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Homework Help: Minimum Mass required to lift up an object from a spring

  1. Aug 8, 2017 #1
    1. The problem statement, all variables and given/known data
    FInd min mass so that block loses contact with ground. Untitled-1.jpg
    Spring is initially in its natural length, the mass attached to spring is on ground(M)
    2. Relevant equations
    Equations of conservation of energy

    3. The attempt at a solution
    Min compression = 2mg/k to lift up object, Normal rxn=0
    ⇒ m=m1

    Can someone tell me if im right or not
  2. jcsd
  3. Aug 8, 2017 #2
    Which mass is on the ground? Is the other mass suddenly released, or is it lowered gradually?
  4. Aug 8, 2017 #3
    Spring is initially in its natural length, the mass attached to spring is on ground
  5. Aug 8, 2017 #4
    Then m1 is suddenly released?
  6. Aug 8, 2017 #5
    Yes sir
  7. Aug 8, 2017 #6
    Then your answer is not correct. From your equation, what is kx equal to?
  8. Aug 8, 2017 #7
    k is spring constant and x is displacement from its natural length
    1/2kx^2 is potential energy stored in spring which is equal to gravitational potential which is M1gx
  9. Aug 8, 2017 #8
    if im wrong please correct my mistake :)
  10. Aug 8, 2017 #9
    Like I said, from your equation, what is kx equal to?
  11. Aug 8, 2017 #10
    i didnt understand your question
  12. Aug 8, 2017 #11
    The equation ##1/2kx^2=m_1gx## is correct. What does this give you for kx?
  13. Aug 8, 2017 #12
    Therefore, M=M1
  14. Aug 8, 2017 #13
    That's not what I get. I get $$kx=Mg=2m_1g$$
    So, ##M=2m_1##
  15. Aug 8, 2017 #14
    The compression in spring should be 2mg/k in order to lift it up, therefore i replaced x by 2mg/k
  16. Aug 8, 2017 #15
    Force balance on M: $$N+kx=Mg$$where N is the upward normal force exerted by the table on M. When N = 0 (M loses just loses contact with the table),
  17. Aug 8, 2017 #16
    so answer should be M/2?
  18. Aug 8, 2017 #17
    Sure. Can you say in words what is happening here physically?
  19. Aug 8, 2017 #18
    You mean more clear question statement?
  20. Aug 8, 2017 #19
    No. I mean mechanistically the reason that, in order for M to lose contact, ##m_1\geq M/2##. Certainly, if m1 were lowered slowly, the condition for loss of contact would be ##m_1\geq M##. So why, in the case where ##m_1## is released quickly, can ##m_1## weigh less and still cause M to lose contact even if it is only half of M? What is the motion of ##m_1## like?
  21. Aug 8, 2017 #20
    If it is released from rest it will produce potential energy energy in the spring, which will further be transferred to our original block M
    and if it is lowered slowly change in potential energy will be negligible
  22. Aug 8, 2017 #21
    Here's my spin on this. After you release m1 quickly, it begins falling, but then overshoots the equilibrium displacement by a factor of 2. So the tension in the cord and spring when m1 reaches its lowest point will be 2x the value at the equilibrium displacement. If, on the other hand, m1 is lowered slowly (say by hand), it will stop moving downward when it is at 1x the equilibrium displacement.
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