I'm not sure if I went about them correctly, but here are the two problems. Any help anyone could provide would be greatly appreciated. Thanks!
--Problem 1--
Find the work done in raising 1,500 pounds of ore from a mine that is 1,500 feet deep. The rope weighs 2 pounds per foot.
1.) The work of the ore is simply W = F * d = 1,500 * 1,500 = 2250000 ft-lbs.
2.) The work of the rope is the integral from 0 to 1,500 of 2x dx = 2250000.
3.) Then add the two together for a total work of 4.5 * 10^6 ft-lbs.
Even if I set Step 2 up like integral[0,1500] of (2)(1,500-x) dx...I still get the same answer.
--Problem 2--
Find the work done in filling an upright cylindrical tank of radius 3 feet and height 10 feet with a liquid that weighs 50 pounds per cubic foot. Assume the liquid is pumped into a hole in the bottom of the tank.
1.) V = pi*r^2*h, so V = 9*pi*delta x
2.) W = integral[0,10] of 50 * 9 * pi * x dx = 450 * pi * x dx = 70,685.83
ft-lbs.
2006-07-26
10:11:17
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1 answers
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asked by
JoeSchmo5819
4