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Work done by a spring

  1. Feb 11, 2012 #1
    1. The problem statement, all variables and given/known data

    We must apply a force of magnitude 80N to hold the block stationary at x = -2.0cm. From that position, we then slowly move the block so that our force does +4.0J of work on the spring-block system; the block is then again stationary. What is the block's position? (Two answers.)


    2. Relevant equations

    F = -kx

    W = (1/2)*k*xi^2 - (1/2)*k*xf^2


    3. The attempt at a solution

    80 = -k(-.02m)
    k = 4000 N/m

    With the spring constant, i plug it into the Work equation to find xf

    4 = .5*4000*(-.02^2) - .5*4000*xf^2

    3.2 = -2000xf^2
    -.0016 = xf^2

    And this is where I am stuck. The answer cannot be an imaginary number, but a real one. I know the equations are correct, so what am I doing wrong? Please explain.
     
  2. jcsd
  3. Feb 11, 2012 #2

    Doc Al

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    Staff: Mentor

    You want final minus initial, not the other way around.
     
  4. Feb 11, 2012 #3
    The problem is that, that is how the book presented us with the equation. It was initial minus final. I understand where you are coming from, but that's how it is.
     
  5. Feb 11, 2012 #4

    Doc Al

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    Staff: Mentor

    Then what does W stand for in your book? The work done by the spring? The work done by you will be opposite to that.
     
  6. Feb 11, 2012 #5
    Doc Al. So what you are telling me is that the Work of +4.0J is caused by me. And its positive because the direction and force are directed towards one way. But the force of a spring is always opposite the displacement. So the work of the spring is -4.0J because its moving one way, while the force is directed towards the opposite?
     
  7. Feb 11, 2012 #6

    Doc Al

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    Staff: Mentor

    Right.
     
  8. Feb 11, 2012 #7
    Thanks for your help Doc. I really appreciate it.
     
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