Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 1347

When an elastic spring is given a displacement of \(10 \mathrm{~mm}\), it gains an potential energy equal to \(\mathrm{U}\). If this spring is given an additional displacement of \(10 \mathrm{~mm}\), then its potential energy will be.............. (A) \(\mathrm{U}\) (B) \(2 \mathrm{U}\) (C) \(4 \mathrm{U}\) (D) \(\mathrm{U} / 4\).

Short Answer

Expert verified
The potential energy when the spring is further displaced by 10 mm is 4 times the initial potential energy. So, the correct answer is (C) \(4 \mathrm{U}\).

Step by step solution

01

Write down Hooke's Law formula for potential energy stored in a spring.

Hooke's Law states that the potential energy (PE) stored in a spring is proportional to the square of its elongation (displacement) from the equilibrium position. The formula for the potential energy is: \[ PE = \frac{1}{2} k x^2 \] Where: PE = potential energy stored in the spring, k = spring constant (a factor specific to the spring), x = displacement of the spring from its equilibrium position.
02

Write the formula for the initial potential energy U.

Using Hooke's Law formula, we'll write down the potential energy (U) when the spring is displaced by 10 mm (0.01 m). \[ U = \frac{1}{2} k(0.01)^2 \]
03

Calculate the potential energy when the spring is further displaced by 10 mm.

Now, we need to find the potential energy when the spring is further displaced by another 10 mm (0.01 m). The total displacement now is 20 mm (0.02 m). Let's call the new potential energy as V. \[ V = \frac{1}{2} k(0.02)^2 \]
04

Simplify and compare the new potential energy V with the initial potential energy U.

Let's rewrite and simplify the equations for U and V: \[ U = \frac{1}{2} k(0.01)^2 = \frac{1}{2} k \times 0.0001 \] \[ V = \frac{1}{2} k(0.02)^2 = \frac{1}{2} k \times 0.0004 \] Now, we can see that V is 4 times greater than U: \[ V = 4U \]
05

Choose the correct answer option.

From the analysis, we have found that the potential energy when the spring is further displaced by 10 mm is 4 times the initial potential energy. So, the correct answer is: (C) \(4 \mathrm{U}\)

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Two wires made up of same material are of equal lengths but their radii are in the ratio \(1: 2\). On stretching each of these two strings by the same tension, the ratio between their fundamental frequency is \(\ldots \ldots \ldots .\) (A) \(1: 2\) (B) \(2: 1\) (C) \(1: 4\) (D) \(4: 1\)

For particles \(\mathrm{A}\) and \(\mathrm{B}\) executing S.H.M., the equation for displacement is given by $\mathrm{y}_{1}=0.1 \sin (100 \mathrm{t}+\mathrm{p} / 3)$ and \(\mathrm{y}_{2}=0.1\) cos pt respectively. The phase difference between velocity of particle \(\mathrm{A}\) with respect to that of \(\mathrm{B}\) is \(\ldots \ldots\) \((\mathrm{A})-(\pi / 3)\) (B) \((\pi / 6)\) (C) \(-(\pi / 6)\) (D) \((\pi / 3)\)

The ratio of frequencies of two waves travelling through the same medium is \(2: 5 .\) The ratio of their wavelengths will be.... (A) \(2: 5\) (B) \(5: 2\) (C) \(3: 5\) (D) \(5: 3\)

If the maximum frequency of a sound wave at room temperature is $20,000 \mathrm{~Hz}\( then its minimum wavelength will be approximately \)\ldots \ldots\left(\mathrm{v}=340 \mathrm{~ms}^{-1}\right)$ (A) \(0.2 \AA\) (B) \(5 \AA\) (C) \(5 \mathrm{~cm}\) to \(2 \mathrm{~m}\) (D) \(20 \mathrm{~mm}\)

Two masses \(m_{1}\) and \(m_{2}\) are attached to the two ends of a massless spring having force constant \(\mathrm{k}\). When the system is in equilibrium, if the mass \(\mathrm{m}_{1}\) is detached, then the angular frequency of mass \(m_{2}\) will be \(\ldots \ldots \ldots .\) (A) \(\sqrt{\left(\mathrm{k} / \mathrm{m}_{1}\right)}\) (B) \(\sqrt{\left(\mathrm{k} / \mathrm{m}^{2}\right)}\) (C) \(\sqrt{\left(k / m_{2}\right)+m_{1}}\) (D) \(\sqrt{\left\\{k /\left(m_{1}+m_{2}\right)\right\\}}\)
See all solutions

Recommended explanations on English Textbooks

Multiple Choice Questions

Read Explanation

Lexis and Semantics

Read Explanation

Pragmatics

Read Explanation

Creative Story

Read Explanation

Research and Composition

Read Explanation

English Grammar

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.

Sign-up for free