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#### alane1994

##### Active member

- Oct 16, 2012

- 126

Here is the problem verbatim.

*A trough has a trapezoidal cross section with a height of 1 m and horizontal sides of width 0.5 m and 1 m. Assume the length of the trough is 11 m.*

How much work is required to pump the water out of the trough (to the level of the top of the trough) when it is full? Use 1000kg/m

for the acceleration due to gravity.

How much work is required to pump the water out of the trough (to the level of the top of the trough) when it is full? Use 1000kg/m

^{3}for the density of water and 9.8 m/s^{2}for the acceleration due to gravity.

I have done some work on the problem.

The equation that I have been taught for this is.

[tex]W=\int^b_a{\rho g A(y)D(y)}dy[/tex]

A(y) is the cross-sectional area of the horizontal slices.

D(y) is the distance the slices must be lifted.

I have figured out this much.

[tex]W=\int^1_0{(1000)(9.8)(A(y))(1-y)}dy[/tex]

What I need help with is figuring out what A(y) is. Any and all help is GREATLY appreciated!