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Relating position and force to speed

  1. Oct 4, 2011 #1
    1. The problem statement, all variables and given/known data

    The force graphed in figure is applied to a 1.8-{\rm kg} box initially at rest at x = 0 on a frictionless, horizontal surface.
    (Graph is attached)

    2. Relevant equations

    F=ma (?)

    3. The attempt at a solution

    Don't know how to relate speed to these two scalars..
     

    Attached Files:

  2. jcsd
  3. Oct 4, 2011 #2

    cepheid

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    You haven't stated what the problem is. At all.
     
  4. Oct 4, 2011 #3
    I need to find the speed when position = 5, 10 and 15cm. Sorry.
     
  5. Oct 4, 2011 #4

    SammyS

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    What level of math goes with your course?
     
  6. Oct 4, 2011 #5
    It's physics 2A, I'm not having trouble with math I'm having trouble with finding the units of m/s out of units like force and position.
     
  7. Oct 4, 2011 #6

    SammyS

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    Well, if it's calculus based, we might give one answer.

    Trig. & algebra another.

    Elementary algebra. ???

    How is work done related to kinetic energy.
     
  8. Oct 4, 2011 #7
    I had to take calculus to get into this course.
     
  9. Oct 4, 2011 #8

    cepheid

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    A simple application of the work-energy theorem should get you on your way.
     
  10. Oct 4, 2011 #9

    SammyS

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    What does area under the graph represent?
     
  11. Oct 4, 2011 #10
    So F=ma can give me the acceleration of the object, I have no units of time to convert to speed here
     
  12. Oct 4, 2011 #11
    The area represents work done?
     
  13. Oct 4, 2011 #12

    SammyS

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    think about force time distance. Look at cepheid's suggestion.
     
  14. Oct 4, 2011 #13

    cepheid

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    Yes.
     
  15. Oct 4, 2011 #14
    I can solve for work done here but that leaves me with units of (kg*m^2)/s^2, I do not know how to get to speed here. This is probably very obvious, but I'm not seeing it.
     
  16. Oct 4, 2011 #15

    SammyS

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    Cap'ncanada,

    What's this theorem?
     
  17. Oct 4, 2011 #16
  18. Oct 4, 2011 #17

    SammyS

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    That's force.

    Look at the units again. remember, you're looking for speed.
     
  19. Oct 4, 2011 #18
    Err, W=KE=1/2mv^2?

    Im having a lot of trouble with physics...
     
  20. Oct 4, 2011 #19

    cepheid

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    No, that is not the work-energy theorem. That is Newton's 2nd Law. The work-energy theorem relates the work done on an object to its change in energy. If you haven't encountered it before (which seems strange) you can always look it up.

    By the way, the work, of course, has dimensions of force*distance, therefore its units are newton-metres = joules.
     
  21. Oct 4, 2011 #20

    cepheid

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    That's pretty much it, but it should be that the work done on the object is equal to its CHANGE in kinetic energy. Therefore W = ΔKE = (1/2)mvf2 - (1/2)mvi2, where vf and vi are the final and initial velocities respectively.
     
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