An inverted conical tank is 3m tall and 1m in diameter at its widest point. The water

is being pumped out of a spout 2m above the top of the tank. Recall that the density
of water is � = 1000kg/m^3

(a) Find the work needed to empty the tank if it is full. Include units.
(b) Find the work to empty half the tank (assuming it is full to begin with).

To find the work needed to empty the tank, we can use the concept of potential energy. The work done is equal to the change in potential energy of the water.

Let's start by calculating the volume of the tank. The tank is in the shape of an inverted cone, so we can use the formula for the volume of a cone:

V = (1/3) * π * r^2 * h

where V is the volume, π is a mathematical constant approximately equal to 3.14159, r is the radius of the widest point of the cone (which is half the diameter), and h is the height of the cone.

Given that the diameter of the tank is 1m, the radius, r, is (1/2)m = 0.5m. The height of the tank, h, is 3m. Plugging in these values:

V = (1/3) * π * (0.5m)^2 * 3m

V = (1/3) * π * 0.25m^2 * 3m

V = (1/3) * π * 0.75m^3

Now, we can calculate the mass of the water in the tank, m, using the density value:

m = ρ * V

Given that the density of water is 1000kg/m^3:

m = 1000kg/m^3 * 0.75m^3

m = 750kg

(a) To find the work needed to empty the full tank, we need to consider the change in potential energy. The change in potential energy is equal to the mass of the water times the gravitational field strength times the change in height:

ΔPE = m * g * h

where ΔPE is the change in potential energy, m is the mass of the water, g is the acceleration due to gravity (approximately 9.8m/s^2), and h is the height from which the water falls.

Given that the water is being pumped out of the spout 2m above the top of the tank, the height from which the water falls is the total height of the tank plus 2m:

h = 3m + 2m

h = 5m

Plugging in the values:

ΔPE = 750kg * 9.8m/s^2 * 5m

ΔPE = 36750 Joules

So, the work needed to empty the tank if it is full is 36750 Joules.

(b) To find the work needed to empty half the tank, we need to consider the change in potential energy for half the volume of water. We can use the same formula, but with half the mass and half the height:

ΔPE = (m/2) * g * (h/2)

Plugging in the values:

ΔPE = (750kg/2) * 9.8m/s^2 * (5m/2)

ΔPE = 45937.5 Joules

So, the work to empty half the tank (assuming it is full to begin with) is 45937.5 Joules.