/*! 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 40 An elastic string of unstretched... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An elastic string of unstretched length \(I\) and force constant \(k\) is stretched by a small length \(x .\) It is further stretched by another small length \(y\). The work done in the second stretching is (a) \(\frac{1}{2} k x^{2}\) (b) \(\frac{1}{2} k\left(x^{2}+y^{2}\right)\) (c) \(\frac{1}{2} k y(2 x+y)\). (d) \(\frac{1}{2} k x(x+2 y)\)

Short Answer

Expert verified
The correct answer is (c): \(\frac{1}{2} ky(2x+y)\).

Step by step solution

01

Understanding Elastic Potential Energy

The work done on an elastic string while stretching it corresponds to the elastic potential energy stored in it. This energy, for a small additional stretch, can be expressed using the formula for change in elastic potential energy.
02

Work Done Formula

The work done during an additional stretch from length \(x\) to \(x+y\) is given by the change in elastic potential energy: \( W = \Delta U = U(x+y) - U(x) \).
03

Elastic Potential Energy Calculations

The elastic potential energy for a stretch \(s\) of a string or spring is given by \( U = \frac{1}{2} k s^2 \).For the initial stretch of \(x\): \[ U(x) = \frac{1}{2} k x^2 \].For the final stretch \(x + y\):\[ U(x+y) = \frac{1}{2} k (x+y)^2 \].
04

Calculate \(U(x+y) - U(x)\)

Substitute elastic potential energies into the work done formula:\[ W = \frac{1}{2} k (x+y)^2 - \frac{1}{2} k x^2 \]Expanding the square:\[ (x+y)^2 = x^2 + 2xy + y^2 \]So,\[ W = \frac{1}{2} k (x^2 + 2xy + y^2) - \frac{1}{2} k x^2 \]This simplifies to:\[ W = \frac{1}{2} k (2xy + y^2) \]
05

Identifying the Correct Option

The expression obtained corresponds to option (c), which is \( \frac{1}{2} k y (2x+y) \).

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.

Work Done
The concept of "work done" in the context of elastic strings is about understanding how much energy we spend to change the state of the string. When you stretch an elastic string, you do work against the restoring force of the string. This work gets stored as elastic potential energy.

Work done can be calculated by considering the change in the potential energy of the spring or elastic string. For small stretches, this means that each additional stretch requires more work, depending on the previous length and the amount of stretch.

The work done when stretching an elastic string from one position to another is given by the formula: \[ W = \Delta U = U(x+y) - U(x) \]Here:
  • \( W \) is the work done.
  • \( \Delta U \) represents the change in elastic potential energy.
  • \( U(x+y) \) and \( U(x) \) are the potential energies at different stretches.
This formula helps us understand how much extra energy is put into the system as you stretch a string or spring further.
Force Constant
The force constant, represented as \( k \), is a measure of the stiffness of the elastic string or spring. It represents how much force is needed to stretch or compress the string by a unit length.

A higher force constant means the string is stiffer and requires more force to stretch. Conversely, a smaller force constant suggests that the string is more flexible.

The formula for elastic potential energy considering the force constant is as follows:\[ U = \frac{1}{2} k s^2 \]Where:
  • \( U \) is the elastic potential energy.
  • \( k \) is the force constant.
  • \( s \) is the amount of stretch or compression.
Understanding the force constant allows us to predict the behavior of different materials under stretch and determine the amount of energy stored at different deformations.
Stretching an Elastic String
When you stretch an elastic string beyond its original length, you need to consider the energy changes involved. Initially, an elastic string is at rest with no stretch, and as you pull it, it resists with a force that depends on how much it's stretched.

Every time you stretch it by a small length, you do additional work, which accumulates in the form of elastic potential energy. For a complete stretch from an initial length \( x \) to a final length \( x + y \), the extra stretch \( y \) adds new energy to the system.

Let's break down the elastic potential energy formula in this context:
  • Initial stretch energy: \[ U(x) = \frac{1}{2} k x^2 \]
  • Final stretch energy: \[ U(x+y) = \frac{1}{2} k (x+y)^2 \]
Expanding the final stretch gives us the energy change:\[ W = \frac{1}{2} k (x^2 + 2xy + y^2) - \frac{1}{2} k x^2 \]This simplifies to:\[ W = \frac{1}{2} k (2xy + y^2) \]It's essential to recognize the incremental increase in energy with additional stretching, reflecting how a once-simple tension grows more complex with each additional pull.

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

The work done by a force \(F=k x^{2}\) acting on a particle at an angle \(60^{\circ}\) with \(x\) -axis to displace it from \(x=2 \mathrm{~m}\) to \(x\) \(=3 \mathrm{~m}\) is (a) \(\frac{5}{6} k\) (b) \(\frac{5}{2} k\) (c) \(\frac{19}{2} k\) (d) \(\frac{19}{6} k\)

A stone of mass \(1 \mathrm{~kg}\) tied to a light inextensible string of length \(L=10 / 3 \mathrm{~m}\) is whirling in a circular path of radius \(L\), in a vertical plane. If the ratio of the maximum tension in the string to the minimum tension is 4 and if \(g\) is taken to be \(10 \mathrm{~m} / \mathrm{s}^{2}\), the speed of the stone at the highest point of the circle is (a) \(20 \mathrm{~m} / \mathrm{s}\) (b) \(10 \sqrt{3} \mathrm{~m} / \mathrm{s}\) (c) \(5 \sqrt{2} \mathrm{~m} / \mathrm{s}\) (d) \(10 \mathrm{~m} / \mathrm{s}\)

A person with a mass of \(M \mathrm{~kg}\) stands in contact against the wall of a cylindrical drum of radius \(r\) rotating with an angular velocity \(\omega\). If the coefficient of friction between the wall and the clothing is \(\mu\), the minimum rotational speed of the cylinder which enables the person to remain stuck to the wall when the floor is suddenly removed, is (a) \(\omega_{\min }=\sqrt{\frac{g}{\mu r}}\) (b) \(\omega_{\min }=\sqrt{\frac{\mu r}{g}}\) (c) \(\omega_{\min }=\sqrt{\frac{2 g}{\mu r}}\) (d) \(\omega_{\min }=\sqrt{\frac{r g}{\mu}}\)

Three identical cars \(A, B\) and \(C\) are moving at the same speed on three bridges. The car \(A\) goes on a plane bridge \(B\) on a bridge convex upwards and \(C\) goes on a bridge concave upwards. Let \(F_{A}, F_{B}\) and \(F_{C}\) be the normal forces exerted by the cars on the bridges when they are at the middle of the bridges. Then (a) \(F_{A}\) is maximum of the three forces (b) \(F_{B}\) is maximum of the three forces (c) \(F_{c}\) is maximum of the three forces (d) \(F_{A}=F_{B}=F_{C}\)

A body of mass \(3.0 \mathrm{~kg}\) moves under the influence of some external force such that its position \(s\) as a function of time \(t\) is given by \(s=6 t^{3}-t^{2}+1\) where \(s\) is in metres and \(t\) is in seconds. The work done by the force in first three seconds is (a) \(18 \mathrm{~J}\) (b) \(1800 \mathrm{~J}\) (c) \(3660 \mathrm{~J}\) (d) \(36504 \mathrm{~J}\)

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.