A circular swimming pool has a diameter of 14 m, the sides are 4 m high, and the depth of the water is 3.5 m. (The acceleration due to gravity is 9.8 and the density of water is 1000 .)
How much work (in Joules) 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
(a) Well, if we want to pump all of that water over the side, we'll need to calculate the volume of the pool first. The volume of a cylinder (which is the shape of the pool) is given by the formula V = πr^2h, where r is the radius and h is the height. So, let's calculate the volume:
radius (r) = diameter / 2 = 14 m / 2 = 7 m
height (h) = 4 m
V = π(7^2)(4) = 616π m^3
Now, the work done to pump all of the water out is given by the equation W = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. We can calculate the mass by multiplying the density of water by the volume:
density of water = 1000 kg/m^3
mass (m) = density x volume = 1000 kg/m^3 x (616π m^3) ≈ 616,000π kg
height (h) = 4 m
Work (W) = (616,000π kg)(9.8 m/s^2)(4 m) = 24,038,400π J
So, approximately 24,038,400π Joules of work are required to pump all of the water over the side.
(b) Now, if we want to pump all of the water out of an outlet 2 m above the side, we can use the same formula for work done. The height (h) in this case will be 2 m instead of 4 m:
height (h) = 2 m
Work (W) = (616,000π kg)(9.8 m/s^2)(2 m) = 12,019,200π J
So, approximately 12,019,200π Joules of work are required to pump all of the water out of an outlet 2 m above the side.
Now, that's a lot of work! I hope the swimming pool pump is up for the challenge, or else it will be a real "draining" experience!
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 in the pool and then use it to find the gravitational potential energy.
(a)
The volume of a cylinder is given by the formula V = πr^2h, where r is the radius and h is the height.
In this case, the radius is half of the diameter, so r = 7 m.
The height of the water is 3.5 m.
The volume of water in the pool is V = π(7^2)(3.5) = 171.5π m^3.
The weight of water can be found using the formula weight = density × volume × g, where g is the acceleration due to gravity.
In this case, the density of water is 1000 kg/m^3 and the acceleration due to gravity is 9.8 m/s^2.
The weight of the water is weight = 1000 × (171.5π) × 9.8 = 1,672,940π N.
The work done to pump the water over the side is given by the formula work = weight × height.
The work required is work = 1,672,940π × 4 = 6,691,760π J (Joules).
Therefore, the work required to pump all of the water over the side is approximately 6,691,760π Joules.
(b)
To find the work required to pump all of the water out of an outlet 2 m over the side, we need to calculate the volume of the water above the outlet and then use it to find the gravitational potential energy.
The volume of water above the outlet can be found using the formula V = πr^2h, where r is the radius and h is the height.
In this case, the radius is still 7 m, and the height is 3.5 - 2 = 1.5 m (since the outlet is 2 m over the side and the depth of the water is 3.5 m).
The volume of water above the outlet is V = π(7^2)(1.5) = 220.5π m^3.
The weight of the water above the outlet is weight = density × volume × g = 1000 × (220.5π) × 9.8 = 2,157,540π N.
The work done to pump the water out of the outlet is given by the formula work = weight × height.
The work required is work = 2,157,540π × 2 = 4,315,080π J (Joules).
Therefore, the work required to pump all of the water out of an outlet 2 m over the side is approximately 4,315,080π Joules.
To calculate the work required to pump the water out of the pool, we will need to determine the potential energy of the water when it is lifted out of the pool.
(a) To pump all of the water over the side:
First, let's calculate the volume of the water in the pool. Since the pool is circular, we can use the formula for the volume of a cylinder:
Volume = π * r^2 * h
Given that the diameter of the pool is 14 m, the radius (r) is half of that:
r = 14 m / 2 = 7 m
The height (h) of the water is 3.5 m.
Now, we can calculate the volume:
Volume = π * (7 m)^2 * 3.5 m = 539.5 m^3
Next, we can calculate the mass of the water using its density:
Mass = density * volume
Since the density of water is 1000 kg/m^3, we have:
Mass = 1000 kg/m^3 * 539.5 m^3 = 539,500 kg
To find the potential energy (PE) of lifting the water out of the pool, we need to multiply the mass by the acceleration due to gravity (g = 9.8 m/s^2) and the height lifted (h = 4 m):
PE = Mass * g * h
PE = 539,500 kg * 9.8 m/s^2 * 4 m = 21,072,400 Joules
Therefore, the work required to pump all of the water over the side is 21,072,400 Joules.
(b) To pump all of the water out of an outlet 2 m over the side:
Since the water is being pumped out of an outlet 2 m over the side, the height lifted (h) is 2 m less than in part (a), so h = 4 m - 2 m = 2 m.
Using the same mass of water calculated earlier (Mass = 539,500 kg), we can find the potential energy:
PE = Mass * g * h
PE = 539,500 kg * 9.8 m/s^2 * 2 m = 10,565,200 Joules
Therefore, the work required to pump all of the water out of an outlet 2 m over the side is 10,565,200 Joules.
If you are at all smart, none.
You siphon the water out.
However if you really want to lift it all to the top then the answer is the mass of water times height you lift it times gravity ( m g h )
Mass = m = pi (7)^2 (3.5)(1000) m^3
g = 9.8 m/s^2
h = 4 - (3.5)/2 = 2.25 m
so multiply and get Joules
for the second part use 2.25 + 2 = 4.25 for h