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
Best Answer
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.