Question

A tank is full of water. Find the work required to pump the water out of the spout. Use the fact that water weighs 62.5 lb/ft3

Answer 1

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 = $\int\limits^4_0$ (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

Learn more about calculus here: brainly.com/question/24430269

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