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Mass on a vertical Spring: finding maximum speed

  1. Oct 15, 2014 #1
    1. The problem statement, all variables and given/known data

    A spring with spring constant k = 60 N/m and unstretched length of L0 = 2.2 is attached to the ceiling. A block of mass m = 1.5 kg is hung gently on the end of the spring.
    1)

    How far does the spring stretch?

    Answer is .245 m

    2)

    Now the block is pulled down until the total amount the spring is stretched is twice the amount found in the above question. The block is then pushed upward with an initial speed vi = 3.7 m/sec. What is the maximum speed of the block?
    2. Relevant equations

    F spring = -kx
    Wnc = change in ME



    3. The attempt at a solution

    I found the first part just by using force, with the force required by the spring to be equal to mg, then found x from there.

    For part 2, however, I'm having problems.

    I took x to be .245 m (from the first part) x 2 = .49 m

    Then, assuming the forces are all conservative, and that the velocity will be maximum at the spring's relaxed position, I used the formula:

    .5 mv02 + .5kx2 = .5 mvf2

    Solving for Vf, I get 4.82, which is incorrect. I've also tried doubling the entire spring, in case that's what the question was asking, and was still incorrect.
     
  2. jcsd
  3. Oct 15, 2014 #2
    So I figured it out, leaving my method of solution for anyone else who may come across this problem.

    I neglected work due to gravity. So what you can do is take the stretched position in problem 1 as the equilibrium position, then use the formula I used before

    .5 mv02 + .5kx2 = .5 mvf2

    except x is now the amount stretched for the already stretched position, in this case, .49 m. Then solve for Vf.


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