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Chapter 6: Problem 87

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 pump does 109760 J lifting work, 129600 J kinetic energy work, requiring 3.99 kW of power.

Step by step solution

01

Calculate the Work Done in Lifting the Water

The work done in lifting the water can be calculated using the formula \[ W = mgh \]where:- \( m = 800 \, \text{kg} \) (the mass of the water)- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)- \( h = 14.0 \, \text{m} \) (height of the well)Substitute in the values:\[ W = 800 \times 9.8 \times 14.0 = 109760 \, \text{J} \]Thus, the work done per minute in lifting the water is 109760 Joules.
02

Calculate the Work Done in Giving the Water Kinetic Energy

The work done in giving the water kinetic energy is found using the kinetic energy formula:\[ KE = \frac{1}{2} mv^2 \]where:- \( m = 800 \, \text{kg} \)- \( v = 18.0 \, \text{m/s} \) (speed of the ejected water)Substitute in the values:\[ KE = \frac{1}{2} \times 800 \times (18.0)^2 = 129600 \, \text{J} \]Therefore, the work done in imparting kinetic energy to the water is 129600 Joules.
03

Calculate the Total Work Done by the Pump

The total work done by the pump per minute is the sum of the work done in lifting the water and the work done in giving it kinetic energy:\[ W_{\text{total}} = 109760 + 129600 = 239360 \, \text{J} \]
04

Calculate the Power Output of the Pump

The power output can be calculated using the formula:\[ P = \frac{W_{\text{total}}}{t} \]where \( t = 60 \, \text{s} \) (since it's per minute).Substitute in the values:\[ P = \frac{239360}{60} = 3989.33 \, \text{W} \]Thus, the power output required from the pump is approximately 3989.33 Watts (or 3.99 kW).

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It's a vital concept in mechanics.
In the exercise, as water is ejected, it gains speed which translates into kinetic energy.
To compute this energy we use the formula for kinetic energy:
  • \( KE = \frac{1}{2} mv^2 \)
Here, \( m \) is the mass of the object (water in this case) and \( v \) is its velocity.
Plugging in the given values \( m = 800 \, \text{kg} \) and \( v = 18.0 \, \text{m/s} \), the water gains a kinetic energy of \( 129600 \, \text{J} \) as it is expelled from the pump.
This calculation highlights how energy is transformed from potential to kinetic when an object moves.
Potential Energy
Potential energy is stored energy due to the position or height of an object.
It becomes kinetic energy when the object moves.
In our water pump scenario, lifting water from a well involves moving it vertically against gravity.
  • The potential energy is calculated using \( PE = mgh \)
Where \( m \) is mass, \( g = 9.8 \, \text{m/s}^2 \) (gravitational acceleration), and \( h \) is height.
For 800 kg of water lifted 14 m, the potential energy or work done is \( 109760 \, \text{J} \).
Understanding how potential energy is stored and converted is key to analyzing energy systems.
Power Calculation
Power measures how quickly work is done or energy is transferred.
It answers the question: How fast can the pump move the water?
  • Formula for power is \( P = \frac{W}{t} \)
Where \( W \) is the total work done, and \( t \) is time taken.
In our case, it takes a minute (60 seconds) for the pump to move the water.
With the total work of \( 239360 \, \text{J} \), calculated by adding the lifting and kinetic work, the power required is \( \frac{239360}{60} = 3989.33 \, \text{W} \).
Power calculations help in determining the efficiency of machines and are crucial for designing systems that require energy.

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Most popular questions from this chapter

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

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

Leg Presses. As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arranged side by side so that they are parallel to each other. When you push the platform, you compress the springs. You do 80.0 \(\mathrm{J}\) of work when you compress the springs 0.200 \(\mathrm{m}\) from their uncompressed length. (a) What magnitude of force must you apply to hold the platform in this position? (b) How much additional work must you do to move the platform 0.200 \(\mathrm{m}\) farther, and what maximum force must you apply?

The mass of a proton is 1836 times the mass of an electron. (a) A proton is traveling at speed \(V\) . At what speed (in terms of \(V )\) would an electron have the same kinetic energy as the proton? (b) An electron has kinetic energy \(K\) . If a proton has the same speed as the electron, what is its kinetic energy (in terms of \(K ) ?\)

A car is stopped in a distance \(D\) by a constant friction force that is independent of the car's speed. What is the stopping distance (in terms of \(D\)) (a) if the car's initial speed is tripled, and (b) if the speed is the same as it originally was but the friction force is tripled? (Solve using the work-energy theorem.)
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