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A tennis ball dropped vertically (Newton's Forces)

  1. Nov 14, 2009 #1
    1. The problem statement, all variables and given/known data

    A tennis ball has a mass of 0.057 kg. After falling 18 m vertically from rest it has a speed of 12 m/s. What is the magnitude of the average resistive force acting on it as it falls?

    My answer for this was .3306N


    mass = 0.057kg

    Vi = 0 m/s

    Vf = 12m/s

    Xi = 18m

    Xf=0m


    I started with w=mg = -0.05586N

    So FN must be 0.05586N since it is dropped vertically. The free body diagram would be FN pointed up and weight pointing straight down.

    I used the equation V^2 = Vi^2 +2a[tex]\Delta[/tex]x

    (12m/s)^2 = 2a(-18m)

    a = -4m/s^2

    F=ma

    F=(.057kg)(-4m/s^2) = -0.228N

    Fnet = Fn + Fresistance = .5586N + (-.228N) = .3306N
     
  2. jcsd
  3. Nov 14, 2009 #2

    Delphi51

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    Looks good!
     
  4. Nov 14, 2009 #3
    What is the normal force due to in this problem? Wouldn't there only be the force of gravity (pointing down) and the resistive force (pointing up)?
     
  5. Nov 14, 2009 #4

    Delphi51

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    There isn't a normal force. Yes, that solution is confusing. Better to write
    mg - F = ma (downward is positive)
    F = m(g-a) = .057(g - 4) = 0.331 N.
     
  6. Nov 14, 2009 #5
    Okay, thank you. I recognized that after i replied/worked the solution out myself.
     
  7. Nov 15, 2009 #6
    So you set the Fy which is mg and F equal to ma? What would the equation look like if you said downward was negative? I worked it out and got the acceleration as negative. Is it because you said downward was positive that -4m/s^2 was now just 4m/s^2?
     
    Last edited: Nov 15, 2009
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