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Spring equilibrium problem

  1. Apr 29, 2009 #1
    1. The problem statement, all variables and given/known data

    A 2kg object is suspended from a vertical spring that has a constant of 180N/m.

    a) How far does the spring stretch from the unstrained length?

    b) If the object is now pulled downward an additional distance of 5cm. What external force was required to do so?

    c) When the object is released from rest, it begins to move upwards. Find the maximum speed that the object reaches and the maximum height above the release point at which the object releases direction.

    2. Relevant equations

    Fspring= kx

    3. The attempt at a solution

    I would be grateful if someone could check my answers to parts a & b and possibly steer me in the right direction on part c?

    a) Fspring= kx
    F=ma
    ma=kx
    x= ma/k
    =(2kg)(9.8 m/s2)/(180 N/m)
    = 0.1089m

    b) Fspring= kx
    =(180 N/m)(0.05cm)
    = 9 N
     
  2. jcsd
  3. Apr 29, 2009 #2

    rl.bhat

    User Avatar
    Homework Helper

    In the case b, find the elastic potential energy stored in the spring.
    After releasing when it crosses the position in case a, it will have the maximum kinetic energy and gravitational potential energy. Apply the conservation of energy to find the maximum velocity.
     
  4. Apr 29, 2009 #3
    Thanks for your help :smile:

    This is what I got so far...

    PEspring= 1/2kw2= 1/2(180 N/m)(0.05m)= 0.225 J

    ... but I'm having trouble understanding the rest.
     
  5. Apr 29, 2009 #4

    rl.bhat

    User Avatar
    Homework Helper

    At a position total energy = 1/2*m*v^2 + mgh
    Find vmax
     
  6. May 6, 2009 #5
    Thank you very much for your help!
     
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