To find the work needed to pump half the water out of the box, we will follow these steps:

1. Volume of the box:
   The box is 2 m long, 1 m wide, and 1 m deep, so its total volume is: $V = \text{length} \times \text{width} \times \text{depth} = 2 \times 1 \times 1 = 2 \text{ m}^3$ Since we are pumping out half the water, the volume to pump is: $V_{\text{half}} = \frac{2}{2} = 1 \text{ m}^3$

2. Mass of water to pump out:
   The density of water is $1000 \, \text{kg/m}^3$. So the mass of the water to pump is: $m = \text{density} \times V_{\text{half}} = 1000 \times 1 = 1000 \, \text{kg}$

3. Work calculation:
   Work is defined as: $W = \int F \, dx$ where $F$ is the force and $dx$ is the distance moved. The force here is due to the weight of the water, given by $F = m \cdot g$, where $g = 9.8 \, \text{m/s}^2$ is the gravitational constant.

   The work required to pump out the water depends on the distance the water needs to be lifted. At any height $x$, the volume of water at that height needs to be lifted a distance of $1 - x$ meters (since the box is 1 meter deep and we're pumping out from the top).

   The elemental volume at a height $x$ is: $dV = \text{length} \times \text{width} \times dx = 2 \times 1 \times dx = 2 \, dx$ The elemental mass of water at height $x$ is: $dm = \text{density} \times dV = 1000 \times 2 \, dx = 2000 \, dx$ The elemental force is the weight of this water: $dF = dm \cdot g = 2000 \cdot 9.8 \, dx = 19600 \, dx$ The distance this elemental volume needs to be lifted is $(1 - x)$.

   So the elemental work is: $dW = dF \cdot (1 - x) = 19600 (1 - x) \, dx$

4. Total work:
   To find the total work, we integrate from $x = 0$ to $x = 0.5$ (since we are pumping out half the water, which occupies the lower half of the box): $W = \int_0^{0.5} 19600 (1 - x) \, dx$ Let's calculate this integral.The total work required to pump half the water out of the box is $7350 \, \text{J}$ (joules).

Would you like further details on any step or have any questions?

Here are some related questions for practice:

1. How would the work change if the box were twice as deep?
2. What would be the work required to pump out all the water instead of half?
3. How does the density of water affect the amount of work required?
4. What if the box had a slanted side instead of being rectangular—how would that impact the calculation?
5. How much work would be required if the gravitational constant were different, say on the Moon?

Tip: When calculating work, always keep in mind that the force varies with the position, especially in fluid problems, because of the changing distance over which the force acts.