Question

A \(5 \mathrm{~m}\) tall cylindrical tank with radius of \(2 \mathrm{~m}\) is filled with 3 m of gasoline, with a mass density of \(737.22 \mathrm{~kg} / \mathrm{m}^{3}\). Compute the total work performed in pumping all the gasoline to the top of the tank.

Short answer

The total work performed is approximately 91484.81 joules.

Step-by-step solution

Understand the Problem

We have a cylinder partially filled with gasoline. The height of the gasoline is 3 m, and we need to calculate the work done to pump all of this gasoline to the top of the 5 m tall tank. Remember, work is calculated as force times distance and we must consider the varying distance each 'slice' of gasoline has to be lifted.

Establish the Formula for Work

The work done to move a small slice of gasoline to the top of the tank involves calculating the force needed (weight of the slice) times the distance to the top. The weight of each slice is the volume of the slice times the density of gasoline times gravity. The force can be expressed as \( F = \delta \cdot V \cdot g \), where \( V = \pi r^2 \Delta h \).

Set Up the Integral

To find the total work, integrate over the height of the gasoline from 0 to 3 m. Each slice at height \( h \) in the tank has to be moved \( 5 - h \) meters (since the tank is 5 m tall). Integrate: \[ W = \int_{0}^{3} \pi (2)^2 \cdot (737.22 \times 9.81) \cdot (5 - h) \, dh \].

Simplify the Constants

Before integrating, simplify the constants in the integral. The expression becomes: \( W = \int_{0}^{3} 8703.22(5-h) \, dh \), where 8703.22 is \( \pi (2)^2 \times 737.22 \times 9.81 \).

Calculate the Integral

Solve \[ \int_{0}^{3} 8703.22(5-h) \, dh \]. This simplifies to \[ 8703.22 \cdot (5 \cdot 3 - \frac{3^2}{2}) = 8703.22 \cdot (15 - 4.5) = 8703.22 \cdot 10.5 \].

Compute the Total Work

Calculate the result of the integral: \[ W = 8703.22 \times 10.5 = 91484.81 \text{ joules} \].

Key concepts

Work and Energy

In physics, the terms "work" and "energy" are closely related. Work is essentially the energy transferred by a force over a distance. When we calculate the work done in moving gasoline, we are looking at the energy required to lift it against gravity.
The formula to determine work is given by \( W = F \times d \), where \( F \) is the force applied, and \( d \) is the distance over which the force is applied. In this context, the force is the weight of the gasoline, which depends on its mass and the acceleration due to gravity.
Understanding this concept helps us realize that for each slice of gasoline, as it is lifted higher, a varying amount of energy is required, leading us to use calculus to compute the total work.

Integrals

Integrals are fundamental tools in calculus that help determine accumulated quantities, like area under a curve or total work done. When dealing with tasks like pumping gasoline, integration allows us to calculate the total work needed by considering each infinitesimally small slice of gasoline separately.
We express the work for each slice as an integral because each slice is lifted a different distance. In the problem, we solve the integral \[ W = \int_{0}^{3} 8703.22(5-h) \, dh \],
where \( h \) ranges from 0 to 3 as the gasoline is lifted. Solving this helps us accumulate the work done for each slice over the entire height of the gasoline.

Mass Density

Mass density is a measure of how much mass is contained in a given volume. It's a crucial factor when calculating work, as it determines the weight of the liquid being lifted.
In this problem, the gasoline's mass density is \( 737.22 \, \text{kg/m}^3 \). This means each cubic meter of gasoline weighs 737.22 kilograms. The density affects the force, and hence the work needed, since the weight of a slice of gasoline is given by \( \text{mass \( \delta \)} \times \text{gravity \( g \)} \).
Understanding this concept allows us to accurately compute the force component of the work formula.

Volume of a Cylinder

Calculating the volume of a cylinder is an essential skill in many physics and engineering problems. For any cylindrical object, the volume \( V \) is given by \( V = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height.
In our exercise, the tank is cylindrical with a radius of 2 meters. Each unit slice of height \( \Delta h \) therefore has a volume expressed as \( V = \pi (2)^2 \Delta h \).
The understanding of the cylinder’s volume formula allows us to determine the exact volume of gasoline each slice represents, aiding our computation of its weight, and consequently the force, needed in the work calculation.