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Work Problem (easy but weird)

  1. Apr 12, 2005 #1
    This seems easy but I'm not getting the right answer!

    The question: A 0.5 kg ball is thrown into the air. At a height of 20m above the ground, it is travelling at 15 m/s.

    a)How much work was done by someone at ground level throwing the ball up into the air

    Ok, for a), Work=Force x Distance

    W = (0.5 kg x 9.81 N/kg) x 20 m
    = 98 J

    But that's not the answer in my textbook! I realize my substitution for "Force" is wrong but I dont know why

  2. jcsd
  3. Apr 12, 2005 #2


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    That work (which is asked) is nothing but the initial KE...Find it.

  4. Apr 12, 2005 #3
    What is the answer your book got?

    Wouldn't your force be [tex]0.5kg[/tex] x [tex]15m/s^2=7.5N[/tex]

    [tex]7.5N[/tex] x [tex]20m=150J[/tex]

    *Note, just a guess. I probably don't know the right answer either. I really should have taken a physics class this year.

    Just want to try to be helpful.
  5. Apr 12, 2005 #4
    Another thing... remember that once the person's hands leave the ball, the only force acting on it is gravity (and air resistance, but that's negligable). Remember the nice rule that [tex]\bigtriangleup{K.E.} = W[/tex]
  6. Apr 12, 2005 #5
    Yes! The answer is 150 J! However, Force=mass x accel
    You said that the accel is 15 m/s/s.....why?? Isnt that the speed? Why is the speed and accel the same?
  7. Apr 12, 2005 #6


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    Shouldn't it be [itex] 156.25 \ \mbox{J} [/itex]...?

  8. Apr 12, 2005 #7
    I got something right? Wow! Confidence boost.

    [tex]15m/s[/tex] would be the speed. However, [tex]15m/s^2[/tex] would be acceleration.

    Speed can be meters per second, but acceleration the change of the the rate of speed, which would be [tex]15m/s^2[/tex].
  9. Apr 12, 2005 #8
    My book says 1.5e2 J...I guess thats to 2 significant digits...but can anyone explain to me why the 15 m/s became a 15 m/s/s?
    Last edited: Apr 12, 2005
  10. Apr 12, 2005 #9
    But....to know accel, wouldnt you need the initial velocity which we dont know? I still dont understand :redface:
  11. Apr 12, 2005 #10


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    If you mean [tex] 1.5 \cdot 10^{\mbox{2}} [/tex] ,then i believe you.

    You have to realize that u're being asked to find the work done to imprime a certain velocity to the ball.And that work is exactly the initial KE of the ball.U'll have to use the theorem of varation of KE (or the law of the total mechanical energy) to find it.

    It's not difficult at all.

  12. Apr 12, 2005 #11
    Oups yeah I meant 1.5e2 ....so

    Ek = 1/2 x m x v^2
    I know the mass, but not the initial velocity!
  13. Apr 12, 2005 #12
    [tex]A=\frac{\delta v}{\delta t}[/tex].

    So your book says [tex]15m/s[/tex], and not [tex]15m/s^2?[/tex]
  14. Apr 12, 2005 #13
    Are you sure thye answer is supposed to be 150J? I get alittle over 154J.
  15. Apr 12, 2005 #14
    Right. It says that at 20m, it is travelling at 15 m/s
  16. Apr 12, 2005 #15
    It's to 2 significant digits (rounding)
    Oh boy I still dont know how to solve this problem :confused:
  17. Apr 12, 2005 #16


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    Lucretius,please.You're iincorrect.As to the OP,i already said what u should be doing...

  18. Apr 12, 2005 #17
    :frown: what'd I do wrong?
  19. Apr 12, 2005 #18
    That's what I wanna know too,,I still dont get it
  20. Apr 12, 2005 #19


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    There's no indication of any force acting on the ball during flight,therefore,it's natural to assume that the only force doing work on the ball while in motion is the force of earth's gravitational attraction.

  21. Apr 12, 2005 #20

    What Lucretius did to solve this makes no sense; he lucked out getting close to the right answer (no offense Lucretius!).

    Dextercioby is steering you the right way, pay attention to him.
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