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Block on a spring problem

  1. Dec 8, 2006 #1
    The force constant of a spring is 400 N/m and the unstretched length is .65 m. A 2.0 kg block is suspended from the spring. An external force slowly pulls the block down, until the spring has been stretched to a length of .78 m. The external force is then removed, and the block rises. In this situation, when the spring has contracted to a length of .66 m, the acceleration of the block, including direction is closest to:
    (multiple choices given)

    2. Relevant equations
    I guess since it's a spring problem, the relevant equations would be potential energy=(1/2)kx^2 and maybe F=-kx.

    3. The attempt at a solution
    This problem is from a test I'm correcting so I know that the answer is 7.8 m/s^2 downward. However, I have no idea where to start. I considered that I would use (1/2)(400)(.78-.65)^2-(1/2)(400)(.66-.65)^2 to find the net work done, but I don't know where to go from there. I'm sorry my attempt at a solution is so minimal, but as I said before, I really don't know where to start. If this isn't sufficient enough please tell me.
    Last edited: Dec 8, 2006
  2. jcsd
  3. Dec 8, 2006 #2
    Stop me if I'm wrong, but if the spring's unstretched length is .65 m, and the spring only coils back to .66 m, shouldn't the block be accelerating upward?

    The final velocity has to be larger than the initial velocity, and when you divide that by the time it took to rise, you'll have a positive acceleration. You shouldn't obtain a negative acceleration until the block passes the .65 m mark.
  4. Dec 8, 2006 #3
    Well, I don't know what to say except that that was the answer my teacher gave me.
  5. Dec 8, 2006 #4
    Hmm. Maybe this is a difficult topic for teachers to grasp? Or at least explain. I'm having the same sort of problem with an ideal spring problem I posted this afternoon - have a look if you want.
  6. Dec 8, 2006 #5
    no, you guys are wrong--the spring will start slowing BEFORE .65--dont forget that it has weight. Therefore it will not naturally stay at .65, but rather at .699m. This is because that is where that force of the spring is equal to the weight.
  7. Dec 8, 2006 #6
    Oh, I see. For some reason when I read the problem I thought that the weight would hang at .65 m.
  8. Dec 8, 2006 #7


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    Newton 2 and a FBD usually works well. Here we have the weight of the 2 kG block acting down on the block, and the spring force, kx, acting up (the spring is still in tension because it is still slightly stretched by .01m beyond its equilibrium position). So x =.01 m. Solve for 'a' using F_net = ma. . What is the direction of the acceleration?
  9. Dec 9, 2006 #8
    I've got it now. Thanks everybody.

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