
Calculating work?
A trough is 9 meters long, 1.5 meters wide, and 2 meters deep. The vertical crosssection of the trough parallel to an end is shaped like an isoceles triangle (with height 2 meters, and base, on top, of length 1.5 meters). The trough is full of water (density 1000 kg/m^3 ). Find the amount of work in joules required to empty the trough by pumping the water over the top. (Note: Use 9/8 m/s^2 as the acceleration due to gravity.)
So I know that work = (force)(distance) = (mass)(density)(distance), but I can't figure out how all the pieces go together.
Can someone please help me figure this out?

I am going to use $\displaystyle 9810 \;\ \frac{N}{m^{3}}$ as the weight density of water.
by similar triangles $\displaystyle \frac{w}{\frac{3}{2}}=\frac{x}{2}$
$\displaystyle w=\frac{3}{4}x$
$\displaystyle \int_{0}^{2}(2x)(9810)(\frac{3x}{4})(9)dx=88290 \;\ J$