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Homework Help: Mass on Vertical Spring stretch

  1. Jun 8, 2010 #1
    1. The problem statement, all variables and given/known data
    A spring with spring constant k = 55 N/m and unstretched length of L0 is attached to the ceiling. A block of mass m = 3 kg is hung gently on the end of the spring.

    a) How far does the spring stretch?
    ----------ANSWER: .53509 m

    This is the hard part...

    **Now the block is pulled down until the total amount the spring is stretched is twice the amount found in part (a). The block is then pushed upward with an initial speed vi = 2 m/s.

    2. Relevant equations

    KE = (1/2) mv^2
    KE = (1/2) kx^2
    KE + PE = 0

    3. The attempt at a solution

    I tried simply doubling the distance found in part a to get x=1.07018, and then plugged that into the formula KE = 1/2 kx^2 to get KE = 31.495.

    I then used that value in KE = 1/2 mv^2 and solved for v to get v=4.58, which is wrong. I also tried to add 2 to that to get v=6.58, but thats wrong too.

    How do I incorporate the Vi = 2 into this?
     
  2. jcsd
  3. Jun 8, 2010 #2

    rock.freak667

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    What are you trying to find?
     
  4. Jun 8, 2010 #3
    oh oops...
    b) What is the maximum speed of the block?
     
  5. Jun 8, 2010 #4

    rock.freak667

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    ah I see. In that case, initially it is stretched to x=1.07018 m, so how much PE is there? Since it is given an initial velocity as well, there is an initial KE as well.How much is this PE?

    The total energy is the sum of the initial PE and KE, at the maximum velocity, v, what should the final PE be?
     
  6. Jun 8, 2010 #5
    so the initial KE would be 1/2 * mv^2, which would give me 6.
    and the initial PE i found to be 31.495 using 1/2 kx^2, with x as 1.07018.

    so i set total Energy (37.495) = 1/2 mv^2 to find the max velocity, and i got v=4.9996, which is still incorrect.

    hmm..
     
  7. Jun 8, 2010 #6

    rock.freak667

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    What is the correct answer?
     
  8. Jun 8, 2010 #7
    They don't give it, unfortunately.
     
  9. Jun 8, 2010 #8

    rock.freak667

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    I think I made an error as to where the equilibrium position is, if you have the spring, and then hang the mass it goes down a distance δ. If you pull it down another distance δ, the spring is stretched 2δ, but from the equilibrium position with the mass, the extension is just δ (0.53509 m). Try it again.
     
  10. Jun 8, 2010 #9
    okay, so i got PE initial = 1/2 kx^2 = 7.8738
    and KE initial = 1/2 mv^2 = 6
    total energy = 13.8738

    so at max velocity, 13.8738 = 1/2 mv^2 (all KE)
    which would give me 3.041 m/s as max velocity.

    That works!
    Thanks a ton, I never would have realized the equilibrium position stayed the same.
     
  11. Jun 8, 2010 #10

    rock.freak667

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    Yep, I was confusing the equilibrium position of the mass with the position of the spring.
     
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