Answer:
Part (i) work required to pump the contents to the rim is 281,913.733 lb.ft
Part (ii) work required to pump the liquid to a level of 5ft above the cone's rim is 426,484.878 lb.ft
Explanation:
The center of mass of a uniform solid right circular cone of height h lies on the axis of symmetry at a distance of h/4 from the base and 3h/4 from the top.
Center mass of the liquid Z = (24-19)ft + 19/4 = 5ft + 4.75ft = 9.75 ft
Mass of liquid in the cone = volume × density (ρ) = ¹/₃.π.r².h.ρ
where;
r is the radius of the liquid surface = [6*(19/24)]ft = 4.75ft
ρ is the density of liquid = 64.4 lb/ft³
h is the height of the liquid = 19 ft
Mass of liquid in the cone = ¹/₃ × π × (4.75)² × 19 × 64.4 = 28,914.229 lbs
Part (i) work required to pump the contents to the rim
Work required = 28,914.229 lbs × 9.75 ft = 281,913.733 lb.ft
Part (ii) work required to pump the liquid to a level of 5 ft above the cone's rim
Extra work required = 28,914.229 lb × 5ft = 144571.145 lb.ft
Total work required = (281,913.733 + 144571.145) lb.ft
= 426,484.878 lb.ft
is the force, 
is the mass, and 
is the acceleration of the object.