an aquarium 2 m long, 1 m wide, and 1 m deep is full full of water. Find work needed to pump half of the water out of the aquarium.

To solve the problem of finding the work needed to pump half of the water out of the aquarium, we'll use the concept of work done in lifting a volume of water to a certain height.

### Step-by-Step Solution:

1. **Determine the Volume of Water to be Pumped Out:**

   The dimensions of the aquarium are:
   - Length (\(L\)) = 2 m
   - Width (\(W\)) = 1 m
   - Height (\(H\)) = 1 m

   The total volume of the aquarium is:
   \[
   V = L \times W \times H = 2 \times 1 \times 1 = 2 \, \text{m}^3
   \]

   Since we are pumping out half of the water, the volume to be pumped out is:
   \[
   V_{\text{pumped}} = \frac{V}{2} = \frac{2}{2} = 1 \, \text{m}^3
   \]

2. **Set Up the Coordinate System:**

   Let's set up a coordinate system where the bottom of the aquarium is at \(y = 0\) and the top is at \(y = 1\) meter. We are interested in pumping the top half of the water, which is between \(y = 0.5\) m to \(y = 1\) m.

3. **Calculate the Work Done in Pumping Out Water:**

   The work done to pump out a small slice of water at a height \(y\) is given by:
   \[
   dW = \text{(Volume of slice)} \times \text{(Density of water)} \times \text{(Gravitational acceleration)} \times \text{(Height)}
   \]

   The volume of a thin slice of water at height \(y\) with thickness \(dy\) is:
   \[
   dV = L \times W \times dy = 2 \times 1 \times dy = 2 \, dy
   \]

   The density of water, \(\rho\), is \(1000 \, \text{kg/m}^3\), and the gravitational acceleration \(g\) is \(9.8 \, \text{m/s}^2\).

   The distance each slice of water must be lifted to the top of the aquarium (to \(y = 1\) m) is \(1 - y\).

   The work required to lift this small slice to the top is:
   \[
   dW = \rho g (2 \, dy) (1 - y)
   \]

   Substituting the values for \(\rho\) and \(g\):
   \[
   dW = 1000 \times 9.8 \times 2 \, (1 - y) \, dy = 19600 (1 - y) \, dy
   \]

4. **Integrate to Find Total Work:**

   To find the total work, integrate \(dW\) from \(y = 0.5\) to \(y = 1\) (the top half of the water):

   \[
   W = \int_{0.5}^{1} 19600 (1 - y) \, dy
   \]

   Let's calculate the integral:

   \[
   W = 19600 \int_{0.5}^{1} (1 - y) \, dy
   \]

   \[
   W = 19600 \left[ y - \frac{y^2}{2} \right]_{0.5}^{1}
   \]

   Now, evaluate the definite integral:

   \[
   W = 19600 \left( \left(1 - \frac{1^2}{2}\right) - \left(0.5 - \frac{0.5^2}{2}\right) \right)
   \]

   \[
   W = 19600 \left( \left(1 - \frac{1}{2}\right) - \left(0.5 - \frac{0.25}{2}\right) \right)
   \]

   \[
   W = 19600 \left( \left(0.5\right) - \left(0.5 - 0.125\right) \right)
   \]

   \[
   W = 19600 \left(0.5 - 0.375 \right)
   \]

   \[
   W = 19600 \times 0.125
   \]

   \[
   W = 2450 \, \text{J}
   \]

### Conclusion:

The work needed to pump half of the water out of the aquarium is **2450 joules**.

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Would you like a more detailed explanation, or do you have any questions about this solution?

### Related Questions:

1. How would the work required change if the aquarium was twice as deep?
2. What is the work required to pump all the water out of the aquarium instead?
3. How would the calculation differ if the density of water was different?
4. Can we calculate the power needed if the water is pumped out over a specific period of time?
5. What if the aquarium was not a rectangular prism but a cylinder?

### Tip:

Remember that the work needed to move a fluid depends not only on the volume but also on the height to which the fluid is lifted.

Learn how to calculate the work needed to pump half of the water out of a 2m x 1m x 1m aquarium. Detailed step-by-step solution using principles of work and energy in physics, including integral calculus. Suitable for advanced high school students and physics enthusiasts.

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