/*! 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 16 An ideal spring of negligible ma... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 16

An ideal spring of negligible mass is 12.00 \(\mathrm{cm}\) long when nothing is attached to it. When you hang a \(3.15-\mathrm{kg}\) weight from it, you measure its length to be 13.40 \(\mathrm{cm}\) . If you wanted to store 10.0 \(\mathrm{J}\) of potential energy in this spring, what would be its total length? Assume that it continues to obey Hooke's law.

Short Answer

Expert verified
The spring length would be 21.52 cm to store 10.0 J of energy.

Step by step solution

01

Initial Setup and Understanding

First, identify the information given:- Original spring length, \( L_0 = 12.00 \, \text{cm} = 0.1200 \, \text{m}\).- Length with 3.15 kg weight, \( L_1 = 13.40 \, \text{cm} = 0.1340 \, \text{m}\).- Weight, \( m = 3.15 \, \text{kg}\).The spring force constant \( k \) needs to be calculated, and then used to find the length where the spring's potential energy is 10.0 J.
02

Calculate the Spring Constant

The force exerted by the weight is equal to \( F = mg \), where \( g = 9.81 \, \text{m/s}^2 \). This force causes a change in the spring's length of \( \Delta x = L_1 - L_0 = 0.1340 - 0.1200 = 0.0140 \, \text{m} \).Using Hooke's Law \( F = k \Delta x \),\[ k = \frac{mg}{\Delta x} = \frac{3.15 \times 9.81}{0.0140} \approx 2204.25 \, \text{N/m}\].
03

Determine the Required Extension for Desired Potential Energy

The potential energy stored in a spring is given by \( U = \frac{1}{2}k(\Delta x')^2 \).We want this energy to be 10.0 J:\[ 10.0 = \frac{1}{2} \times 2204.25 \times (\Delta x')^2 \].Solve for \( \Delta x' \):\[ 20.0 = 2204.25 \times (\Delta x')^2 \].\[ (\Delta x')^2 = \frac{20.0}{2204.25} \approx 0.009073 \].\[ \Delta x' \approx \sqrt{0.009073} \approx 0.0952 \, \text{m}\].
04

Calculate the Total Length of the Spring

The total length \( L_2 \) of the spring when storing 10.0 J of potential energy can be found using:\[ L_2 = L_0 + \Delta x' = 0.1200 + 0.0952 = 0.2152 \, \text{m} \].Convert to cm:\[ L_2 = 21.52 \, \text{cm} \].

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.

Spring Potential Energy
Spring potential energy is an essential concept in physics, particularly when studying springs and mechanical systems. This energy is essentially the elastic potential energy stored in a spring when it is either compressed or stretched. According to Hooke's Law, the potential energy in a spring is represented by the formula:\[ U = \frac{1}{2} k \Delta x^2 \]Where:
  • \( U \) is the potential energy
  • \( k \) is the spring constant
  • \( \Delta x \) is the displacement of the spring from its equilibrium position
In practical terms, this means how far the spring is either compressed or stretched from its original length. A higher spring constant signifies a stiffer spring that requires more force to stretch or compress. Conversely, a smaller displacement results in less energy stored. Therefore, understanding the relationship between these variables helps us predict how much energy can be stored in a spring. This study is vital for designing systems where controlling energy storage and release is necessary, such as in automotive suspensions or measuring devices.
Spring Constant Calculation
The spring constant, denoted as \( k \), measures the stiffness of a spring. It defines how much force is required to stretch or compress the spring by a unit distance. One common method to calculate the spring constant is by using Hooke's Law:\[ F = k \Delta x \]Where:
  • \( F \) is the force applied to the spring
  • \( \Delta x \) is the change in length of the spring
  • \( k \) is what we want to find
In the exercise, we calculate \( k \) by measuring the force exerted by a known weight and the corresponding change in length of the spring. Utilizing the formula \( k = \frac{F}{\Delta x} \), we determine the spring constant by dividing the force (due to the weight) by the stretched length. The spring constant helps us understand how resistant a spring is to deformation鈥攈igher values mean the spring is more resistant.
Force Exerted by a Weight
The force exerted by a weight is fundamental when dealing with springs because it drives the change in spring length. This force is the gravitational force acting on a mass, calculated by:\[ F = mg \]Where:
  • \( m \) is the mass of the object
  • \( g \) is the acceleration due to gravity (approximately 9.81 m/s虏 on Earth)
So, if you have a mass suspended from a spring, the weight's force pulls the spring, causing it to extend. The extent of this extension depends on both the force and the spring constant. By understanding the force exerted due to weight, you can predict how much a spring will stretch, calculate spring constants, and even design systems to handle specific forces. This forms the basis for applications ranging from weighing scales to complex load-bearing systems.

