theres a box 2 m long, 1 m wide, and 1 m deep full fo water. Find work needed to pump half the water out of the box, density of water is 1000 kg/m^3

To calculate the work required to pump half of the water out of the box, we need to use the concept of gravitational potential energy. Work in this context is the energy required to lift the water to the top of the box.

### Given Data:
- Length of the box: \( L = 2 \, \text{m} \)
- Width of the box: \( W = 1 \, \text{m} \)
- Depth of the box: \( D = 1 \, \text{m} \)
- Density of water: \( \rho = 1000 \, \text{kg/m}^3 \)
- Gravitational acceleration: \( g = 9.8 \, \text{m/s}^2 \)

### 1. Volume and Mass of Water:
The total volume of water in the box is:
\[
V_{\text{total}} = L \times W \times D = 2 \, \text{m} \times 1 \, \text{m} \times 1 \, \text{m} = 2 \, \text{m}^3
\]
The mass of water in the full box is:
\[
m_{\text{total}} = \rho \times V_{\text{total}} = 1000 \, \text{kg/m}^3 \times 2 \, \text{m}^3 = 2000 \, \text{kg}
\]
Half of this mass would be:
\[
m_{\text{half}} = \frac{2000}{2} = 1000 \, \text{kg}
\]

### 2. Work Calculation:
We will calculate the work needed to pump out half the water. Since the water at different depths requires different amounts of work, we'll integrate to find the total work required.

Let \( x \) be the depth of water below the top of the box. The force needed to lift a thin layer of water at depth \( x \) to the top is given by the mass of that layer multiplied by gravity:
\[
dF = \rho g A dx
\]
where \( A \) is the area of the box's base, \( A = 2 \, \text{m}^2 \), and \( dx \) is a small thickness of the layer. The work to lift this small layer to the top is:
\[
dW = \text{Force} \times \text{Distance} = dF \times x = \rho g A x \, dx
\]

To find the total work to pump half the water, we integrate from \( 0 \, \text{m} \) (bottom of the box) to \( 0.5 \, \text{m} \) (half the depth):
\[
W = \int_0^{0.5} \rho g A x \, dx
\]

Substituting the values \( \rho = 1000 \, \text{kg/m}^3 \), \( g = 9.8 \, \text{m/s}^2 \), and \( A = 2 \, \text{m}^2 \):
\[
W = \int_0^{0.5} 1000 \times 9.8 \times 2 \times x \, dx = 19600 \int_0^{0.5} x \, dx
\]

Evaluating the integral:
\[
W = 19600 \left[ \frac{x^2}{2} \right]_0^{0.5} = 19600 \times \frac{(0.5)^2}{2} = 19600 \times \frac{0.25}{2} = 19600 \times 0.125 = 2450 \, \text{J}
\]

Thus, the work needed to pump half the water out of the box is **2450 joules**.

---

Would you like further details or have any questions? Here are some related topics you could explore:

1. How to compute work for non-uniform density fluids.
2. Impact of varying gravitational acceleration on work.
3. Work required to pump all the water from the box.
4. Energy comparisons for different volumes of water.
5. Concept of potential energy in other contexts.

**Tip:** Always consider the effect of depth when calculating work needed to lift a fluid—lower layers require more work to lift.

theres a box 2 m long, 1 m wide, and 1 m deep full of water. Find work needed to pump half the water out of the box, density of water is 1000 kg/m^3

Learn how to calculate the work needed to pump half the water out of a box. This problem involves gravitational potential energy, integration, and fluid physics. Suitable for students in Grades 11-12.

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