Problem 62 - Physics

Question

A fountain sends a stream of water straight up into the air to a maximum height of 5.00 \(\mathrm{m} .\) The effective cross-sectional area of the pipe feeding the fountain is \(5.00 \times 10^{-4} \mathrm{~m}^{2}\) Neglecting air resistance and any viscous effects, determine how many gallons per minute are being used by the fountain. (Note: \(1 \mathrm{gal}=3.79 \times 10^{-3} \mathrm{~m}^{3}\).)

Step-by-Step Solution

Answer

The fountain uses approximately 78.6 gallons per minute.

Step 1: Understand the problem

We need to find the flow rate in gallons per minute of water being used by the fountain, given its maximum height and the cross-sectional area of the pipe.

Step 2: Use energy conservation to find velocity

At the maximum height, the kinetic energy is zero and potential energy is maximized. Use the conservation of energy equation: \( \frac{1}{2}mv^2 = mgh \). Solve for the initial velocity \( v \) as \( v = \sqrt{2gh} \), where \( g = 9.81 \, \text{m/s}^2 \) and \( h = 5.00 \, \text{m} \).

Step 3: Calculate velocity

Substitute \( g = 9.81 \, \text{m/s}^2 \) and \( h = 5.00 \, \text{m} \) into the equation: \[ v = \sqrt{2 \times 9.81 \times 5.00} \approx 9.90 \, \text{m/s} \].

Step 4: Find flow rate in cubic meters per second

The flow rate \( Q \) can be calculated as \( Q = A v \), where \( A = 5.00 \times 10^{-4} \, \text{m}^2 \). Substitute \( v = 9.90 \, \text{m/s} \), \[ Q = 5.00 \times 10^{-4} \times 9.90 = 4.95 \times 10^{-3} \, \text{m}^3/ ext{s}. \]

Step 5: Convert flow rate to gallons per minute

Convert cubic meters per second to gallons per minute. First, find gallons per second: \[ Q_{gal/s} = \frac{4.95 \times 10^{-3} \, \text{m}^3/s}{3.79 \times 10^{-3} \, \text{m}^3/ ext{gal}} \approx 1.31 \, \text{gal/s}. \]Then convert to gallons per minute: \[ Q_{gal/min} = 1.31 \, \text{gal/s} \times 60 \approx 78.6 \, \text{gal/min}. \]

Key Concepts

Conservation of Energy

Conservation of Energy is a fundamental principle in physics. It states that energy in an isolated system remains constant. For our fountain problem, we use this concept to relate the water's initial kinetic energy to its potential energy at the highest point of the stream. When water is shot upwards, it initially has kinetic energy due to its velocity. As it rises, this energy converts into potential energy due to gravity. At the top of its arc, all kinetic energy is converted, and the potential energy is at its maximum. To calculate the initial velocity, we set the kinetic energy equation \( \frac{1}{2}mv^2 \) equal to the potential energy equation \( mgh \).Since mass \( m \) cancels out, we solve for velocity \( v \) using the formula \[ v = \sqrt{2gh} \],where \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity, and \( h = 5.00 \, \text{m} \) is the maximum height reached. This results in a calculated velocity for the water stream.

Flow Rate Calculation

Flow rate calculation is crucial for determining the volume of fluid passing through a point per unit time. In this scenario, it helps us know how much water the fountain uses. Flow rate \( Q \) is calculated by the equation:\[ Q = A v \],where \( A \) is the cross-sectional area of the fountain pipe, and \( v \) is the velocity of the water. In the exercise, the area \( A = 5.00 \times 10^{-4} \, \text{m}^2 \)and the velocity has been determined as approximately \( 9.90 \, \text{m/s} \).So, the flow rate in cubic meters per second is:\[ Q = 5.00 \times 10^{-4} \times 9.90 \approx 4.95 \times 10^{-3} \, \text{m}^3/s \].Finally, to find gallons per minute (since our problem requires this unit), we first convert cubic meters to gallons using the conversion \( 1 \, \text{gal} = 3.79 \times 10^{-3} \, \text{m}^3 \). Calculate:

• \( Q_{gal/s} = \frac{4.95 \times 10^{-3} \, \text{m}^3/s}{3.79 \times 10^{-3} \, \text{m}^3/\text{gal}} \approx 1.31 \, \text{gal/s} \)
• Convert to gallons per minute: \( Q_{gal/min} = 1.31 \, \text{gal/s} \times 60 \approx 78.6 \, \text{gal/min} \)

Kinematics

Kinematics involves the study of motion without considering the forces that cause this motion. It's crucial in understanding how objects move in terms of speed and trajectory, making it a fundamental part of physics studies. In our fountain problem, kinematics helps describe how the water stream moves upwards and downwards.The motion of the water in the fountain is a straight line, going up to a certain height and then coming back down. When dealing with this vertical motion, equations of kinematics provide expressions that describe the velocity and position of the water at any time during its path. Important aspects to consider include:

• Initial Velocity (\( v_0 \)): It is the velocity we calculated using conservation of energy, as it’s the speed the water has right at the fountain's outlet.
• Maximum Height (\( h \)): This is the highest point the water reaches, given as 5 meters in the exercise. At this point, the water's velocity is zero before it starts descending.
• Acceleration due to Gravity (\( g \)): Gravity acts downward, decelerating the water as it rises and accelerating it downwards as it falls, and is quantified as \( 9.81 \, \text{m/s}^2 \).

By combining kinematic equations with energy principles, one can gain a full picture of the fluid's motion in different stages of its trajectory.