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Tossed ball - Work done

  1. Feb 20, 2010 #1
    Hey all, I'm an idiot when it comes to algebra/physics/anything with numbers so you'll probably be seeing me around alot until the end of the semester, but I'll try to lend aid when I can
    Anyway, this is what I've got:

    1. The problem statement, all variables and given/known data
    A 0.267 kg ball is thrown straight up from 2.33 m above the ground. Its initial vertical speed is 12.00 m/s. A short time later, it hits the ground. Calculate the total work done by the force of gravity during that time.

    2. Relevant equations

    Force due to gravity = m * g

    Height of trajectory = V^2 / 2 * g

    Force = mass * distance (No Cos(theta) here because it's a straight up and down trajectory, right?)

    3. The attempt at a solution

    So, I figured the force of gravity working on the ball:
    F = mg = 2.62 N = .267 * 9.81

    Trajectory height = V^2 / 2 * g = 7.74866 = 12.33^2/(2*9.81)

    Then I added 2.33 to the (7.75*2) (that's the distance up and down) which ended up being 17.83
    Then used Force = Mass * Distance to get 46.689 Joules for an answer and apparently it's wrong, and I have no clue what to do.

    I'm sure there's just a hole in my logic somewhere. Thanks for any help.
  2. jcsd
  3. Feb 20, 2010 #2
    [tex]Work = F \cdot s[/tex]

    The dot product makes all the difference. Recall also the definition of work: the work done by a force on an object is the product of the force and the displacement of the body in the direction of the force.

    Thus, when the ball is moving upwards, gravity does negative work on the ball. When it is moving downwards, gravity does positive work on the ball.
  4. Feb 20, 2010 #3
    Fantastic, I put 0 as total work done once, forgetting about the 2.33 M that it fell past where it started. Got the answer, thanks for your help!
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