A circular swimming pool has a diameter of 20 meters, the sides are 3 meters high, and the depth of the water is 2.5 meters. The acceleration due to gravity is 9.8 m/s2 and the density of water is 1000 kg/m3.
How much work is required to:
(a) pump all of the water over the side?
(b) pump all of the water out of an outlet 2 m over the side?
Well, it seems like we have to do some heavy lifting!
(a) To pump all of the water over the side, we need to figure out the volume of the pool first. Since it's a circular pool with a diameter of 20 meters, we can use the formula for the volume of a cylinder: V = πr^2h, where r is the radius (half of the diameter) and h is the height. So, the volume of the pool is V = π(10)^2(3) = 300π cubic meters.
Now, to pump all of this water over the side, we need to determine the weight of the water. The formula for weight is W = mg, where m is the mass and g is the acceleration due to gravity. Since the density of water is 1000 kg/m^3 and the pool's volume is 300π cubic meters, the mass of the water is m = density x volume = 1000 x 300π = 300,000π kg.
Therefore, the work required to pump all of the water over the side is W = mgd, where d is the depth of the water. So, W = (300,000π kg) x (9.8 m/s^2) x (2.5 m) = 7,350,000π joules.
(b) Now, let's say we're pumping the water out of an outlet 2 meters over the side. This means we need to work against gravity and the additional height.
To find the total height the water needs to be lifted, we add the depth of the water (2.5 m) and the height of the outlet (2 m): 2.5 m + 2 m = 4.5 meters
The work required to pump the water out of the outlet is W = mgd. Using the same mass (300,000π kg) and acceleration due to gravity (9.8 m/s^2), the work becomes W = (300,000π kg) x (9.8 m/s^2) x (4.5 m) = 13,230,000π joules.
So, the work required to pump all of the water out of an outlet 2 meters over the side is 13,230,000π joules. Just remember to stretch those pump muscles before starting!
To find the work required to pump all of the water over the side of the circular swimming pool, we need to calculate the volume of the water first.
(a) Volume of water:
The circular swimming pool is like a cylinder with a radius of 10 meters (diameter/2) and a height of 2.5 meters. The formula for the volume of a cylinder is V = πr^2h.
V = π(10^2)(2.5)
V = π(100)(2.5)
V = 785.3981634 m^3
The density of water is given as 1000 kg/m^3. So, the mass of the water is:
mass = density × volume
mass = 1000 × 785.3981634
mass = 785,398.1634 kg
Now, we can calculate the work required using the formula:
work = force × distance
The force required to lift the water is equal to the weight of the water, which is given by:
weight = mass × acceleration due to gravity
weight = 785,398.1634 × 9.8
weight = 7,689,897.959 kg·m/s^2 (or N)
To lift the water over the side, we need to pump the water up to the height of the sides, which is 3 meters. Therefore, the distance is 3 meters.
work = 7,689,897.959 × 3
work = 23,069,693.877 Joules
Therefore, the work required to pump all of the water over the side of the circular swimming pool is approximately 23,069,693.877 Joules.
(b) To pump all of the water out of an outlet 2 meters over the side, we can use a similar approach.
The distance to pump the water is now 2 meters.
work = 7,689,897.959 × 2
work = 15,379,795.918 Joules
Therefore, the work required to pump all of the water out of an outlet 2 meters over the side is approximately 15,379,795.918 Joules.
To calculate the work required to pump the water over the side of the swimming pool, we need to calculate the potential energy of the water.
(a) Pumping all of the water over the side:
To calculate the potential energy of the water, we need to determine the volume of the water first. The circular swimming pool can be considered as a cylinder. The volume (V) of a cylinder is given by the formula V = πr^2h, where r is the radius of the cylinder and h is the height.
Given that the diameter of the pool is 20 meters, the radius (r) is half the diameter, so r = 20/2 = 10 meters.
The height (h) of the cylinder is given as 3 meters.
Using the formula, the volume of water (V) is:
V = π(10^2)(3)
= 300π cubic meters (approx.)
The mass (m) of the water can be calculated using the density (ρ) and the volume (V). The formula is m = ρV, where ρ is the density of water.
Given that the density of water is 1000 kg/m^3, the mass (m) of the water is:
m = (1000)(300π)
= 300,000π kg (approx.)
The potential energy (PE) of an object can be calculated using the formula PE = mgh, where g is the acceleration due to gravity and h is the height.
Given that the height (h) of the water is 2.5 meters and the acceleration due to gravity (g) is 9.8 m/s^2, the potential energy (PE) of the water is:
PE = (300,000π)(9.8)(2.5)
= 7,350,000π joules (approx.)
Therefore, the work required to pump all of the water over the side is approximately 7,350,000π joules.
(b) Pumping all of the water out of an outlet 2 meters over the side:
To calculate the work required in this case, we need to find the difference in potential energy between the initial and final positions of the water.
Given that the water is pumped out of an outlet 2 meters over the side, the height difference (h) is 2 meters. The potential energy difference (ΔPE) can be calculated as ΔPE = mgΔh.
Using the same mass (m) of water calculated previously (300,000π kg), the potential energy difference (ΔPE) is:
ΔPE = (300,000π)(9.8)(2)
= 5,880,000π joules (approx.)
Therefore, the work required to pump all of the water out of an outlet 2 meters over the side is approximately 5,880,000π joules.
the volume of water is
10^2 * π * 2.5 = 785.4 m^3
That much water weighs
785.4 * 1000 * 9.8 = 7696920 N
The center of mass is 1.75m below the top of the pool, so the work required to lift it all is
7696920N * 1.75m = 13,470,000 J
But, if you want to use calculus, consider a thin sheet of water of thickness dx. Its volume is
100π dx m^3
So, its weight of water is
100π*1000*9.8 = 3078761 N
So, add up the work required to lift each sheet up to the top:
∫[0,2.5] 3078761(3-x) dx = 3,470,000