/*! 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 113 A spring of spring constant \(5 ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 113

A spring of spring constant \(5 \times 10^{3} \mathrm{~N} / \mathrm{m}\) is stretched initially by \(5 \mathrm{~cm}\) from the unstretched position. Then the work required to stretch it further by another \(5 \mathrm{~cm}\) is \([\mathbf{2 0 0 3}]\) (A) \(12.50 \mathrm{~N} / \mathrm{m}\) (B) \(18.75 \mathrm{~N} / \mathrm{m}\) (C) \(25.00 \mathrm{~N} / \mathrm{m}\) (D) \(6.25 \mathrm{~N} / \mathrm{m}\)

Short Answer

Expert verified
The short answer is: The work required to stretch the spring further by another 5cm is \(18.75 \mathrm{~N} / \mathrm{m}\). The correct answer is (B) \(18.75 \mathrm{~N} / \mathrm{m}\).

Step by step solution

01

Identify the given values

We are given the spring constant (k) as \(5 \times 10^{3} \mathrm{~N} / \mathrm{m}\), the initial stretched length as \(5 \mathrm{~cm}\) and the length when the spring is stretched further by another \(5 \mathrm{~cm}\) as 10cm. Converted these lengths to meters: Initial length: 0.05m (since 1cm = 0.01m), Final length: 0.10m
02

Apply the formula for work done

We can use the formula for the work done on a spring, W = (1/2)k(x2^2 - x1^2), where k is the spring constant, x1 is the initial stretch length and x2 is the final stretch length. W = (1/2) * \(5 \times 10^{3} \mathrm{~N} / \mathrm{m}\) * (0.10^2 - 0.05^2)
03

Solve for the work done

Now, we can plug in the values and solve for the work done W: W = (1/2) * \(5 \times 10^{3} \mathrm{~N} / \mathrm{m}\) * (0.01 - 0.0025) W = (1/2) * \(5 \times 10^{3} \mathrm{~N} / \mathrm{m}\) * 0.0075 W = 0.5 * (5000) * 0.0075 W = 2500 * 0.0075 W = 18.75 The work required to stretch it further by another 5cm is \(18.75 \mathrm{~N} / \mathrm{m}\). The correct answer is (B) \(18.75 \mathrm{~N} / \mathrm{m}\).

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!

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

During inelastic collision between two bodies, which of the following quantities always remain conserved? (A) Total kinetic energy (B) Total mechanical energy (C) Total linear momentum (D) Speed of each body

A position-dependent force \(F=7-2 x+3 x^{2} N\) acts on a small body of mass \(2 \mathrm{~kg}\) and displaces it from \(x=0\) to \(x=5 \mathrm{~m} .\) The work done in joule is (A) \(70 \mathrm{~J}\) (B) \(270 \mathrm{~J}\) (C) \(35 \mathrm{~J}\) (D) \(135 \mathrm{~J}\)

A particle is moving with kinetic energy \(E\), straight up an inclined plane with angle \(\alpha\), the co-efficient of friction being \(\mu\). The work done against friction, up to when the particle comes to rest, is (A) \(\frac{E \mu \cos \alpha}{\sin \alpha+\mu \cos \alpha}\) (B) \(\frac{E \cos \alpha}{\sin \alpha+\mu \cos \alpha}\) (C) \(\frac{E}{\sin \alpha+\mu \cos \alpha}\) (D) \(\frac{E}{g(\sin \alpha+\mu \cos \alpha)}\)

The potential energy function for the force between two atoms in a diatomic molecule is approximately given as \(U(x)=\frac{a}{x^{12}}-\frac{b}{x^{6}}\), where \(a\) and \(b\) are constants and \(x\) is the distance between the atoms. If the dissociation energy of the molecule is \(D=\left[U(x=\infty)-U_{\text {at equilibrium }} D\right.\) is \(\quad\) [2010] (A) \(\frac{b^{2}}{2 a}\) (B) \(\frac{b^{2}}{12 a}\) (C) \(\frac{b^{2}}{4 a}\) (D) \(\frac{b^{2}}{6 a}\)

A bullet losses \(19 \%\) of its kinetic energy when passes through an obstacle. The percentage change in its speed is (A) Reduced by \(10 \%\) (B) Reduced by \(19 \%\) (C) Reduced by \(9.5 \%\) (D) Reduced by \(11 \%\)
See all solutions

Recommended explanations on Physics Textbooks

Force

Read Explanation

Physics of Motion

Read Explanation

Oscillations

Read Explanation

Kinematics Physics

Read Explanation

Thermodynamics

Read Explanation

Modern Physics

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.