How far must you stretch a spring with k = 1000 N/m to store 200 J of energy?

Short Answer

Expert verified

As the all we solve above from that we have , the spring's stretched distance is0.63m.

Step by step solution

01

Introduction

You must exert effort in order to compress or stretch a spring. You must apply a force to the spring that is the same magnitude as the force the spring applies to you, but in the opposite direction. Fext = kx is the force you apply in the direction of the displacement x from its equilibrium position.

02

Explanation

In a compressed or extended spring, the potential energy stored is,

Us=12k(Δs)2

We have, k= spring constant andΔsis the length of the spring compressed or stretched.

Rewriting equation Δs,

Δs=2Usk

Putting 200Jfor Usand 1000N/mfor k, solve for Δs.

Δs=2(200J)1000N/m

=0.63m

As the all we solve above from that we have , the spring's stretched distance is 0.63m.

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 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

In a physics lab experiment, a compressed spring launches a 20gmetal ball at a 30°angle. Compressing the spring 20cmcauses the ball to hit the floor 1.5mbelow the point at which it leaves the spring after traveling 5.0mhorizontally. What is the spring constant?

A system in which only one particle can move has the potential energy shown in FIGUREEX10.31. What is the y-component of the force on the particle at y=0.5mand 4m?

In Problems 66through68 you are given the equation used to solve a problem. For each of these, you are to

a. Write a realistic problem for which this is the correct equation.

b. Draw the before-and-after pictorial representation.

c. Finish the solution of the problem.

12(0.20kg)(2.0m/s)2+12k(0m)2=12(0.20kg)(0m/s)2+12k(-0.15m)2

Protons and neutrons (together called nucleons) are held together in the nucleus of an atom by a force called the strong force. At very small separations, the strong force between two nucleons is larger than the repulsive electrical force between two protons—hence its name. But the strong force quickly weakens as the distance between the protons increases. A well-established model for the potential energy of two nucleons interacting via the strong force is

U=U01-e-x/x0

where x is the distance between the centers of the two nucleons, x0 is a constant having the value xo=2.0×10-15m, and Uo=6.0×10-11J.

Quantum effects are essential for a proper understanding of nucleons, but let us innocently consider two neutrons as if they were small, hard, electrically neutral spheres of mass 1.67×10-27kgand diameter 1.0×10-15m. Suppose you hold two neutrons 5.0×10-15mapart, measured between their centers, then release them. What is the speed of each neutron as they crash together? Keep in mind that both neutrons are moving.

You have a ball of unknown mass, a spring with spring constant 950N/m, and a meter stick. You use various compressions of the spring to launch the ball vertically, then use the meter stick to measure the ball’s maximum height above the launch point. Your data are as follows:

Use an appropriate graph of the data to determine the ball’s mass.

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.

Sign-up for free