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Easy question newtons 1st law of motion, applying force to a hockey puck

  1. Oct 3, 2009 #1
    ** sorry. this would be newtons 2nd law of motion. oops.

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

    Hockey puck with mass 0.160kg at rest at x=0 and t=0 on horizontal frictionless surface. hockey player applies force of 0.250 N until t = 2.00s. ... ... if the same force is applied again at t = 5.00 s what is the position and speed of the puck at t = 7.00s?

    2. Relevant equations

    x(final) = x(initial)+v(initial)t+(1/2)a(t^2)
    v(final) = v(initial)+at

    3. The attempt at a solution

    at t = 2 seconds, the puck is at x = 3.125m and the velocity is 3.125m/s. the acceleration (from F = ma) is 1.5625 m/s^2.



    how do I incorportate reapplying the force? i imagine since the surface is frictionless and the acceleration is constant that the puck keeps increasing in velocity right?
     
    Last edited: Oct 3, 2009
  2. jcsd
  3. Oct 4, 2009 #2

    tiny-tim

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    Hi offbeatjumi! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    I'm not sure what you're asking. :confused:

    The puck will have zero acceleration, and therefore the same velocity, until the force is reapplied.

    So find the new x at the end of that constant velocity, and use that as the initial x in the same equation as before. :smile:
     
  4. Oct 4, 2009 #3
    im sorry i worded that wierdly

    this is essentially the question: if the same force is applied again at t = 5.00 s what is the position and speed of the puck at t = 7.00s?

    but what you said helps, ill try it
     
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