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 ρ = 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 and the principle of work-energy equivalence.
(a) Work needed to empty a full tank:
Step 1: Calculate 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)πr^2h, where r is the radius and h is the height of the cone.
Given: height (h) = 3m, diameter = 1m (which means the radius is 0.5m)
Substituting these values into the formula, we get V = (1/3) * π * (0.5m)^2 * 3m = π * (0.25m^2) * 3m = 0.75π m^3 (volume of the tank)
Step 2: Calculate the potential energy of the water in the tank.
Potential energy (PE) = mgh, where m is the mass of the water, g is the acceleration due to gravity, and h is the height the water is lifted.
Given: density (ρ) of water = 1000kg/m^3, height (h) = 3m, gravitational acceleration (g) = 9.8m/s^2
To find the mass (m) of the water, we can multiply the density by the volume: m = ρV = 1000kg/m^3 * 0.75π m^3 = 750π kg
Substituting the values into the potential energy formula, we get PE = (750π kg) * (9.8m/s^2) * (3m) = 22050π J (Joules)
So, the work needed to empty the full tank is 22050π Joules.
(b) Work to empty half the tank:
Since we are emptying half the tank, the height of the water column is halved.
Step 1: Calculate the new volume of the water.
Now, the height of the water (h) becomes 3m/2 = 1.5m.
Using the formula for the volume of a cone, V = (1/3)πr^2h, we replace h with 1.5m to find the new volume: V = (1/3) * π * (0.5m)^2 * 1.5m = 0.125π m^3
Step 2: Calculate the potential energy of the water in the tank.
Using the formula PE = mgh, the mass (m) is found by multiplying the density (ρ) by the new volume: m = ρV = 1000kg/m^3 * 0.125π m^3 = 125π kg
Substituting the values into the potential energy formula, we get PE = (125π kg) * (9.8m/s^2) * (2m) = 2450π J (Joules)
So, the work needed to empty half the tank is 2450π Joules.