Having much difficulty with this word problem:
A water well has an 8-inch diameter and is 175 feet deep. Assume that water enters the well at a rate of 4 gallons per minute and the pump works at a rate of 12 gallons per minute. If the water is 25 feet from the top of the well, determine the amount of work done in pumping it dry.
I know the weight of the water is 62.4.
The work formula is W = Integral( F(x) ) from -175 to -25.
I know I need the volume of the dy disc: pi*4^2 and the weight of that disc to be: 62.4*pi*4^2.
I can't figure out how to factor in the amount of water coming into the well or the rate of water being pump.
I take my numbers for the units from the following web sites:
The starting volume of water is V_0 = pi * (1/3 ft)^2 * (175 - 25) ft ~= 52.34 ft^3.
In gallons it is V_0 = 52.34 ft^3 * 7.58 gallon/ ft^3 ~= 796.7 gallons
The mass of one gallon of water is 8.338 pounds
1 kg = 2.2 lb
1 ft = 0.3048 m
The work needed to raise one gallon of water one foot up is
Wgf = 8.338 lb/gal / (2.2 lb/kg) * 9.8 m/s^2 * 0.3048 m/ft = 11.32 J /(gal * ft)
This number ...
The solution discusses the question regarding the water that jumped from the well.