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Initial Velocity Question

  1. Oct 26, 2011 #1
    1. The problem statement, all variables and given/known data
    A 20.00kg curling stone is released at the hog line and moves 28.35m [W] to sit on the button and score a point. If the coefficient of kinetic friction between the stone and the ice is 0.0020000, what was the initial speed of the stone ?

    So we know:
    m=20.00kg
    d=28.35m[W]
    Fk=0.0020000
    Vi=?


    2. Relevant equations



    3. The attempt at a solution
    I'm stuck on what I have to find first to end up finding the velocity. Once I know what I need to find first I can figure it on from there. I am not asking for any answers, just what I have to find first. Any help is appreciated.
     
  2. jcsd
  3. Oct 26, 2011 #2
    What principle relates work to energy?
     
  4. Oct 26, 2011 #3
    Force of friction ?
     
  5. Oct 26, 2011 #4
    Or are you studying distance, acceleration equations?
     
  6. Oct 26, 2011 #5
    Distance and acceleration equations. We haven't learned anything with energy yet.
     
  7. Oct 26, 2011 #6
    This problem can be solved by a velocity, distance, acceleration approach or by energy considerations. Which is it?
     
  8. Oct 26, 2011 #7
    Ok, our posts crossed. Do you know of an equation that relates velocity to distance and acceleration?
     
  9. Oct 26, 2011 #8
    The only equation I know that has velocity, acceleration and distance is :
    Δd=v1Δt + 1/2 aΔt^2
     
  10. Oct 26, 2011 #9
    You have d = 1/2 * a * t^2. You also have V = a * t. Using both equations, eliminate the time variable.
     
  11. Oct 26, 2011 #10
    t=v/a
    2d/a=t^2
     
  12. Oct 26, 2011 #11
    Plug the expression you have for t into the distance equation. That gives you a relationship between velocity, distance, and acceleration. You can get the acceleration from Newton's Law considering friction and mass and gravity.

    I have to be signing off now. If you are still having difficulty, hang in there. There are plenty of competent helpers on this forum who will be glad to help out.
     
  13. Oct 26, 2011 #12
    Ahhh, okay. Thank you for your help !
     
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