To solve this problem, we need to calculate the work required to pump half of the water out of the aquarium. Here's the step-by-step approach:

Given Data:

• Dimensions of the aquarium: 2 m long, 1 m wide, and 1 m deep.
• Density of water: $1000 \, \text{kg/m}^3$.
• Gravitational acceleration: $g = 9.8 \, \text{m/s}^2$ (assumed standard value).

Step 1: Volume of water in the aquarium

The volume of the aquarium when full is: $\text{Volume} = \text{Length} \times \text{Width} \times \text{Depth} = 2 \, \text{m} \times 1 \, \text{m} \times 1 \, \text{m} = 2 \, \text{m}^3$ So, when full, the aquarium contains 2 cubic meters of water. Half of this is $1 \, \text{m}^3$.

Step 2: Mass of water

The mass of the water being pumped out is: $\text{Mass} = \text{Density} \times \text{Volume} = 1000 \, \text{kg/m}^3 \times 1 \, \text{m}^3 = 1000 \, \text{kg}$

Step 3: Work to pump water

To find the total work needed, we need to consider that the force required to lift each small volume of water varies with height. The work to move a small slice of water at depth $y$ from the top is: $\text{Work} = \text{Force} \times \text{Distance}$ The force is given by: $\text{Force} = \text{Weight of slice} = \text{Volume of slice} \times \text{Density of water} \times g$ For a small slice $dy$ at depth $y$, the volume is: $dV = \text{Length} \times \text{Width} \times dy = 2 \times 1 \times dy = 2 \, dy$ So, the force for that slice is: $dF = 1000 \times g \times 2 \, dy = 2000g \, dy$

Step 4: Integrating over half of the tank

Since we are pumping out only the upper half of the water (from depth 0 to 0.5 m), the total work $W$ is the integral of force times distance, with distance being $y$ (the distance from each slice to the top): $W = \int_0^{0.5} 2000g \, y \, dy$ Let's compute this integral.The total work required to pump half of the water out of the aquarium is 2450 joules.

Would you like further clarification or more details? Here are some related questions to expand on this topic:

1. How would the work change if the aquarium were deeper or larger?
2. How is the work affected if we pump all the water instead of half?
3. What if the water density or gravity were different (e.g., on another planet)?
4. Could this method be applied to other fluids, and how would their properties influence the result?
5. How does the relationship between force and distance contribute to the varying work as depth increases?

Tip: For similar problems involving work and fluids, always break down the problem by considering the small slices of volume and integrating the force over the distance.