/*! 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 20 How far must you stretch a sprin... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 20

How far must you stretch a spring with \(k=1000 \mathrm{N} / \mathrm{m}\) to store \(200 \mathrm{J}\) of energy?

Short Answer

Expert verified
The stretch required to store 200 J of energy in the spring is roughly \(x = \sqrt{0.4} \, \mathrm{m}\) or approximately 0.632 meters.

Step by step solution

01

Identify the given values

The given values are the spring constant \(k = 1000 \mathrm{N}/\mathrm{m}\) and the desired stored energy \(U = 200 \mathrm{J}\).
02

Apply the formula for energy stored in a spring

The principle of potential energy stored in a spring can be described by \(U = \frac{1}{2} k x^2\), where \(U\) is the stored energy, \(k\) is the spring constant, and \(x\) is the stretch or compression of the spring. You need to solve this equation for \(x\).
03

Solve for x

You can solve for \(x\) by rearranging the formula to: \(x = \sqrt{\frac{2U}{k}}\). Substituting \(k = 1000 \mathrm{N}/\mathrm{m}\) and \(U = 200 \mathrm{J}\) into the equation gives:\(x = \sqrt{\frac{2 \times 200 \mathrm{J}}{1000 \mathrm{N}/\mathrm{m}}}\)
04

Calculate

The calculation gives \(x = \sqrt{\frac{400}{1000}} = \sqrt{0.4} \, \mathrm{m}\). This is the stretch required to store the given energy in the spring.

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 boy reaches out of a window and tosses a ball straight up with a speed of \(10 \mathrm{m} / \mathrm{s}\). The ball is \(20 \mathrm{m}\) above the ground as he releases it. Use energy to find a. The ball's maximum height above the ground. b. The ball's speed as it passes the window on its way down. c. The speed of impact on the ground.

At what speed does a \(1000 \mathrm{kg}\) compact car have the same kinetic energy as a \(20,000 \mathrm{kg}\) truck going \(25 \mathrm{km} / \mathrm{hr} ?\)

It's your birthday, and to celebrate you're going to make your first bungee jump. You stand on a bridge 100 m above a raging river and attach a 30 -m-long bungee cord to your harness. A bungee cord, for practical purposes, is just a long spring, and this cord has a spring constant of 40 N/m. Assume that your mass is 80 kg. After a long hesitation, you dive off the bridge. How far are you above the water when the cord reaches its maximum elongation?

A block of mass \(m\) slides down a frictionless track, then around the inside of a circular loop-the-loop of radius \(R\) From what minimum height \(h\) must the block start to make it around the loop without falling off? Give your answer as a multiple of \(R\)

A 10 kg box slides 4.0 m down the frictionless ramp shown then collides with a spring whose spring constant is \(250 \mathrm{N} / \mathrm{m}\) a. What is the maximum com- pression of the spring? b. At what compression of the spring does the box have its maximum speed?
See all solutions

Recommended explanations on Physics Textbooks

Modern Physics

Read Explanation

Electromagnetism

Read Explanation

Particle Model of Matter

Read Explanation

Thermodynamics

Read Explanation

Electricity

Read Explanation

Quantum 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.