(1 point) Water in a cylinder of height 11 ft and radius 3 ft is to be pumped ou
ID: 2880146 • Letter: #
Question
(1 point) Water in a cylinder of height 11 ft and radius 3 ft is to be pumped out. The density of water is 62.4 lb/ft3. Find the work required if
(a) The tank is full of water and the water is to be pumped over the top of the tank.
Work =
(include units)
(b) The tank is full of water and the water must be pumped to a height 6 ft above the top of the tank.
Work =
(include units)
(c) The depth of water in the tank is 8 ft and the water must be pumped over the top of the tank.
Work =
(include units)
Explanation / Answer
Consider a thin "disk" of water of thickness dx, which is a distance "x" below the place where it needs to be
pumped up to. The work required to raise the disk by the required distance "x" is:
Work = (disk's weight)(x)
= (disk's volume)(density of water)(x)
= (r²dx)()(x) (note: "" stands for density of water)
= r²xdx = 25*62.4 xdx = 4900.9 x dx
To get the total work for ALL the disks of water, integrate that from the x=(top of the water) to x=(bottom of water).
Need more hints?
For (a), integrate from x=0 to x=11ft (because the "top disk" needs to be raised 0 ft., and the "bottom disk" needs
to be raised 11 ft.)
W = integral[x=0 to 11 ] 4900.9 x dx = 4900.9 x^2/2 [x=0 to 11 ] = 296.504 k J
For (b), integrate from x = 6ft to x = 17ft. (because "top disk" needs to be raised 6ft and "bottom disk" needs to be
raised 17ft).
W = integral[x=6 to 17 ] 4900.9 x dx = 4900.9 x^2/2 [x=6 to 17 ] = 619.964 k J
For (c), integrate from x = 8ft to x = 14ft. (because...etc.)
W = integral[x=8 to 14 ] 4900.9 x dx = 4900.9 x^2/2 [x=8 to 14 ] = 323.459 k J
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