Water is being pumped out of a triangular trough that is 6m long, 2m tall, and 3m wide along the top. Water must be pumped out the top end of a spout which is another 1m tall. (See image for clarification). How much work must be done to pump all of the water out of the spout.
My force slice = (9.8)(1000)(6)(3-3/7x) My Displacement = (1+x) My integral is for 0 to 2 of (9.8)(1000)()(3-3/7x)(1+x)dx
My Final answer for it is 294000J. Does Any one know if this is correct? I have spent a good amount of time on this problem and wanted to see If I have mastered it or if it needs more work!
This is the video I followed along with if this helps! https://www.youtube.com/watch?v=fJtxJv5sdqo
You can work this out without calculus.
Pumping out the water is equivalent to moving the water's center of mass from its current position in the trough to a height of 1 meter above the trough. The center of mass (centroid) of an isosceles triangle is at $\frac 1 3$ the height of the triangle, i.e. at $\frac 1 3 \cdot 2 = \frac 2 3$ m below the top of the trough. The total height the center of mass must be moved is therefore $H= 1\frac 2 3$ m.
The volume of the trough is the area of the triangle times the length of the trough, or $$V = \frac 1 2 \cdot 2 \cdot 3 \cdot 6 = 18 \ \text {m}^3$$
The work done is the change in potential energy, i.e. $W = m \cdot g \cdot H$ or $$W = 18 \ \text {m}^3 \cdot 1000 \ \frac{\text{kg}}{\text{m}^3} \cdot 9.8 \ \frac{\text{m}}{\text{s}^2} \cdot 1\frac 2 3 \ \text{m} = 294,000 \ \text{J}$$
So your answer looks good.