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Spring and energy problem

  1. Jun 30, 2015 #1
    1. The problem statement, all variables and given/known data
    The block in the figurelies on a horizontal frictionless surface, and the spring constant is 44 N/m. Initially, the spring is at its relaxed length and the block is stationary at position x = 0. Then an applied force with a constant magnitude of 2.0 N pulls the block in the positive direction of the x axis, stretching the spring until the block stops. When that stopping point is reached, what are (a) the position of the block, (b) the work that has been done on the block by the applied force, and (c) the work that has been done on the block by the spring force? During the block's displacement, what are (d) the block's position when its kinetic energy is maximum and (e) the value of that maximum kinetic energy?

    http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/ch0/33.gif



    2. Relevant equations
    Fs=-kx , k=x2/2



    3. The attempt at a solution
    i'm trying to first solve for final x postion xf so that i can find work. which will be equivalent to final kinetic energy.
    20150630_171115_zpsxlshgqpp.jpg
     
  2. jcsd
  3. Jun 30, 2015 #2

    Nathanael

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    When the block is at x=2/45 meters then the acceleration of the block will be zero, but the problem asks for where the block stops, which implies the velocity is zero.
     
  4. Jun 30, 2015 #3

    SammyS

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    You really should explain your reasoning for the various steps you're taking.

    I see you solved an equation to find some value for x. What is significant about this value you found for x.
     
  5. Jun 30, 2015 #4
    K=1/2mv^2 where v=0 . but i was not given mass. so i cannot solve for K here.
     
  6. Jun 30, 2015 #5
    i erased that equation. i realized that Fs is the spring force not the constant force, so it does not apply to hookes law equation.
     
  7. Jun 30, 2015 #6

    Nathanael

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    Does the mass make a difference?
     
  8. Jun 30, 2015 #7
    i guess not. so Kf=0 also
     
  9. Jun 30, 2015 #8
    i dont see how to relate the kinetic energies to solving the displacement of the object.
     
  10. Jun 30, 2015 #9

    Nathanael

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    If the kinetic energy is zero, what happened to all the energy?
     
  11. Jun 30, 2015 #10
    its all potential energy?
     
  12. Jun 30, 2015 #11

    Nathanael

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    Talk to me in math :smile:
     
  13. Jun 30, 2015 #12
    U=potential engery=work=∫F dx
    work=2d for the 2N force
    work=∫Fsdx, is Fs=44N/m here? i thought that was the sprink constant k in hookes law?
     
  14. Jun 30, 2015 #13

    Nathanael

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    Right, this is the work done by each force. For KE to be zero, they must ____ ?


    On one side you have units of newtons and on the other side the units are newtons/meter, so thats a dead giveaway that the equation can't be correct.
     
  15. Jun 30, 2015 #14
    ok, that wouldnt make sense.
    so we have two work equations which are simultaneous equations at final x position? it wouldnt make sense to set them equal to eachother though.
     
  16. Jun 30, 2015 #15

    Nathanael

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    Well in order for the kinetic energy to be zero, they must add to zero (so that way the net work done on the block is zero).

    Or you can look at it in different words: (but the idea is the same)
    The work done by the external force must equal the change in (KE+PE) of the system. (PE is potential energy from the spring.)
     
  17. Jun 30, 2015 #16
    i got -1/11 but the correct answer is positive. did i do the math right?
     
  18. Jun 30, 2015 #17

    Nathanael

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    It should be positive. Can you show your work?
     
  19. Jun 30, 2015 #18
    2d=-22d2
    0=-22d2 - 2d
    0= 22d2 +2d
    0= 2d(11d+1)
    0=11d+1
    -1=11d
    d=-1/11
     
  20. Jun 30, 2015 #19

    Nathanael

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    From the equation Wspring+Wexternal.force=0 you get Wspring=-Wexternal.force
    However the force of the spring acts in the opposite direction as the displacement, thus it is doing negative work (so your equation is missing an extra negative sign)

    Other than this you're good.
     
  21. Jun 30, 2015 #20
    ok thanks.
    i dont understand what part d is asking, because the kinetic energy is zero.
     
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