Emptying a partially filled swimming pool If the water in the swimming pool in Exercise 35 is 2 m deep, then how much work is required to pump all the water to a level 3 m above the bottom of the pool?

Verified step by step guidance

1. Identify the shape and dimensions of the swimming pool from Exercise 35, as these are necessary to set up the integral for work. Typically, the pool's cross-sectional area as a function of depth is needed.
2. Express the volume of a thin horizontal slice of water at depth \(y\) (measured from the bottom) as \(dV = A(y) \, dy\), where \(A(y)\) is the cross-sectional area at depth \(y\) and \(dy\) is the thickness of the slice.
3. Calculate the weight of this thin slice of water using \(dW_{weight} = \rho g \, dV\), where \(\rho\) is the density of water and \(g\) is the acceleration due to gravity.
4. Determine the distance the water slice must be lifted to reach 3 m above the bottom, which is \$3 - y\( meters, and express the work done to lift this slice as \)dW = dW_{weight} \times (3 - y)$.
5. Set up the integral for the total work by integrating \(dW\) from the bottom of the water level (\(y=0\)) to the current water depth (\(y=2\)), i.e., \(W = \int_0^2 \rho g A(y) (3 - y) \, dy\), and prepare to evaluate this integral.

Key Concepts

Work Done by a Variable Force

Work is calculated as the integral of force over distance. When pumping water from different depths, the force varies with the volume and height of water being moved, requiring integration to sum the work done over all layers.

Density and Weight of Water

The weight of water depends on its density (approximately 1000 kg/m³) and gravitational acceleration (9.8 m/s²). This weight determines the force needed to lift each volume element of water to the target height.

Setting up the Integral for Pumping Work

To find total work, divide the water into thin horizontal slices, calculate the work to lift each slice to the target height, and integrate over the depth. The distance each slice is lifted varies with its depth.