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Determining the position from the velocity

  1. Jun 1, 2008 #1
    1. The problem statement, all variables and given/known data

    An object (masse m) is placed on a slope with an angle a at a height of h above the ground. There is no friction on the inclined plane. The inclined plane leads to a horizontal plane 2 meters long that ends with a spring with a constant of k. There is friction on the horizontal plane (but not the inclined plane). Where is the object when its velocity is equal to 1m/s?

    angle a= 30 degrees

    2. Relevant equations

    W = E(cin final) - E(cin initial)

    3. The attempt at a solution

    W = 0.5*mv² - 0.5*m(0)²
    =0.5*mv² = F*d

    but I don't know how to add in the energie given by the spring and the energy lost from friction.

    Any help would be appreciated thanks
  2. jcsd
  3. Jun 1, 2008 #2


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    Was the force of friction given as 1N, or is that your assumption?

    Insofar as it was given, remember the work-energy formula.
  4. Jun 1, 2008 #3
    Yes, it's given. And sorry my attempt is a bit poor, I just need a bit of help starting off in the right direction. Thanks
  5. Jun 1, 2008 #4
    I seriously doubt that you need all the info (but maybe I'm doing something wrong):

    [tex] \frac{1}{2}mv^2 =mgh [/tex]

    solving for v:

    [tex] v = \sqrt{2gh} [/tex]

    Inserting data I get 5 cm lower w.r.t initial height?
  6. Jun 1, 2008 #5


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    Hi zakare! :smile:

    The energy absorbed by the spring is 1/2*kx², where x is the decrease in length of the spring;

    and the energy lost from friction equals the work done against friction. :smile:
  7. Jun 1, 2008 #6
    It will also reach 1 m/s when its a bit in the spring

    edit: first post and im late :(
  8. Jun 1, 2008 #7


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    Welcome to PF!

    Hi Multicol ! Welcome to PF! :smile:
    :smile: that's ok … spring was early this year! :smile:
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