, but I'm restricted by the rules of this site. See what you need to do is to balance forces, good that you developed it thus far. This allows me to help you further. The work done is done by the resultant force and not just by the buoyant force, since you apply the resultant force since the weight of the object is already helping you. $F_{resultant}=\rho_f ga^2x-\rho_wga^3$.
Now try to get the work done by using $\int F_{resultant}\mathrm{d}x$ within the limits you specified.
I hope, this does not qualify as an answer. I'm just clearing a part of a misconception.
Disclaimer: I am adding this part after the OP showed his work. 
The work done is calculated as $$\int\mathrm{d}W=\int F_{resultant}\mathrm{d}x=\int_{0.8a}^{a}(\rho_f g a^2x-\rho_wga^3)\mathrm{d}x$$ 
A: An object in general describes a cross-sectional area $A(\ell)$ where $\ell$ is the distance that it has been pushed into the water; a good rule of thumb is $\int_0^L d\ell~A(\ell) = V,$ the total volume of the object, when it is fully submerged at depth $L$. Of course in your case $A(\ell)$ is just a constant $A$.
In order to calculate the work needed to submerge a floating object into the water, you indeed need to solve first for the equilibrium floating depth $h$, $$\rho_\text{obj}~V= \rho_\text{fluid} ~\int_0^hd\ell ~A(\ell),$$and then you need to submerge it further. As you submerge it further the force needed will increase with the amount of water displaced,$$F(x) = \rho_\text{fluid} ~g~\int_h^{h+x} d\ell ~A(\ell).$$Because this force is not constant to get the work you must integrate over the distance you push the item again,$$W=\int dx~F(x) = \rho_\text{fluid} ~g~ \int_0^{L-h}dx~\int_h^{h+x}d\ell~A(\ell).$$
For a constant area you should just get $A~x$ for the first integral and therefore $\frac12 A (L-h)^2$ for the second. You seem to have gotten this but not realized that it was a work, not a force.
A more realistic scenario for approximating various boats is to assume $A(\ell) = \gamma~\ell$ for some side-steepness factor $\gamma$, this corresponds to a boat which is a triangular prism. Or you could try to model one as a prism made by a parabola if you like.
