A tank has a rectangular base 3 m × 4 m, the vertical sides are 6 m high, and the
depth of the water is 2 m. How much work is required to pump all of the water out over the
side? The density of water is 1000 kg/m3
and gravitational acceleration is 9.8 m/s2
.
Find the mass of the water
3*4*2 * 1000 kg = 24,000 kg
Find the height of the center of mass of the water
1 meter
lift it 6 - 1 = 5 meters
work in = potential gained = m g h = 24,000 * 9.8 * 5 Joules
Now if you think to use a siphon, it takes zero :)
To calculate the work required to pump all the water out of the tank, we need to find the weight of the water first.
The weight of an object can be calculated using the formula: weight = mass × gravitational acceleration.
1. Calculate the volume of the water:
The volume of the water is equal to the base area multiplied by the depth of the water. In this case, the base area is 3m × 4m = 12m^2, and the depth of the water is 2m. So the volume of the water is 12m^2 × 2m = 24m^3.
2. Calculate the mass of the water:
The mass of the water can be calculated using the formula: mass = density × volume. The density of water is 1000 kg/m^3, and the volume of the water is 24m^3. So the mass of the water is 1000 kg/m^3 × 24m^3 = 24000 kg.
3. Calculate the weight of the water:
The weight of the water can be calculated using the formula: weight = mass × gravitational acceleration. The gravitational acceleration is given as 9.8 m/s^2. So the weight of the water is 24000 kg × 9.8 m/s^2 = 235200 N (Newtons).
4. Calculate the work required:
The work formula is: work = force × distance.
In this case, the force required is equal to the weight of the water, which is 235200 N. The distance is the height of the tank, which is 6m.
So the work required to pump all the water out over the side is 235200 N × 6m = 1,411,200 joules (J).