Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

INT

A vertical spring with mass is at equilibrium. The mass is then pushed up so that the spring is unstretched. The mass is dropped from this distance d from the equilibrium point. prove that the maximum distance that the spring stretches is 2d.

  • one year ago
  • one year ago

  • This Question is Closed
  1. furnessj
    Best Response
    You've already chosen the best response.
    Medals 0

    Is it definiely 2d and not \[\sqrt{2} d\]? If you use GPE as \[mgd\] then the systems total energy can be written as \[mgd = \frac{ 1 }{ 2 }mv ^{2} + \frac{ 1 }{ 2 }kx ^{2}\]. When the spring was stretched, at equilibrium position, the Force, mg caused the spring to stretch to d. So \[F = mg = kd\] (I.e Hookes law). Now, to get the maximum x, the velocity will be zero (spring fully stretched out). This means the KE is zero in the big equation, plus use Hooke's law to substitute out the k, then solve for x. I get sqrt(2) x d...

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.