1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Finding the value of k (spring constant)

  1. Dec 10, 2014 #1
    During a test experiment, the engineer finds an ideal very light spring has already been stretched 1.4m from its original length. He also finds that he needs to perform 100J of work to further stretch this spring an additional 1.5m. What is the value of the spring constant for this spring?

    2. Relevant equations

    W = F*d

    F = -kx

    So I wrote out that F1 = -kx to stretch the spring the initial 1.4m and then to stretch it the additional 1.5m it would take:

    F2 +F1 = -2.9*k

    I also wrote out:

    F1 = -1.5k

    Then I used 100 = F2*1.5

    so F2 = 66.67 although I'm not sure if you can use this equation for a spring

    Then 66.67 -1.5k = -2.9k and I tried solving for k but that didn't work so I'm not sure what else to try.
  2. jcsd
  3. Dec 10, 2014 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    There are two problems with that approach. First, your F2 is the additional force to get the extra 1.5m extension, but while extending it that 1.5m the total force applied is between F1and F1+F2. Therefore F1is an important part of the work done.
    Secondly, the force only reaches F1+F2 at full extension, so neither can you write 100 = (F1+F2)*1.5.

    In terms of k, what work was done to stretch it 1.4m? What total work was done to stretch it 1.4+1.5m?
  4. Dec 10, 2014 #3
    So W1 = 1/2k*(1.4)^2 = 0.98k
    W2 = 100
    WorkTotal = 1/2k(2.9)^2 = 4.205k

    Then I put it together 4.205k = 0.98k + 100J

    so k = 31 N/M ?

    Also, I thought 1/2kx^2 was for potential energy of a spring. So I'm kind of confused about the difference between the two now
  5. Dec 10, 2014 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Looks right.
    The work done to stretch a spring (from slack) equals the potential energy stored in the spring (assuming no losses). Where's the confusion?
  6. Dec 10, 2014 #5
    Thanks. I guess I just need to review this chapter some more
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted