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Chapter 7: Problem 86

\(\bullet\) \(\bullet\) 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?

Short Answer

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

Step by step solution

01

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.
02

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: \[ W_1 = 750 \times 9.8 \times 14 = 102900 \mathrm{J}. \]
03

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: \[ W_2 = \frac{1}{2} \times 750 \times 18^2 = 121500 \mathrm{J}. \]
04

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, \[ W_{total} = W_1 + W_2 = 102900 + 121500 = 224400 \mathrm{J}. \]
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

\(\bullet\) \(\bullet\) The aircraft carrier \(U S S\) George Washington has mass \(1.0 \times 10^{8} \mathrm{kg} .\) When its engines are developing their full power of \(260,000\) hp, the George Washington travels at its top speed of 35 knots \((65 \mathrm{km} / \mathrm{h}) .\) If 70\(\%\) of the power output of the engines is applied to pushing the ship through the water, what is the magnitude of the force of water resistance that opposes the carrier's motion at this speed?

\(\bullet\) \(\bullet\) A 20.0 kg rock slides on a rough horizontal surface at 8.00 \(\mathrm{m} / \mathrm{s}\) and eventually stops due to friction. The coefficient of kinetic friction between the rock and the surface is \(0.200 .\) What average thermal power is produced as the rock stops?

\(\bullet\) \(\bullet\) At the site of a wind farm in North Dakota, the average wind speed is \(9.3 \mathrm{m} / \mathrm{s},\) and the average density of air is 1.2 \(\mathrm{kg} / \mathrm{m}^{3} .\) (a) Calculate how much kinetic energy the wind contains, per cubic meter, at this location. (b) No wind turbine can capture all of the energy contained in the wind, the main reason being that capturing all the energy would require stop- ping the wind completely, meaning that air would stop flowing through the turbine. Suppose a particular turbine has blades with a radius of 41 \(\mathrm{m}\) and is able to capture 35\(\%\) of the avail- able wind energy. What would be the power output of this tur- bine, under average wind conditions?

\(\bullet\) To stretch a certain spring by 2.5 \(\mathrm{cm}\) from its equilibrium position requires 8.0 \(\mathrm{J}\) of work. (a) What is the force constant of this spring? (b) What was the maximum force required to stretch it by that distance?

\(\bullet\) The total height of Yosemite Falls is 2425 ft. (a) How many more joules of gravitational potential energy are there for each kilogram of water at the top of this waterfall com- pared with each kilogram of water at the foot of the falls? (b) Find the kinetic energy and speed of each kilogram of water as it reaches the base of the waterfall, assuming that there are no losses due to friction with the air or rocks and that the mass of water had negligible vertical speed at the top. How fast (in \(\mathrm{m} / \mathrm{s}\) and mph) would a 70 \(\mathrm{kg}\) person have to run to have that much kinetic energy? (c) How high would Yosemite Falls have to be so that each kilogram of water at the base had twice the kinetic energy you found in part (b); twice the speed you found in part (b)?
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