Problem 86 - College Physics

Question

\(\bullet\) \(\bullet\) A pump is required to lift 750 liters of water per minute from a well 14.0 \(\mathrm{m}\) deep and eject it with a speed of 18.0 \(\mathrm{m} / \mathrm{s}\) . How much work per minute does the pump do?

Step-by-Step Solution

Answer

The pump does 224,400 J of work per minute.

Step 1: Understanding Work Done by the Pump

The work done by the pump in this context involves two main components: lifting the water to a certain height and giving the water a final kinetic energy. We'll calculate both separately and sum them up for a total work.

Step 2: Calculate the Work to Lift the Water

The work required to lift water is calculated using the formula for gravitational work, which is \( W_1 = mgh \), where \( m \) is the mass of water, \( g \) is the acceleration due to gravity (\( 9.8 \mathrm{m/s^2} \)), and \( h \) is the height (14.0 meters). First, convert the volume of water to mass, knowing the density of water is \( 1000 \mathrm{kg/m^3} \). Thus, \( 750 \mathrm{liters} = 0.75 \mathrm{m^3} \) equals \( m = 0.75 \times 1000 = 750 \mathrm{kg} \). Then, substitute into the formula: \[ W_1 = 750 \times 9.8 \times 14 = 102900 \mathrm{J}. \]

Step 3: Calculate the Work to Eject the Water

The work required to give the water speed involves kinetic energy, which is \( W_2 = \frac{1}{2}mv^2 \), where \( m \) is mass (750 kg) and \( v \) is velocity (18 m/s). Substitute into the equation: \[ W_2 = \frac{1}{2} \times 750 \times 18^2 = 121500 \mathrm{J}. \]

Step 4: Calculate Total Work Per Minute

The total work done per minute by the pump is the sum of the work to lift the water and the work to eject with given speed. Thus, \[ W_{total} = W_1 + W_2 = 102900 + 121500 = 224400 \mathrm{J}. \]

Step 5: Conclusion

The pump does \( 224400 \mathrm{J} \) of work per minute to lift and eject 750 liters of water from a 14-meter well at 18 m/s.

Key Concepts

Gravitational Work

In the context of lifting water from a well, gravitational work refers to the energy required to lift the water against the force of gravity. This can be understood using the formula for gravitational work, \( W_1 = mgh \), where \( m \) is the mass of the water, \( g \) is the acceleration due to gravity (approximated as \( 9.8 \ \mathrm{m/s}^2 \)), and \( h \) is the height from which the water is being lifted. To apply this formula:

• First, convert the given volume of water in liters to cubic meters, as 1 cubic meter equals 1000 liters. For the 750 liters in this problem, this becomes \( 0.75 \ \mathrm{m}^3 \).
• Next, recognize that the density of water is \( 1000 \ \mathrm{kg/m}^3 \). Therefore, the mass of the water is \( 0.75 \times 1000 = 750 \ \mathrm{kg} \).
• Finally, use the gravitational work formula to find: \[ W_1 = 750 \times 9.8 \times 14 = 102900 \ \mathrm{J} \]

Gravitational work is a key concept in physics, describing how energy is expended when an object is moved vertically against gravity.

Kinetic Energy

Kinetic energy pertains to the energy that an object possesses due to its motion. For the case of water being ejected at a speed of \( 18 \ \mathrm{m/s} \), we must calculate the energy required to achieve such velocity. The formula for kinetic energy is \( W_2 = \frac{1}{2}mv^2 \), where \( m \) is the mass of the water and \( v \) is its velocity. Let's break down what these symbols mean:

• The mass \( m \) of the water is already calculated as \( 750 \ \mathrm{kg} \).
• The velocity \( v \), given as \( 18 \ \mathrm{m/s} \), is the speed you want the water to reach after being pumped.

Substituting into the formula gives us:\[ W_2 = \frac{1}{2} \times 750 \times (18)^2 = 121500 \ \mathrm{J} \]Kinetic energy is an essential part of our everyday experiences, explaining the energy transformations involved in motion.

Mass Conversion

Mass conversion, in the context of this problem, involves changing the volume of a fluid into its corresponding mass for further calculations. This process is critical because most formulas for work and energy computations require the mass, not volume.Here's how it's done:

• First, know the conversion between volume units, understanding that 1000 liters equal 1 cubic meter. In this instance, \( 750 \ \mathrm{liters} \) converts to \( 0.75 \ \mathrm{m}^3 \).
• Water has a known density of \( 1000 \ \mathrm{kg/m}^3 \). Thus, to convert the volume to mass, multiply the cubic meter volume by this density, resulting in \( 750 \ \mathrm{kg} \) of water from \( 0.75 \ \mathrm{m}^3 \).

These conversions are not unique to water and apply to various substances, each with its specific density value.

Physics Problem-Solving

Physics problem-solving is a systemic approach that requires applying known principles to new situations. Let's focus on solving work and energy problems as in the example of the water pump. Key steps in effective problem-solving include:

• Understanding the problem: Identify what is being asked. Here, the total work done by the pump involved lifting and ejecting water.
• Identify necessary concepts: Recognize which physics concepts are pertinent. In this case, gravitational work and kinetic energy are used.
• Break down the problem: Attack the problem in parts. Calculate gravitational work and kinetic energy separately before adding them.
• Convert units: Ensure consistency by converting all measurements to suitable units, like mass in kilograms and volume in cubic meters.
• Perform calculations with accuracy: Double-check math as errors can lead to wildly incorrect conclusions.

Problem-solving in physics enhances logical thinking skills and deepens understanding of how the world works through quantifiable principles.