Answer:
18816π ≈ 59112 Joules
Step-by-step explanation:
Draw a picture (like the one attached). The cone has a radius R and height H. The spout has height a. The tank is filled with water to height h.
Cut a thin slice from the volume of the water. This slice is a cylindrical disc with a radius r, a thickness dy, and a height y.
Using similar triangles, we can say:
r / y = R / H
The work required to lift this slice up to the spout is:
dW = dm g (H + a − y)
where dm is the mass of the slice and g is the acceleration due to gravity.
Mass is density times volume, so:
dW = dV ρ g (H + a − y)
Substituting the volume of the cylindrical disc:
dW = dy π r² ρ g (H + a − y)
From our similar triangles equation, we know that r = R/H y, so:
dW = dy π (R/H y)² ρ g (H + a − y)
Rearranging:
dW = π (R/H)² ρ g y² (H + a − y) dy
dW = π (R/H)² ρ g ((H + a)y² − y³) dy
The work to lift all the slices between y=0 and y=h is:
W = ∫ dW
W = π (R/H)² ρ g ∫₀ʰ ((H + a)y² − y³) dy
Integrating:
W = π (R/H)² ρ g (⅓(H + a)y³ − ¼y⁴) |₀ʰ
W = π (R/H)² ρ g (⅓(H + a)h³ − ¼h⁴)
Given:
R = 2 m
H = 5 m
a = 1 m
h = 2 m
g = 9.8 m/s²
ρ = 1000 kg/m³
W = π (2/5)² (1000) (9.8) (⅓(5 + 1)(2)³ − ¼(2)⁴)
W = 18816π
W ≈ 59112 Joules
It takes approximately 59.1 kJ of work.