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 \(60.0-\mathrm{kg}\) skier starts from rest at the top of a ski slope 65.0 \(\mathrm{m}\) high. (a) If frictional forces do \(-10.5 \mathbf{k J}\) of work on her as she descends, how fast is she going at the bottom of the slope? (b) Now moving horizontally, the skier crosses a patch of soft snow, where \(\mu_{\mathrm{k}}=0.20\) If the patch is 82.0 \(\mathrm{m}\) wide and the average force of air resistance on the skier is 160 \(\mathrm{N}\) , how fast is she going after crossing the patch? (c) The skier hits a snowdrift and Penetrates 2.5 \(\mathrm{m}\) into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?

A Hooke's law force \(-k x\) and a constant conservative force \(F\) in the \(+x\) -direction act on an atomic ion. (a) Show that a possible potential-energy function for this combination of forces is \(U(x)=\frac{1}{2} k x^{2}-F x-F^{2} / 2 k\) . Is this the only possible function? Explain. (b) Find the stable equilibrium position. (c) Graph \(U(x)\) (in units of \(F^{2} / k )\) versus \(x\) (in units of \(F / k )\) for values of \(x\) between \(-5 F / k\) and 5\(F / k\) . (d) Are there any unstable equilibrium positions? (e) If the total energy is \(E=F^{2} | k,\) what are the maximum and minimum valnes of \(x\) that the ion reaches in its motion? If the ion has mass \(m,\) find its maximum speed if the total energy is \(E=F^{2} / k .\) For what value of \(x\) is the speed maximum?

A 0.60\(\cdot \mathrm{kg}\) book slides on a horizontal table. The kinetic friction force on the book has magnitude 1.2 \(\mathrm{N}\) . (a) How much work is done on the book by friction during a displacement of 3.0 \(\mathrm{m}\) to the left? (b) The book now slides 3.0 \(\mathrm{m}\) to the right, returning to its starting point. During this second 3.0 \(\mathrm{m}\) displacement, how much work is done on the book by friction? (c) What is the total work done on the book by friction during the complete round trip? (d) On the basis of your answer to part (c), would you say that the friction force is conservative or nonconservative? Explain.

A \(75-\mathrm{kg}\) roofer climbs a vertical 7.0 -m ladder to the flat roof of a house. He then walks 12 \(\mathrm{m}\) on the roof, climbs down another vertical \(7.0-\mathrm{m}\) ladder, and finally walks on the ground back to his starting point. How much work is done on him by gravity (a) as he climbs up; \((b)\) as he climbs down; (c) as he walks on the roof and on the ground? (d) What is the total work done on him by gravity during this round trip? On the basis of your answer to part (d), would you say that gravity is a conservative or nonconservative force? Explain.

A force parallel to the \(x\) -axis acts on a particle moving along the \(x\) -axis. This force produces potential energy \(U(x)\) given by \(U(x)=\alpha x^{4},\) where \(\alpha=1.20 \mathrm{J} / \mathrm{m}^{4} .\) What is the force (magnitude and direction) when the particle is at \(x=-0.800 \mathrm{m} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Physical Quantities And Units

Read Explanation

Thermodynamics

Read Explanation

Electrostatics

Read Explanation

Torque and Rotational Motion

Read Explanation

Mechanics and Materials

Read Explanation

Electromagnetism

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.