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Energy conservation with tension

  1. Oct 6, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-10-6_10-13-44.png

    2. Relevant equations
    Conservative of energy

    mg(y2-y1) +1/2 k (s2-s02) = 1/2 mv12 +1/2 mv22

    3. The attempt at a solution
    v1 = 0 at rest
    y2 = 0 bottom

    What I got is v2 = 8.20 m/s but not correct,
    I don't know how I can take into account the tension..
    Fspring = -ks = -4000 N/m * 0.2 m = 800 N at position 2, isn't it? why it says 500 N?
     
  2. jcsd
  3. Oct 6, 2016 #2

    TSny

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    In the formulas F = -ks and U = (1/2)ks2, does s represent the length of the spring or does it represent how much the spring is stretched?
     
  4. Oct 6, 2016 #3
    U is for how much the string stretches.
    And the force is also, but it does not say whether the string is unstretch when it is in position 2
     
  5. Oct 6, 2016 #4
    If it is of lowest tension in pos 2, then in pos 1, the tension and U will be higher than it usually is.
     
  6. Oct 6, 2016 #5

    TSny

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    If the spring were unstretched in position 2 then it would exert no tension force in position 2.
     
  7. Oct 6, 2016 #6
    you're right! and would it be 0.125 m for the length when it is unstretched? cuz I found that T= ks => 500 N= 4000 N/m * s
    SO s = 0.125 (the string is stretched for 0.125 m in pos 2)
     
  8. Oct 6, 2016 #7
    Oh no, should be 0.2 - 0.125 as s0 (unstretch)
     
  9. Oct 6, 2016 #8

    TSny

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    Just to be clear what you are saying, what are your answers for the following
    (1) What is the natural (unstretched) length of the spring?
    (2) How much is the spring stretched beyond its natural length in postion 2?
     
  10. Oct 6, 2016 #9
    (1) 0.075 m
    (2) 0.125 m

    and so in pos 1, the length stretched will be length of the spring - 0.075 m
    Correct ? Thanks
     
  11. Oct 6, 2016 #10

    TSny

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    Yes, that's right.
     
  12. Oct 7, 2016 #11
    Would the equation be:

    mgy1 + 1/2 mv12 + 1/2k(s12-s02) = mgy2 + 1/2 mv22+ 1/2k(s22-s02)

    For the bold part, will it be zero, cuz it is perpendicular to the motion?
     
  13. Oct 7, 2016 #12

    TSny

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    The expressions for the potential energy of the spring at positions 1 and 2 are not correct. When the spring is in position 2, the spring is stretched a certain amount from its natural length. So, the potential energy of the spring at position 2 is not zero.

    The fact that the spring is perpendicular to the motion at point 2 does not affect the potential energy at that point. The force of the spring is perpendicular to the motion at position 2, which means that at that instant the spring force is not doing any work. So, the potential energy of the spring is not changing at that one instant, but the spring does have nonzero potential energy at that instant.
     
  14. Oct 7, 2016 #13
    mgy1 + 1/2 mv12 + 1/2k(s12-s02) = mgy2 + 1/2 mv22+ 1/2k(s22-s02)

    So I can assume the height y2 be zero, and y1 be 0.25 m
    Given v1 is 0, v2 is what we are finding.

    But do we need to know the spring at unstretch state? Cuz we need the change of energy?
    ΔUweight + ΔUspring = ΔKE
    mg(y2 -y1) + 1/2 k (s22-s12) = 1/2 m (v22-v12) ??
    Loss in Uweight + Loss in Uspring = Gain in KE
     
  15. Oct 7, 2016 #14

    TSny

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    You are not calculating the potential energy of the spring correctly. Hooke's law is F = -kx and the potential energy of the spring is (1/2)kx2 where x is the amount by which the spring is stretched from its unstretched length. If s0 is the unstretched length of the spring and if s is the length of the spring when it is stretched, how would you express x in terms of s and s0? Then how would you express the potential energy in terms of s and s0,
     
  16. Oct 7, 2016 #15


    mgy1 + 1/2 mv12 + 1/2k(x12) = mgy2 + 1/2 mv22+ 1/2k(x22)
    where x1 is length of spring in pos 1 - unstretch
    x2 is length of spring in pos 2 - unstretch?
     
  17. Oct 7, 2016 #16
    So x1 = 0.245 and x2 = 0.125?
    where the equation now be
    mg(y2 - y1)+ 1/2k(x22-x12) = 1/2 mv22

    (4)(-9.81)(0-0.25) + 1/2 (-4000) (0.1252-0.2452) = 1/2 (4) v2

    9.81 +2000(0.444) = 2 v2
    v = 7.02 m/s

    Correct?
     
  18. Oct 7, 2016 #17

    TSny

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    You got the right answer, but there are sign errors in your first equation above.
    g and k are positive numbers.

    You wrote the energy equation correctly in post #15, but you did not rearrange it correctly when you solved for 1/2 mv22.
     
  19. Oct 7, 2016 #18
    So on the left hand side, 1 - 2 then on the right hand side 1/2 m (v2^2 - v1^2), this case 2 -1 ??
     
  20. Oct 7, 2016 #19

    TSny

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    Yes
     
  21. Oct 7, 2016 #20
    Great thanks!
    I think what I was off is the calculation of x1 and x2...
    I thought it was 1/2 k (S12 - S02) + the other energy = 1/2 k (S22 - S02) + other energy

    where S0 is the un-stretched distance
     
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