Problem 21 - Fundamentals of Physics

Question

Two identical cylindrical vessels with their bases at the same level each contain a liquid of density \(1.30 \times 10^{3}\) \(\mathrm{kg} / \mathrm{m}^{3}\). The area of each base is \(4.00 \mathrm{~cm}^{2}\), but in one vessel the liquid height is \(0.854\) \(\mathrm{m}\) and in the other it is \(1.560 \mathrm{~m}\). Find the work done by the gravitational force in equalizing the levels when the two vessels are connected.

Step-by-Step Solution

Answer

Work done is obtained by subtracting final potential energy from initial potential energy.

Step 1: Understanding the problem

We are asked to find the work done by the gravitational force in equalizing the levels of liquid in two connected vessels. One vessel has liquid height of 0.854 m and the other has 1.560 m. Both liquids have a density of \(1.30 \times 10^{3}\, \text{kg/m}^3\) and the base area of each vessel is \(4.00 \times 10^{-4} \text{ m}^2\).

Step 2: Calculate the initial potential energy

The potential energy in each vessel is given by \(PE = mgh\), where \(m\) is the mass, \(g\) is the acceleration due to gravity (\(9.81 \, \text{m/s}^2\)), and \(h\) is the height. For each vessel, \(m = \rho \cdot V = \rho \cdot A \cdot h\). Calculate the initial potential energy for both vessels.

Step 3: Calculate the average height after equalization

After the vessels are connected, the liquid will level out. Find the average height of the liquid by taking the mean of the two original heights: \(h_{\text{average}} = \frac{h_1 + h_2}{2}\).

Step 4: Calculate the final potential energy

After equalization, the height of the liquid in both vessels will be \( h_{\text{average}} \). Calculate the total final potential energy for both vessels using \( PE_{\text{final}} = 2 \cdot (\rho \cdot A \cdot h_{\text{average}} \cdot g \cdot h_{\text{average}}/2)\) since the center of mass is at half of the height \( h_{\text{average}} \).

Step 5: Find the work done by gravitational force

Work done is the change in potential energy. Subtract the final potential energy from the initial potential energy: \( W = PE_{\text{initial}} - PE_{\text{final}} \).

Step 6: Calculate numerical values

Substitute given values into the equations and solve for the numerical result. Convert units where necessary, e.g. area from \( \text{cm}^2 \) to \( \text{m}^2 \): \(4.00 \, \text{cm}^2 = 4.00 \times 10^{-4} \, \text{m}^2\). Calculate initial energy for each vessel: \(PE_1\) and \(PE_2\), use \(h_1 = 0.854 \, \text{m}\) and \(h_2 = 1.560 \, \text{m}\). Calculate \(h_{\text{average}}\), then \(PE_{\text{final}}\) and \(W\).

Key Concepts

Potential Energy in Fluids

Potential energy is the energy that is stored in an object due to its position or height. In the context of fluids, potential energy is related to the height of the liquid in a container. The formula for potential energy is given by \( PE = mgh \), where:

• \( m \) is the mass of the liquid.
• \( g \) is the acceleration due to gravity, which is approximately \( 9.81 \, \text{m/s}^2 \).
• \( h \) is the height of the liquid.

To understand how potential energy works in our exercise, consider two vessels with different liquid heights. Although they have the same density and area, the potential energy differs due to the height. When the liquids reach the same level after connecting the vessels, their potential energy changes accordingly. The work done during this process is the transfer of energy from one form to another as the liquid levels balance out.
This change in potential energy is crucial for calculating the work done by the gravitational force.

Density and Its Role in Fluids

Density is a measure of how much mass is contained in a given unit volume. It is often represented by the symbol \( \rho \), and its unit is kilograms per cubic meter (\( \text{kg/m}^3 \)). In the exercise, both vessels contain a liquid with a density of \( 1.30 \times 10^{3} \, \text{kg/m}^3 \). Density plays a significant role in determining the mass of the liquid and thus, impacts the potential energy.

• The relationship between mass and volume is given by \( m = \rho \cdot V \)
• Since \( V = A \cdot h \) for a cylindrical vessel, where \( A \) is the base area and \( h \) is the height, we have \( m = \rho \cdot A \cdot h \)

In our scenario, the density, combined with the height and area of the liquid, allows us to calculate the initial and final potential energy of the liquids. Understanding density is key to grasping how variations in height and volume affect the system's overall energy.

Work Done by Gravitational Force

Work done by a force is defined as the force applied over a distance. In this exercise, when the liquid levels in the vessels equalize, the gravitational force does work by moving the fluid from one vessel to another until balance is reached.

• The formula to calculate work done is \( W = PE_{\text{initial}} - PE_{\text{final}} \)
• \( PE_{\text{initial}} \) is the total potential energy when both vessels have different heights.
• \( PE_{\text{final}} \) is the potential energy when vessels have the same liquid height.

Calculating the work involves finding the difference between the initial and final potential energies. This calculation considers how energy changes as the heights change and liquid levels stabilize.
Gravitational force essentially aids in redistributing potential energy to bring about equilibrium, providing an excellent demonstration of energy conservation in physics.