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How do i work out the magnitude of a force

  1. Dec 15, 2009 #1
    how do i work out the magnitude of the force that has to be applied to compress a spring?
     
  2. jcsd
  3. Dec 15, 2009 #2

    berkeman

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    Welcome to the PF. You use the spring constant and Hooke's Law:

    http://en.wikipedia.org/wiki/Spring_(device [Broken])

    .
     
    Last edited by a moderator: May 4, 2017
  4. Dec 15, 2009 #3
    is it 0.5kxsquared or just -kx?
     
  5. Dec 15, 2009 #4

    berkeman

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    The restorative force is -kx. The energy stored in the spring is the 1/2 kx^2. What course is this for? Or is it for a project of yours?
     
  6. Dec 15, 2009 #5
    predicting motion course, struggling abit. i even worked out the spring constant on the question before. now its asking me to work out the speed which means combining equations at a guess. its an open university course, so alot of reading involved which doesn't suit me really. As soon as someone explains it in a different way i normally understand it. advice is good on here though!
     
  7. Dec 15, 2009 #6

    berkeman

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    Glad to help. I'm going to go ahead and move this thread to the Homework Help forums, since it's about schoolwork. Let us know if you have follow-up questions. As always, it's best to show us as much of your work as possible, since we can often guide you better after we see how you are working out the problems.
     
  8. Dec 15, 2009 #7
    everything......................!
    its about a small toy attached to a spring with a suction device on the bottom.
    the total compression of the spring is 1.2 cm
    when the toy pops off the device it reaches a maximum height of 34 cm
    the mass of the toy is 0.65g
    the spring is ideal and air resistance and other losses are negligible.

    the first part was to work out the constant of the spring,
    i used these equations...
    Strain potential energy: E = 0.5kx2 and gravitational potential energy: E = mgh
    Re-arranging this equation to give k (spring constant) as the subject i get:
    K = 2 mgh/ x2
    Inserting these values into the equation i get:
    K = 2 x(6.5 x 10-4 kg) x 9.81 m s-2 x 0.34 m / (0.012 m)2
    giving the value of the spring constant as 30.1 N m


    the next part was to work out the magnitude of the force that has to be applied in order to compress the spring sufficiently so the toy can be pressed down to the table.

    F = Kx
    F = 30.1 kg m2 s-2 x 0.012 m
    giving 0.36 N as the answer.

    the next part is to calculate the speed in which the toy leaves the table, working on that bit now, hope i'm right so far!!
     
  9. Dec 15, 2009 #8

    berkeman

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    Looking pretty good so far, but the units on K are not quite right. Should be N/m, not Nm. Double check that you worked the equation right, and check your units as you carry them along (and cancel them just like you do real numbers).
     
  10. Dec 15, 2009 #9
    To calculate the speed with which the toy leaves the table i will have used the following equation:
    E = ½ m v2
    Re-arranged to give:
    V =√ E / 0.5 x m
    V =√ 0.36 N / 0.5 x (6.5 x 10-4)
    V = 33.3 m s-1
    So the speed is 33.3 m s-1

    This seems quite fast!?
     
  11. Dec 15, 2009 #10

    berkeman

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    The energy in the compressed spring is not 0.36N (wrong number, wrong units).
     
  12. Dec 15, 2009 #11
    not sure now. thought i was doing ok! do i work out the magnitude of the force like the following or am i off the mark?!

    F = Kx
    F = 30.1 kg m2 s-2 x 0.012 m
    F = 0.36 N (is N supposed to be in joules?)

    or do i use the 0.5kx-2 equation.
     
  13. Dec 15, 2009 #12
    are you online tomorrow evening to help at all?
     
  14. Dec 16, 2009 #13
    are you ok to help? got another question, or do i put that on a different message?
     
  15. Dec 16, 2009 #14

    berkeman

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    The potential energy PE in the compressed spring is converted to kinetic energy KE as the spring is released. Say that all of the PE is converted to KE, what does that give you for the takeoff velocity?
     
  16. Dec 16, 2009 #15
    i really don't understand. getting confused with it.
     
  17. Dec 16, 2009 #16
    You have your toy on its spring, the spring is compressed and the toy is not moving.

    All the energy in the system is stored as potential energy in the spring. As you release this toy the spring starts to "uncompress" itself. As the spring uncompresses itself your toy begins to move upwards.

    So what's happening behind the scenes in terms of energy conservation?

    We know that all energy is conserved.

    So the energy is going from potential energy (stored in the COMPRESSED spring), to kinetic energy (the toy moving upward). The moment your spring is relaxed (x=0) you could say that all the potential energy in the spring has been converted to kinetic energy.

    We can also predict the energy transfer after all the energy is become kinetic energy. Since all the energy is now stored in kinetic energy the toy is going to be moving upwards (kinetic energy is not 0, therefore it has a NONZERO VELOCITY) when the toys reachs its maximum height it won't be moving(VELOCITY=0).

    All the energy is now gravitational potential energy!

    PE Compressed Spring -> Kinetic -> Gravitational PE

    Hope this helps.
     
  18. Dec 16, 2009 #17

    PhanthomJay

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    Questions unrelated to this topic should be posted separately. Otherwise, they are quickly going to get lost in the abyss (black hole).
     
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