/*! 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 15 Two springs \(A\) and \(B\left(k... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 15

Two springs \(A\) and \(B\left(k_{A}=2 k_{B}\right)\) are stretched by applying forces of equal magnitudes at the ends. If the energy stored in \(A\) is \(E\), then energy stored in \(B\) is (A) \(\frac{E}{2}\) (B) \(2 E\) (C) \(E\) (D) \(\frac{E}{4}\)

Short Answer

Expert verified
\(U_B = 2 E\)

Step by step solution

01

Express the applied force in terms of the spring constants and displacements

We are given that the forces applied to both springs are equal. Hooke's Law states that \(F = kx\), where \(F\) is the force, \(k\) is the spring constant, and \(x\) is the displacement from the equilibrium position. For spring A, the force is given by \(F = k_A x_A\), and for spring B, the force is given by \(F = k_B x_B\). Since the forces are equal, we can equate the two expressions: \(k_A x_A = k_B x_B\)
02

Substitute the given relationship between the spring constants

We are also given that the spring constant of spring A is twice that of spring B: \(k_A = 2k_B\). Substituting this into the previous equation, we get: \(2k_B x_A = k_B x_B\)
03

Solve for the relationship between the displacements

Next, we need to solve the equation for the relationship between the displacements of the springs: \(\frac{x_A}{x_B} = \frac{k_B}{2k_B}\) Simplifying, we get: \(x_A = \frac{1}{2}x_B\)
04

Use the energy formula to find the energy stored in spring A and B

Now, we will use the formula for the energy stored in a spring, \(U = \frac{1}{2}kx^2\). The energy stored in spring A is given by: \(U_A = \frac{1}{2}k_A x_A^2\) The energy stored in spring B is given by: \(U_B = \frac{1}{2}k_B(x_B)^2\)
05

Substitute the relationships between the spring constants and displacements

Now, we substitute the relationships we found in Steps 2 and 3 into the energy formulas: \(U_A = \frac{1}{2}(2k_B)(\frac{1}{2}x_B)^2\) \(U_B = \frac{1}{2}k_B (x_B)^2\)
06

Compare the energy stored in spring A and B

Let's simplify and compare the expressions of the energy stored in spring A and B: \(U_A = k_B(\frac{1}{4}x_B^2)\) \(U_B = \frac{1}{2}k_B (x_B)^2\) Now, we can express the energy stored in spring A in terms of the energy stored in spring B: \(\frac{U_A}{U_B} = \frac{k_B(\frac{1}{4}x_B^2)}{\frac{1}{2}k_B (x_B)^2} = \frac{1}{2}\) Therefore, the energy stored in spring B is twice the energy stored in spring A: \(U_B = 2 U_A\) So, the correct answer is (B) \(2 E\).

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

A block of mass \(m\) is pulled by a constant power \(P\) placed on a rough horizontal plane. The friction co-efficient between the block and surface varies with its speed \(v\) as \(\mu=\frac{1}{\sqrt{1+v}} .\) The acceleration of the block when its speed is \(3 \mathrm{~m} / \mathrm{s}\) will be (A) \(\frac{P}{3 m}-\frac{g}{2}\) (B) \(\frac{P}{3 m}+\frac{g}{2}\) (C) \(\frac{P}{3 m}\) (D) \(\frac{g}{2}\)

A uniform chain of length \(2 \mathrm{~m}\) is kept on a table such that a length of \(60 \mathrm{~cm}\) hangs freely from the edge of the table. The total mass of the chain is \(4 \mathrm{~kg}\). What is the work done in pulling the entire chain on the table? (A) \(12 \mathrm{~J}\) (B) \(3.6 \mathrm{~J}\) (C) \(7.2 \mathrm{~J}\) (D) \(1200 \mathrm{~J}\)

A particle of mass \(2 \mathrm{~kg}\) starts moving in a straight line with an initial velocity of \(2 \mathrm{~m} / \mathrm{s}\) at a constant acceleration of \(2 \mathrm{~m} / \mathrm{s}^{2}\). The rate of change of kinetic energy is (A) Four times the velocity at any moment. (B) Two times the displacement at any moment. (C) Four times the rate of change of velocity at any moment. (D) Constant throughout.

The co-efficient of friction between the block and plank is \(\mu\) and its value is such that block becomes stationary with respect to plank before it reaches the other end. Then (A) the work done by friction on the block is negative. (B) the work done by friction on the plank is positive. (C) the net work done by friction is negative. (D) net work done by the friction is zero.

Two blocks \(A\) and \(B\) of equal mass \(m=1 \mathrm{~kg}\) are lying on a smooth horizontal surface as shown in Fig. \(4.27\). A spring of force constant \(K=200 \mathrm{~N} / \mathrm{m}\) is fixed at one end of block \(A\). Block \(B\) collides with another end of the spring with velocity \(v_{0}=2 \mathrm{~m} / \mathrm{s}\). What will be the maximum compression of the spring? [in decimeter \(]\)
See all solutions

Recommended explanations on Physics Textbooks

Modern Physics

Read Explanation

Scientific Method Physics

Read Explanation

Fundamentals of Physics

Read Explanation

Space Physics

Read Explanation

Fluids

Read Explanation

Rotational Dynamics

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.