Atank has a shape of a cone with a radius at the top of 2 m and a height of 5 m. the tank also has a 1 m spout at the top of the tank. the tank is filled with water up to a height of 2 m. find the work needed to pump all the water out the top of the spout. (use 9.8 m/s2 for g and the fact that the density of water is 1000 kg/m3.)

The work needed to pump all the water out the top of the spout of a tank which is shape in a cone is 130.800 kJ.

Work done is the force applied on a body to move it over a distance. Work done for the pump can be given as,

Here (F) is the magnitude of force applied on the pump and (d) is the distance covered by the water.

From the second law of motion, the force is the product of mass times acceleration. Thus the above equation can be written as,

Here, (g) is the acceleration force due to gravity.

The tank has a shape of a cone with a radius at the top of 2 m and a height of 5 m. The density of water is 1000 kg/m³.

As the mass of the water is the density of water times volume of the tank. Thus the mass can be calculated as,

The tank is filled with water up to a height of 2 m. Thus, put the values in the above formula as,

Thus, the work needed to pump all the water out the top of the spout is 130.800 kJ.

Learn more about the work done here;

link

18816π ≈ 59112 Joules

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.

he kicked it 10 feet straight up into the air? i think this works.

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