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A pump is required to lift \(800 \mathrm{~kg}\) of water (about 210 gallons) per minute from a well \(14.0 \mathrm{~m}\) deep and eject it with a speed of \(18.0 \mathrm{~m} / \mathrm{s}\). (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
The work done per minute in lifting the water is approximately \( 1.1 \times 10^5 \mathrm{~J} \). The work done in giving the water kinetic energy is approximately \( 1.3 \times 10^5 \mathrm{~J} \). The total work done by the pump is therefore about \( 2.4 \times 10^5 \mathrm{~J} \). Thus, the power output of the pump must be about \( 4.0 \times 10^3 \mathrm{~W} \) or \( 4.0 \mathrm{~kW} \).

Step by step solution

01

Calculate the work done in lifting the water

Work done is given by the equation \( W = mgh \) for lifting up an object, where \( m \) is the mass, \( g \) the acceleration due to gravity, and \( h \) the height. We have \( m = 800 \mathrm{~kg} \), \( g = 9.81 \mathrm{~m/s^2} \) (approximate value of acceleration due to gravity on Earth's surface), and \( h = 14.0 \mathrm{~m} \). Plug these values into the formula and calculate.
02

Calculate the work done in giving the water kinetic energy

The work done in this case is equal to the kinetic energy of the water. Kinetic energy is given by the formula \( KE = 0.5mv^2 \), where \( m \) is the mass and \( v \) the velocity. We have \( m = 800 \mathrm{~kg} \) and \( v = 18.0 \mathrm{~m/s} \). Plug these values into the formula and calculate.
03

Calculate the total work done

The total amount of work done by the pump is the work done in lifting the water plus the work done in giving the water kinetic energy. Add the results of Steps 1 and 2.
04

Calculate the power output of the pump

Power is the work done per time, given by the formula \( P = W/t \), where \( W \) is the work done and \( t \) the time duration. In this case, we know that the pump does this work every minute, so \( t = 1 \mathrm{~minute} = 60 \mathrm{~seconds} \). Plug in the total work done from Step 3 and the value for time into the formula and calculate.

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Key Concepts

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

Power
Power is a fundamental concept in physics, representing the rate at which work is done or energy is transferred over time.
Understanding power helps us measure how quickly tasks are accomplished. For example, a powerful engine performs mechanical work rapidly.

The formula for calculating power is:
  • \[ P = \frac{W}{t} \]
Where:
  • \( P \) is power
  • \( W \) is the work done
  • \( t \) is the time over which the work is done
Combining this with the problem at hand, after calculating the work performed by the pump per minute, you apply this formula to find the power output necessary.
The unit of power is watts (W), equivalent to joules per second, capturing how much energy is shifted or transformed every second.
Keen students will explore how power factors in real-world issues, like specifying household appliances, which affect energy consumption and can lead to cost savings.
Kinetic Energy
Kinetic Energy is the energy that an object possesses due to its motion.
When considering objects in motion, such as water being ejected by a pump, this form of energy becomes vital.

To calculate kinetic energy, you use the formula:
  • \[ KE = \frac{1}{2} mv^2 \]
Where:
  • \( KE \) is the kinetic energy
  • \( m \) is the mass of the object
  • \( v \) is the velocity of the object
This formula shows that both the mass and speed of an object determine how much kinetic energy it holds.
In our exercise, when calculating the energy of water being ejected at a given speed, these principles help assess how much work has been transformed into motion.
Students should grasp that increasing either the mass or velocity "significantly" increases kinetic energy, leading to greater power demands from machinery, like pumps.
Gravitational Potential Energy
Gravitational Potential Energy (GPE) is the energy stored in an object due to its position relative to Earth.
Elevating an object within Earth's gravitational field stores energy that can potentially convert back to kinetic energy when the object is released.

To determine the gravitational potential energy, the formula is as follows:
  • \[ GPE = mgh \]
Where:
  • \( GPE \) is gravitational potential energy
  • \( m \) is the mass of the object
  • \( g \) is the acceleration due to gravity (approximately \( 9.81 \text{ m/s}^2 \))
  • \( h \) is the height above the reference point
In the textbook exercise, this concept applies when analyzing the energy needed to lift water from a well.
Understanding GPE is crucial for recognizing how energy must be applied to raise objects, impacting how much work a device like a pump must perform.Gravitational potential energy is ubiquitous in real-world scenarios, helping solve problems involving lifting or elevating materials.

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Most popular questions from this chapter

\(\mathrm{A} 6.0 \mathrm{~kg}\) box moving at \(3.0 \mathrm{~m} / \mathrm{s}\) on a horizontal, frictionless surface runs into one end of a light horizontal spring of force constant \(75 \mathrm{~N} / \mathrm{cm}\) that is fixed at the other end. Use the work-energy theorem to find the maximum compression of the spring.

BIO Power of the Human Heart. The human heart is a powerful and extremely reliable pump. Each day it takes in and discharges about \(7500 \mathrm{~L}\) 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 ( \(1.63 \mathrm{~m}\) ). The density (mass per unit volume) of blood is \(1.05 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). (a) How much work does the heart do in a day? (b) What is the heart's power output in watts?

A loaded grocery cart is rolling across a parking lot in a strong wind. You apply a constant force \(\vec{F}=(30 \mathrm{~N}) \hat{\imath}-(40 \mathrm{~N}) \hat{\jmath}\) to the cart as it undergoes a displacement \(\vec{s}=(-9.0 \mathrm{~m}) \hat{\imath}-(3.0 \mathrm{~m}) \hat{\jmath}\). How much work does the force you apply do on the grocery cart?

A force \(\vec{F}\) that is at an angle \(60^{\circ}\) above the horizontal is applied to a box that moves on a horizontal frictionless surface, and the force does work \(W\) as the box moves a distance \(d\). (a) At what angle above the horizontal would the force have to be directed in order for twice the work to be done for the same displacement of the box? (b) If the angle is kept at \(60^{\circ}\) and the box is initially at rest, by what factor would \(F\) have to be increased to double the final speed of the box after moving distance \(d ?\)

constant eastward acceleration of \(a=2.80 \mathrm{~m} / \mathrm{s}^{2}\). A worker assists the cart by pushing on the crate with a force that is eastward and has magnitude that depends on time according to \(F(t)=(5.40 \mathrm{~N} / \mathrm{s}) t .\) What is the instantaneous power supplied by this force at \(t=5.00 \mathrm{~s} ?\)

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