V = 3.14*3^2*10 = 282.60 ft^3 = 3.05^3*282.60 = 8018 dm^3 = 8018 kg
average height = 3.05/2 m
W = mgh = 8018/1000*3.05/2*9.8 = 120 kjoule
You need not to resolve to calculus to solve the problem.
Sorry, I am not familiar with imperial units.
You find the weight of the water, and consider lifting a cylinder of water a distance of 5 ft.
Work done = weight of the water * 5 ft
The weight of each horizontal slab is
(62.4 lb/ft^3)(pi*9 ft^2) dz,
and the distance it must be raised is (10 ft - z).
Let's dispense with the units for a moment:
Integral from z = 0 to z = 10 of
561.6 (10 - z) dz
= (561.6)(10z - (1/2)z^2) to be evaluated at z=10
= (561.6) (100 - 50) = 28080 ft-lb
Kasab's answer is correct.
Tank volume = πr2×h = π(3ft)2×10ft = 282.74 ft3
Mass of water = 282.74ft3 × 62.4lbm/ft3 = 17,643.18 lbm
The average distance lifted = 10ft/2 = 5ft
Work = force × distance = 17,643.18lbf×5ft = 88,215.92 ftlbf
(Work/time = power, but time is not involved in Work).
Check this former answer for the equations and how solution was found:
http://answers.yahoo.com/question/index?...
Find the work done in pumping the water out of top of cylindrical tank of radius 3 ft and height 10 ft. The tank is full and the density of water is 62.4lb/ft^3
