The work required to pump the water out of the spout is 5000ft⋅lb
Given
The volume V=12(4ft)(5ft)(6ft)=60ft3,
Then the weight is w = γV = (62.5lb /ft³) (60ft³) = 3750lb.
Since work is force × distance, you need a distance here, and the appropriate one is the distance you must raise the center of mass to get the H2O out of the spout.
Since the center of mass of a uniform triangular area is 13 of the way from any side, the distance is (4ft)/3 so the work is (3750lb) (4ft)/3=5000 ft-lb.
In calculus, the area of a horizontal section yft from the lowest point is (5ft) × (6ft) × y / (4ft) = 7.5y ft2.
We have to lift that slice 4ft−y, so work is
w = [tex]\int\limits^4_0[/tex] (62.5)(4 − y) (7.5y)
dy = (62.5)[15y2−2.5y3]40 = 5000ft⋅lb
The work required to pump the water out of the spout is 5000ft-lb
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