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Spring: work and speed

  1. Nov 27, 2007 #1
    Question:
    A 0.2 kg mass is attached to a spring k = 10 N/m) and hangs vertically near the earth's surface (g = 9.81 m/s2 ). The mass makes contact with a wall as it moves vertically and a constant frictional force of magnitude 5N acts on the mass as it moves.

    a) Calculate the amount of work required to pull the spring down by 1 m.
    b) Calculate the speed of the mass as it passes through the equilibrium position after being pulled down by 1 m.



    Eq'n
    U = (1/2)k x^2


    Attempt:

    a) W = U = (1/2)k x^2
    (1/2)(10 N/m) (1m)^2
    = 5 N*m
    = 5 J

    I am lost on part b, someone suggested

    v = sqrt(k/m) * x

    ... But I have no clue where they derived this equation...

    I tried
    K = 1/2 m v^2
    v = sqrt(2K/m)

    ...But I believe this is incorrect... suggestions?
     
  2. jcsd
  3. Nov 27, 2007 #2
    all the elastic potential energy is converted back into kinetic. so equate Eelastic - Ethermal = Ekinetic and solve. but you have to also consider energy lost due to friction, so put that in the equation too.
     
  4. Nov 27, 2007 #3
    wait... is this assuming that I did part a correctly? or...
     
  5. Nov 27, 2007 #4

    Astronuc

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    Staff: Mentor

    Part a must include the work associated with the frictional force of 5 N over 1 m, in addition to the mechanical energy in the spring.

    In part b, when the spring recoils, the friction is again present, so not all the spring energy will be transformed into kinetic energy.
     
  6. Nov 27, 2007 #5
    So I am assuming that its:

    W = Us + Ug + 5Nm

    However how would I find Ug without knowing it's height?
     
  7. Nov 27, 2007 #6

    Astronuc

    User Avatar

    Staff: Mentor

    The mass changes elevation by 1 m, from the equilibrium position, and since the mass is going down, it's gravitational potential energy (GPE) decreases. When the mass goes up, it's GPE increases.
     
  8. Nov 27, 2007 #7
    New attempt:
    Us = W = U = (1/2)k x^2
    (1/2)(10 N/m) (1m)^2
    = 5 N*m

    Ug = (0.1 kg)(1m)(-9.8 m/s^2)
    = -.98 J


    W= Us + Ug+ (-5J)
    5J + (-98 J) + (-5J)
    = -.98 J

    b) W = E - Wnc (Work energy Theorem for Systems)
    W = 0 Because there is no internal nonconservative forces.. I still havent a clue where to start with this one...
    Therefore E:
     
  9. Nov 27, 2007 #8

    Astronuc

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    Staff: Mentor

    OK, when the spring is being pulled down, the friction and spring force act in the same direction. Part of the work being done goes into the stored mechanical energy in the spring and part is used to overcome friction.

    When the spring recoils, the spring pulls (acts up) but friction is acting down.
     
  10. Dec 1, 2007 #9
    New attempt:

    Us = W = U = (1/2)k x^2
    (1/2)(10 N/m) (1m)^2
    = 5 N*m

    Ug = (0.1 kg)(1m)(-9.8 m/s^2)
    = -.98 J

    Uf = 5J


    W= Us - Ug + (5J)
    5J -.98 J + 5J
    = 9.8 J

    So far, is this correct?
     
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