Not sure what you are doing here. What's the moment of inertia of a cube about one edge? (Depending on what you know, you may need to use the parallel axis theorem.)

What's the change in gravitational PE as the cube falls from its initial position to its final position? (Hint: What happens to the center of mass?)

ok it seems im wrong from the word go..i only have the equation of innertia for an axis about the centre of the cube parallel to an axis (which is the one i used).

so using the parallel axis thoerem (which oddly enough we've used for previous questions but i've never recalled being taught it) i get..

MI = 2/3 Ma^2 + Ma^2
= 5/3 Ma^2

that sound right?

as for the second part I'm not sure what you mean. the PE would fall to 0 as the KE rises to equal the PE?

ah right..a is the distance to a side ,not an edge...so its a triangle you need to make...

...so the distance from centre to an edge is

d^2 = a^2 + a^2
=2a^2
d = root (2a^2)
= root 2 a?

again with part 2 im not sure how you mean. the height of the centre to the table will change reletive to the angle the edge of the box makes with the table, and the length of the side. should i try to find the distance it travels?

To find the change in PE you need to know how far the center of mass of the box fell. Realize that if you measure with respect to the table surface, the PE of the box goes from [itex]mgh_1[/itex] to [itex]mgh_2[/itex] where h is the height of center of mass above the table. (It's not zero at any point.)

but i dont know how to find the height above the table of the centre of the cube, or how that is even relevant. Wouldn't it depend on the angle made by the edge of the box? since no angle is given, do you just use a variable to represent it or asume its at 45 degrees?