/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 85 A pump is required to lift 800 k... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 85

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 \(3,991.2 \ \text{W}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

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/s虏 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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

One end of a horizontal spring with force constant 76.0 N/m is attached to a vertical post. A 2.00-kg block of frictionless ice is attached to the other end and rests on the floor. The spring is initially neither stretched nor compressed. A constant horizontal force of 54.0 N is then applied to the block, in the direction away from the post. (a) What is the speed of the block when the spring is stretched 0.400 m? (b) At that instant, what are the magnitude and direction of the acceleration of the block?

A small block with a mass of 0.0600 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (\(\textbf{Fig. P6.71}\)). The block is originally revolving at a distance of 0.40 m from the hole with a speed of 0.70 m/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.10 m. At this new distance, the speed of the block is 2.80 m/s. (a) What is the tension in the cord in the original situation, when the block has speed \(\upsilon = 0.70\) m/s? (b) What is the tension in the cord in the final situation, when the block has speed \(\upsilon = 2.80\) m/s? (c) How much work was done by the person who pulled on the cord?

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.80 \(\times\) 10\(^6\) N, one 14\(^\circ\) west of north and the other 14\(^\circ\) east of north, as they pull the tanker 0.75 km toward the north. What is the total work they do on the supertanker?

To stretch a spring 3.00 cm from its unstretched length, 12.0 J of work must be done. (a) What is the force constant of this spring? (b) What magnitude force is needed to stretch the spring 3.00 cm from its unstretched length? (c) How much work must be done to compress this spring 4.00 cm from its unstretched length, and what force is needed to compress it this distance?

Use the work\(-\)energy theorem to solve each of these problems. You can use Newton's laws to check your answers. Neglect air resistance in all cases. (a) A branch falls from the top of a 95.0-m-tall redwood tree, starting from rest. How fast is it moving when it reaches the ground? (b) A volcano ejects a boulder directly upward 525 m into the air. How fast was the boulder moving just as it left the volcano?
See all solutions

Recommended explanations on Physics Textbooks

Electromagnetism

Read Explanation

Waves Physics

Read Explanation

Physical Quantities And Units

Read Explanation

Kinematics Physics

Read Explanation

Force

Read Explanation

Radiation

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.