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A pump is required to lift 800 kg of water (about 210 gallons) per minute from a well 14.0 m deep and eject it with a speed of 18.0 m/s. (a) How much work is done per minute in lifting the water? (b) How much work is done in giving the water the kinetic energy it has when ejected? (c) What must be the power output of the pump?

Short Answer

Expert verified
(a) 109,872 J, (b) 129,600 J, (c) 3,991.2 W.

Step by step solution

01

Determine the Work Done in Lifting the Water

The work done to lift the water can be calculated using the formula for gravitational work:\[ W_1 = mgh \]where \( W_1 \) is the work done lifting the water, \( m \) is the mass of the water, \( g \) is the acceleration due to gravity (approximately \(9.81 \ \text{m/s}^2\)), and \( h \) is the height lifted.Substitute the values:\[ W_1 = 800 \ \text{kg} \times 9.81 \ \text{m/s}^2 \times 14.0 \ \text{m} \]\[ W_1 = 109,872 \ \text{J} \] So, the work done per minute in lifting the water is \(109,872 \ \text{J}\).
02

Calculate the Work Done in Giving the Water Kinetic Energy

The work done to give the water kinetic energy can be calculated using the kinetic energy formula:\[ W_2 = \frac{1}{2} mv^2 \]where \( W_2 \) is the work done, \( m \) is the mass of the water, and \( v \) is the velocity.Substitute the values:\[ W_2 = \frac{1}{2} \times 800 \ \text{kg} \times (18.0 \ \text{m/s})^2 \]\[ W_2 = \frac{1}{2} \times 800 \ \times 324 \]\[ W_2 = 129,600 \ \text{J} \]Hence, the work done in giving the water the kinetic energy is \(129,600 \ \text{J}\).
03

Calculate the Total Work Done

The total work done by the pump per minute is the sum of the work done to lift the water and the work done to give it kinetic energy:\[ W_{\text{total}} = W_1 + W_2 \]\[ W_{\text{total}} = 109,872 \ \text{J} + 129,600 \ \text{J} \]\[ W_{\text{total}} = 239,472 \ \text{J} \] Thus, the total work done is \(239,472 \ \text{J}\).
04

Determine the Power Output of the Pump

Power output of the pump is calculated by dividing the total work done by the time duration, which is 1 minute or 60 seconds:\[ P = \frac{W_{\text{total}}}{t} \]\[ P = \frac{239,472 \ \text{J}}{60 \ \text{s}} \]\[ P = 3,991.2 \ \text{W} \]Thus, the power output of the pump must be \(3,991.2 \ \text{W}\).

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Key Concepts

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

Gravitational Work
When it comes to lifting objects against gravity, the concept of gravitational work is crucial. In simple terms, gravitational work refers to the energy required to lift a mass to a certain height against the gravitational force. The formula to calculate this work is given by: \[ W = mgh \]
Here, \( W \) represents the work done, \( m \) is the mass of the object, \( g \) is the acceleration due to gravity (approximately 9.81 m/s² on Earth), and \( h \) is the height to which the object is lifted.
  • Mass: The amount of matter in the object.
  • Gravity: A constant that represents the force pulling objects toward the Earth's center.
  • Height: The vertical distance the object is moved against gravity.

In the original problem, lifting 800 kg of water to a height of 14 meters requires a certain amount of energy. Using the formula, you can calculate this as approximately 109,872 Joules. This energy expenditure is what keeps the water elevated and ready for its next phase of movement.
Kinetic Energy
Once the water is lifted, it must be ejected with velocity. Here is where kinetic energy comes into play. Kinetic energy is the energy that an object possesses due to its motion. The formula for kinetic energy is: \[ KE = \frac{1}{2} mv^2 \]
Where \( KE \) is the kinetic energy, \( m \) is the mass of the object, and \( v \) is the velocity at which the object is moving.
  • Mass: The weight of the water being ejected.
  • Velocity: The speed at which the water exits the pump.

For the given problem, water is ejected at a speed of 18 m/s. Hence, the kinetic energy given to the water can be calculated to be about 129,600 Joules. This energy is what propels the water as it exits, allowing it to move freely away from the pump.
Power Output
Power is defined as the rate at which work is done or energy is transferred over time. It helps us understand how quickly energy is used or generated. The formula to determine power is: \[ P = \frac{W}{t} \]
Where \( P \) is power, \( W \) is the total work done, and \( t \) is the time over which the work is done. In this case, the total work includes both the gravitational work and the kinetic energy.
  • Work: Total energy expended to lift and move the water.
  • Time: Duration of 60 seconds, equivalent to one minute.

By calculating the power output for the pump lifting and ejecting water, we find it to be approximately 3,991.2 Watts. Knowing the power output is essential for assessing how effectively and efficiently the pump operates over the set period.

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Most popular questions from this chapter

While hovering, a typical flying insect applies an average force equal to twice its weight during each downward stroke. Take the mass of the insect to be 10 g, and assume the wings move an average downward distance of 1.0 cm during each stroke. Assuming 100 downward strokes per second, estimate the average power output of the insect.

All birds, independent of their size, must maintain a power output of 10\(-\)25 watts per kilogram of body mass in order to fly by flapping their wings. (a) The Andean giant hummingbird (\(Patagona gigas\)) has mass 70 g and flaps its wings 10 times per second while hovering. Estimate the amount of work done by such a hummingbird in each wingbeat. (b) A 70-kg athlete can maintain a power output of 1.4 kW for no more than a few seconds; the \(steady\) power output of a typical athlete is only 500 W or so. Is it possible for a human-powered aircraft to fly for extended periods by flapping its wings? Explain.

A 4.80-kg watermelon is dropped from rest from the roof of an 18.0-m-tall building and feels no appreciable air resistance. (a) Calculate the work done by gravity on the watermelon during its displacement from the roof to the ground. (b) Just before it strikes the ground, what is the watermelon's (i) kinetic energy and (ii) speed? (c) Which of the answers in parts (a) and (b) would be \(different\) if there were appreciable air resistance?

The human heart is a powerful and extremely reliable pump. Each day it takes in and discharges about 7500 L of blood. Assume that the work done by the heart is equal to the work required to lift this amount of blood a height equal to that of the average American woman (1.63 m). The density (mass per unit volume) of blood is \(1.05 \times 10^3 \, \mathrm{kg/m}^3\). (a) How much work does the heart do in a day? (b) What is the heart's power output in watts?

Two blocks are connected by a very light string passing over a massless and frictionless pulley (\(\textbf{Fig. E6.7}\)). Traveling at constant speed, the 20.0-N block moves 75.0 cm to the right and the 12.0-N block moves 75.0 cm downward. How much work is done (a) on the 12.0-N block by (i) gravity and (ii) the tension in the string? (b) How much work is done on the 20.0-N block by (i) gravity, (ii) the tension in the string, (iii) friction, and (iv) the normal force? (c) Find the total work done on each block.

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