A rectangular tank that is 2 feet long, 3 feet wide and 6 feet deep is filled with a heavy liquid that weighs 80 pounds per cubic foot. How much work is done pumping all of the liquid out over the top of the tank?How much work is done pumping all of the liquid out of a spout 5 feet above the top of the tank?How much work is done pumping two-thirds of the liquid out over the top of the tank?How much work is done pumping two-thirds of the liquid out of a spout 5 feet above the top of the tank?I thought that math was hard enough and now they added physics..
Work is force x distance, which is conveniently pounds x feet. Since we are only interested in the height here (as that is the direction we are pumping), we only need to integrate along that path. The differential weight/force being pumped is the volume (LxWxH = 2 x 3 x dH) times the density (80) and the distance is the height that volume is pumped (6 - H). So W = integral( 6 * 80 * dH * (6-H)) from H=0 to 6 W = integral ((2880 - 480H) dH) = 2880H - 240*H^2 = 2880(6) - 240*(6)^2 = 8640 foot-pounds of work. If you are pumping from the spout, replace the 6-H with 11-H, and the answer becomes 23,040 foot-pounds, which makes sense since you have almost tripled the averge height you were lifting. For the 2/3, change the range of the integration to H = 2 - 6, since H = 0 - 2 represents the water at the bottom of the tank that will still be there later. This gives you answers of 3840 and 13,440 foot-pounds respectively, showing that the last couple of feet are the hardest to pump out (because they have the farthest to go...)
No work required, just hook-up a hose and siphon it out.
