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Newton's II frcitionless surface, constant velocity, angle of incline

  1. Jul 2, 2009 #1
    1. The problem statement, all variables and given/known data

    A physics student playing with an air hockey table (a frictionless surface) finds that if she gives the puck a velocity of 3.79 m/s along the length ( 1.84 m) of the table at one end, by the time it has reached the other end the puck has drifted a distance 2.47 cm to the right but still has a velocity component along the length of 3.79 m/s. She concludes correctly that the table is not level and correctly calculates its inclination from the above information.


    2. Relevant equations

    x=x0+vx0t+1/2at2, since acceleration is constant
    [tex]\Sigma[/tex]Fy=ma



    3. The attempt at a solution

    vx0=3.8m/s
    x=1.75m, x0=0
    vx=x/t

    y=.025m

    This is as far as I get. Setting up the problem is my main issue with this.I actually know the answer to this because I purchased the solutions manual. But, I do not understand it. Working on my own, I am still unclear about the mathematical relationship between this problem and Newton's II.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 2, 2009 #2

    LowlyPion

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    Homework Helper

    Welcome to PF.

    You know the time that it drifts by is 1.84/3.79 seconds.

    So using that time and the drift ...

    .0247 m = 1/2*a*t2

    The angle then you know from gravity makes a = g*sinθ
     
  4. Jul 2, 2009 #3
    Thanks...i appreciate the help. it's starting to make sense. You are using the acceleration of g and the lateral acceleration of the drift to find the angle. Then substituting those values into the constant acceleration equation. Hopefully, I will get to the point where I'm able to make these connections between equations.
     
  5. Jul 2, 2009 #4

    LowlyPion

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    Homework Helper

    There's no real trick to it. The vertical dimension is accelerated. The horizontal ones are not, as long as you are moving along equipotential surfaces, i.e. like at the same height. Time of course runs the same in all dimensions. The rest is just breaking things down and applying the proper considerations and solving for what you need.

    Regardless though Good luck with it.
     
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