/*! 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 42 An ingenious bricklayer builds a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 42

An ingenious bricklayer builds a device for shooting bricks up to the top of the wall where he is working. He places a brick on a vertical compressed spring with force constant \(k=450 \mathrm{N} / \mathrm{m}\) and negligible mass. When the spring is released, the brick is propelled upward. If the brick has mass 1.80 \(\mathrm{kg}\) and is to reach a maximum height of 3.6 \(\mathrm{m}\) above its initial position on the compressed spring, what distance must the bricklayer compress the spring initially? (The brick loses contact with the spring when the spring returns to its uncompressed length. Why?)

Short Answer

Expert verified
The spring must be compressed approximately 0.53 meters.

Step by step solution

01

Understand Potential Energy and Conservation of Energy

We start by recognizing that the problem involves energy transformation: potential energy in the spring is transformed into gravitational potential energy at the brick's maximum height. The energy stored in the compressed spring will convert to the gravitational potential energy of the brick as it reaches its peak height. The gravitational potential energy at maximum height is given by \( U_g = mgh \), where \( m = 1.80 \, \text{kg} \), \( g = 9.81 \, \text{m/s}^2 \), and \( h = 3.6 \, \text{m} \).
02

Calculate Gravitational Potential Energy at Maximum Height

Calculate the gravitational potential energy when the brick reaches 3.6 m. \[ U_g = mgh = 1.80 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 3.6 \, \text{m} = 63.5088 \, \text{J} \]
03

Use Spring Potential Energy Formula

The energy stored in the spring when compressed is given by the spring potential energy formula: \[ U_s = \frac{1}{2}kx^2 \] where \( k = 450 \, \text{N/m} \) and \( x \) is the compression in meters. Since all spring potential energy converts to gravitational potential energy at the top, we equate: \( U_s = U_g \).
04

Set Up the Equation and Solve for x

Equating the stored potential energy in the spring to the gravitational potential energy: \[ \frac{1}{2}kx^2 = mgh \]Substitute the known values into the equation: \[ \frac{1}{2} \times 450 \, \text{N/m} \times x^2 = 63.5088 \, \text{J} \]\[ 225x^2 = 63.5088 \]\[ x^2 = \frac{63.5088}{225} \]\[ x^2 = 0.28226 \]\[ x = \sqrt{0.28226} \approx 0.5315 \, \text{m} \]
05

Conclusion

The initial compression of the spring needed for the brick to reach a height of 3.6 m is approximately 0.5315 meters.

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.

Potential Energy
Potential energy is stored energy available to do work. In this scenario, the brick has potential energy due to its position when placed on a compressed spring. Think of potential energy like a loaded bow. Energy is stored as tension, and when released, it propels the arrow.
This stored energy transforms into another form when released.
  • Elastic Potential Energy: Specifically for the spring, potential energy depends on how much the spring is compressed or extended.
  • Gravitational Potential Energy: This is relevant when the brick reaches a certain height due to gravity.
Understanding these types of potential energy helps us solve problems related to energy conversion.
Spring Constant
The spring constant, denoted as \( k \), is a measure of a spring's stiffness. It tells us how much force is needed to compress or stretch the spring by a unit length. In the exercise, we have a spring constant of 450 N/m.
This means that for every meter the spring is compressed or stretched, 450 newtons of force is exerted. The spring constant is crucial in calculating the elastic potential energy using the formula:
  • Spring Potential Energy: \[ U_s = \frac{1}{2}kx^2 \]
  • Where \( x \) is the displacement from its natural position.
A higher spring constant indicates a stiffer spring, requiring more force for compression or extension.
Gravitational Potential Energy
Gravitational potential energy (\(U_g\)) is the energy an object possesses because of its position in a gravitational field. It's calculated when an object is lifted to a height, where gravity's pull can do work on it.
This energy is given by the formula:
  • \( U_g = mgh \)
  • \( m \): mass of the object (1.80 kg for the brick)
  • \( g \): acceleration due to gravity (9.81 m/s虏)
  • \( h \): height (3.6 m in this case)
This formula shows how the gravitational potential energy increases with greater height and mass. As the brick moves upwards, converting spring energy into gravitational potential energy, it reaches the maximum energy at its highest point.
Energy Transformation
Energy transformation involves changing energy from one form to another. In our exercise, the energy starts as elastic potential energy in the compressed spring and changes into gravitational potential energy as the brick is propelled upward.
The whole process is an example of the conservation of energy principle. This means that the total energy before release (in the spring) equals the total energy afterward (at max height).
While the spring pushes the brick, all of its potential energy converts into kinetic energy, moving the brick upwards. At the brick's highest point, this energy turns entirely into gravitational potential energy.
  • Key Equation: \( U_s = U_g \)
This indicates how potential energy supports the brick's motion. It demonstrates energy never disappears but transforms into useful forms.

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

A little red wagon with mass 7.00 \(\mathrm{kg}\) moves in a straight line on a frictionless horizontal surface. It has an initial speed of 4.00 \(\mathrm{m} / \mathrm{s}\) and then is pushed 3.0 \(\mathrm{m}\) in the direction of the initial velocity by a force with a magnitude of 10.0 \(\mathrm{N}\) . (a) Use the work- energy theorem to calculate the wagon's final speed. (b) Calculate the acceleration produced by the force. Use this acceleration in the kinematic relationships of Chapter 2 to calculate the wagon's final speed. Compare this result to that calculated in part (a).

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 \(K ) ?\)

On an essentially frictionless, horizontal ice rink, a skater moving at 3.0 \(\mathrm{m} / \mathrm{s}\) encounters a rough patch that reduces her speed by 45\(\%\) due to a friction force that is 25\(\%\) of her weight. Use the work- energy theorem to find the length of this rough patch.

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 \(\mathrm{m}\) into the air. How fast was the boulder moving just as it left the volcano? (c) A skier moving at 5.00 \(\mathrm{m} / \mathrm{s}\) encounters a long, rough horizontal patch of snow having coefficient of kinetic friction 0.220 with her skis. How far does she travel on this patch before stopping? (d) Suppose the rough patch in part (c) was only 2.90 \(\mathrm{m}\) long? How fast would the skier be moving when she reached the end of the patch? (e) At the base of a frictionless icy hill that rises at \(25.0^{\circ}\) above the horizontal, a toboggan has a speed of 12.0 \(\mathrm{m} / \mathrm{s}\) toward the hill. How high vertically above the base will it go before stopping?

Varying Coefficient of Friction. A box is sliding with a speed of 4.50 \(\mathrm{m} / \mathrm{s}\) on a horizontal surface when, at point \(P\) . it encounters a rough section. On the rough section, the coefficient of friction is not constant, but starts at 0.100 at \(P\) and increases linearly with distance past \(P\) , reaching a value of 0.600 at 12.5 \(\mathrm{m}\) past point \(P .(\text { a) Use the work-energy theorem to find how far this box }\) slides before stopping. \((b)\) What is the coefficient of friction at the stopping point? (c) How far would the box have slid if the friction coefficient didn't increase but instead had the constant value of 0.100\(?\)
See all solutions

Recommended explanations on Physics Textbooks

Electromagnetism

Read Explanation

Famous Physicists

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Waves Physics

Read Explanation

Electricity and Magnetism

Read Explanation

Electricity

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.