/*! 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 44 Half of a Spring. (a) Suppose yo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Half of a Spring. (a) Suppose you cut a massless ideal spring in half. If the full spring had a force constant \(k\) , what is the force constant of each half, in terms of \(k ?\) (Hint: Think of the original spring as two equal halves, each producing the same force as the entire spring. Do you see why the forces must be equal? (b) If you cut the spring into three equal segments instead, what is the force constant of each one, in terms of \(k ?\)

Short Answer

Expert verified
Each half has a force constant of \( 2k \), each third has a constant of \( 3k \).

Step by step solution

01

Understanding Spring's Force Constant

The force constant, also known as spring constant, of a spring is a measure of its stiffness. For a spring with a constant \( k \), when it is stretched or compressed by a distance \( x \), it exerts a force \( F = kx \).
02

Analyze the Full Spring

Imagine the original spring as composed of two equal halves when it is intact. The force produced by the full spring can be expressed as the sum of the forces exerted by each of these halves. So, for the whole spring, \( F = kx \).
03

Analyze Cutting Spring in Two Halves

When the spring is cut into two equal halves, each half exerts the same force \( F = kx \) to maintain the same stiffness as the original spring. However, because these halves are shorter, each needs a greater force constant to exert the same force: \( F = 2kx_{1/2} \), thus the new force constant for each half becomes \( 2k \).
04

Verify by Equating Forces

For each half-spring to produce the same stiffness as the whole spring, \( 2k \cdot \frac{x}{2} = kx \). This ensures that the half-spring needs a force constant of \( 2k \).
05

Cutting Spring into Three Equal Segments

Similarly, if the spring is divided into three equal segments, the stiffness needed for each segment to maintain the total stiffness of the original is increased due to the reduced length. The force exerted by each needs to be \( 3k \cdot \frac{x}{3} = kx \). Hence the force constant for each segment is \( 3k \).
06

Conclude with Results

For each half of the spring, the force constant is \( 2k \) and for each third of the spring, it is \( 3k \). This is consistent with the understanding that a shorter spring has a higher force constant.

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 91Ó°ÊÓ!

Key Concepts

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

Hooke's Law
Hooke's Law is a fundamental principle in physics that describes the behavior of springs. It defines the relationship between the force exerted by a spring and the amount it is stretched or compressed.

This relationship is expressed with the formula:
  • \(F = kx\)
where:
  • \(F\) is the force exerted by the spring,
  • \(k\) is the spring force constant, and
  • \(x\) is the displacement from the spring's natural, unstressed length.
Hooke's Law tells us that the force produced by a spring is directly proportional to its displacement.
This means that as you stretch or compress a spring more, the force it exerts increases. Understanding this concept is crucial, as it lays the foundation for analyzing any spring system, whether it's a toy, a vehicle suspension, or the spring in a pen.
Spring Stiffness
Spring stiffness, quantified by the spring force constant \(k\), is a measure of a spring's resistance to deformation. A spring with a high stiffness will require more force to stretch or compress than a spring with lower stiffness.

When a spring is cut into segments, its stiffness changes. For example, if you cut a spring into two halves:
  • Each half will be shorter and thus stiffer.
  • The force constant \(k\) of each half becomes \(2k\).
This happens because a shorter spring requires more force to achieve the same displacement as the original longer spring.
This concept reflects the calculation method described when cutting a spring into multiple equal parts, showing why the force constant increases with shorter segments.
Understanding spring stiffness is key for tasks such as designing mechanical systems, where precise control over force and movement is needed.
Ideal Spring
An ideal spring is a theoretical model that assumes the spring follows Hooke's Law perfectly without any deviations. This means:
  • There is no energy loss due to heat or internal friction.
  • The spring will not deform permanently, regardless of how much it's compressed or stretched, as long as it's within its elastic limit.
This idealization simplifies the study of springs and helps predict their behavior under different forces and displacements.

In real-world applications, all springs deviate slightly from the ideal spring model due to materials' imperfections and operating conditions.
However, understanding an ideal spring is vital because it provides a baseline from which real-world springs can be analyzed, allowing engineers and scientists to make necessary adjustments and predict behaviors in more complex 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

Working Like a Horse. Your job is to lift 30 -kg crates a vertical distance of 0.90 m from the ground onto the bed of a truck. (a) How many crates would you have to load onto the truck in 1 minute for the average power output you use to lift the crates to equal 0.50 \(\mathrm{hp} ?\) (b) How many crates for an average power output of 100 \(\mathrm{W} ?\)

A 5.00 -kg block is moving at \(v_{0}=6.00 \mathrm{m} / \mathrm{s}\) along a frictionless, horizontal surface toward a spring with force constant \(k=500 \mathrm{N} / \mathrm{m}\) that is attached to a wall (Fig. P6.85). The spring has negligible mass. \begin{equation} \begin{array}{l}{\text { (a) Find the maximum distance the spring will be compressed. }} \\ {\text { (b) If the spring is to compress by no more than } 0.150 \mathrm{m}, \text { what }} \\\ {\text { should be the maximum value of } v_{0} ?}\end{array} \end{equation}

CPA small block with a mass of 0.0900 \(\mathrm{kg}\) is attached to a cord passing through a hole in a frictionless, horizontal surface (Fig. P6.75). The block is originally revolving at a distance of 0.40 \(\mathrm{m}\) from the hole with a speed of 0.70 \(\mathrm{m} / \mathrm{s} .\) The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.10 \(\mathrm{m} .\) At this new distance, the speed of the block is observed to be 2.80 \(\mathrm{m} / \mathrm{s}\) . (a) What is the tension in the cord in the original situation when the block has speed \(v=0.70 \mathrm{m} / \mathrm{s} ?\) (b) What is the tension in the cord in the final situation when the block has speed \(v=2.80 \mathrm{m} / \mathrm{s} ?\) (c) How much work was done by the person who pulled on the cord?

Rescue. Your friend (mass 65.0 \(\mathrm{kg} )\) is standing on the ice in the middle of a frozen pond. There is very litle friction between her feet and the ice, so she is unable to walk. Fortunately, a light rope is tied around her waist and you stand on the bank holding the other end. You pull on the rope for 3.00 s and accelerate your friend from rest to a speed of 6.00 \(\mathrm{m} / \mathrm{s}\) whileyou remain at rest. What is the average power supplied by the force you applied?

A ski tow operates on a \(15.0^{\circ}\) slope of length 300 \(\mathrm{m} .\) The rope moves at 12.0 \(\mathrm{km} / \mathrm{h}\) and provides power for 50 riders at one time, with an average mass per rider of 70.0 \(\mathrm{kg} .\) Estimate the power required to operate the tow.

See all solutions

Recommended explanations on Physics Textbooks

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.