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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
    1/2kx2=m1gx
    ⇒ 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
    1/2kx2=m1gx
    ⇒kx=2m1g
    ⇒k2mg/k=2m1g
    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),
    $$kx=Mg$$
     
  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
     
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