Problem 28 - Calculus Single Variable

Question

A tank in the shape of a box has square base of side \(5 \mathrm{~m}\). It is \(4 \mathrm{~m}\) deep and is being pumped dry. The pump floats on the surface of the water and pumps the water to the top of the tank, at which point the water runs off. How much work is done in emptying the tank?

Step-by-Step Solution

Answer

Work done is 1,960,000 joules.

Step 1: Determine the volume of water

The volume of water in the tank is the product of the base area and the depth. Since the base is a square with side length 5 meters, the area is \(5 \times 5 = 25 \mathrm{~m}^2\). The depth is given as 4 meters. Thus, the volume of water is \(25 \times 4 = 100 \mathrm{~m}^3\).

Step 2: Understand how work is calculated

Work is calculated by integrating the force over a distance. The force on each slice of water is the weight of that slice, which depends on its volume and the density of water. The distance that slice needs to move to reach the top of the tank is determined by its depth below the top.

Step 3: Define work for a slice

Consider a slice of water at depth \(y\) meters from the top. The thickness of each slice is \(dy\). The volume of the slice is \(25 \ dy \ m^3\) and its weight is \(25 \ dy \ m^3 \times 1000 \ kg/m^3 \times 9.8 \ m/s^2\). Therefore, the force on the slice is \(245000 \ dy\). The slice needs to be moved up \(y\) meters to reach the top.

Step 4: Set up the integral for total work

The total work \(W\) is the integral of the force (weight of the slice) times the distance from the top for each slice from 0 to 4 meters. Use the integral \( W = \int_0^4 245000y \, dy \) to calculate the work.

Step 5: Solve the integral

Taking the integral of the function \(245000y\) over \(y\) from 0 to 4 gives: \[W = \int_0^4 245000y \, dy = 245000 \left[ \frac{y^2}{2} \right]_0^4.\]Evaluate this to get \[W = 245000 \left( \frac{4^2}{2} - \frac{0^2}{2} \right) = 245000 \times 8 = 1,960,000 \ J.\]

Step 6: Interpret the result

The calculation shows that the total work done in pumping the water out of the tank is \(1,960,000\) joules.

Key Concepts

Integral Calculus

Integral calculus is a branch of mathematics that deals with the accumulation of quantities. It is essentially the reverse process of differentiation. In the context of work calculations, integration helps us find the total effect, like total work done, by adding up small contributions over a continuous range.
In this exercise, integration is used to calculate the work done in pumping water out from a tank. We calculated the work by integrating the force needed to move small slices of water a certain distance. By setting up an integral that represents the total work done, we can aggregate these tiny bits of work into a coherent whole.
Integral calculus takes small, manageable pieces - slices of water in this case - and sums them up over an interval. This makes it ideal for solving real-world problems that involve accumulation, like work, volume, or area.

Work Calculation

Work in physics is defined as force applied over a distance. The formula is given by:

• Work = Force imes Distance

In settings involving variable force or distance, like the problem with a tank, integration becomes the tool we use to calculate work.
This exercise involves removing water from a tank through work calculations. We applied the principle of work by integrating the force exerted on water slices, multiplied by the distance each slice traveled to the tank's top.
When presented as a mathematical expression, this boils down to integrating the function that describes the force relative to a variable representing distance. This enables the computation of total work required for the task.

Force and Distance

Understanding the relationship between force and distance is key to work-related problems. Force is the push or pull exerted on an object, characterized by its magnitude and direction. Meanwhile, distance is the measure of how far the object moves under the influence of the force.
In the tank problem, the force needed to lift each slice of water depends on the slice's weight, which is a product of its volume and water density. Each slice also has to travel the distance to the tank's top, which varies depending on its starting position.
The force exerted upward and the distance each slice was moved are both variable quantities. This necessitates the use of calculus, specifically integration, to find the sum of work done over the changing distance for each incremental slice of water.

Volume Calculation

Volume calculation is another crucial part of this exercise, mostly involved at the problem's initial stage. We needed to calculate the volume of water in the tank to determine the force for each slice.
Because the tank is box-shaped with a square base, the volume calculation is straightforward: multiply the area of the base by the depth.
Given a base side of 5 meters and a depth of 4 meters, the volume is calculated as:

• Volume = Base Area imes Depth = 5 imes 5 imes 4

This volume calculation gives us a starting point to figure out the total force on each water slice, as each part contributes proportionally to the overall weight, influencing the work needed to remove the water from the tank.