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korosh23

  • one year ago

Bob throws a 0.50 kg baseball straight up. At a height of 30.0 m it is moving at 12m/s a) what is the ball's kinetic energy? b) What is the gravitational potential energy of the ball? c) How much work was done by Bob to get the ball up into the air?

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  1. anonymous
    • one year ago
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    m=0.50 kg h=30 m v=12m/s a. apply the kinetic energy formula \(\sf KE= \frac{1}{2}mv^2\) b. apply the potential energy formula \(\sf PE= mgh \), where g is the acceleration due to gravity. c. apply work energy theorem: \(\sf W=\Delta E\)

  2. korosh23
    • one year ago
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    Thank you for mentioning the formulas. I already know how to use them, I did part a and b, but not c. Could you please explain me how to do it?

  3. korosh23
    • one year ago
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    @Michele_Laino

  4. korosh23
    • one year ago
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    I do not know how to do C.

  5. Michele_Laino
    • one year ago
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    I'm pondering...

  6. korosh23
    • one year ago
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    Ok. My teacher gave me the answer. He did not show the work. Do you want the answer? Maybe it can help you to find how?

  7. Michele_Laino
    • one year ago
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    ok!

  8. korosh23
    • one year ago
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    46) a) 36J b) 150J c) 180J

  9. Michele_Laino
    • one year ago
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    part a) \[KE = \frac{1}{2}m{v^2} = \frac{1}{2} \times \frac{1}{2} \times 144 = 36Joules\]

  10. korosh23
    • one year ago
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    Yes

  11. Michele_Laino
    • one year ago
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    part b) \[PE = mgh = \frac{1}{2} \times 10 \times 30 = 150Joules\] here g=10 m/sec^2

  12. korosh23
    • one year ago
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    Yes

  13. korosh23
    • one year ago
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    It is challengic! :D

  14. Michele_Laino
    • one year ago
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    yes! Lol! for part c) we can write this equation: \[\Large \frac{1}{2}mv_{INITIAL}^2 - \frac{1}{2}mv_{FINAL}^2 = mgh\]

  15. korosh23
    • one year ago
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    No,

  16. Michele_Laino
    • one year ago
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    where v_FINAL= 12 m/sec

  17. korosh23
    • one year ago
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    ok go on

  18. Michele_Laino
    • one year ago
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    and v_INITIAL is the starting speed of your ball

  19. korosh23
    • one year ago
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    Starting speed is o m/s

  20. korosh23
    • one year ago
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    I think

  21. Michele_Laino
    • one year ago
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    it is impossible, since v_INITIAL =0 then the ball can not go away

  22. korosh23
    • one year ago
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    Then how the answer is 180J? You are tight it is impossible.

  23. korosh23
    • one year ago
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    One more question? it is important for me to know.

  24. korosh23
    • one year ago
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    easy for you

  25. Michele_Laino
    • one year ago
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    I think that it is an approximated value, since the work done by Bob has to be equal to the initial kinetic energy

  26. Michele_Laino
    • one year ago
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    namely: \[W = \frac{1}{2}mv_{FINAL}^2 + mgh = 36 + 150\]

  27. korosh23
    • one year ago
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    W is change in kinetic energy, but not mechanical energy. It is fine, we can move to another question.

  28. korosh23
    • one year ago
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    May I ask you my another question?

  29. Michele_Laino
    • one year ago
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    W is the work done by Bob. more explanation: if the ball is thrown by a compressed spring, then we can say that the potential energy of the spring is equal to the kinetic energy of the ball, so the work done by the spring on the ball is equal to the initial kinetic energy of our ball

  30. Michele_Laino
    • one year ago
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    the same reasoning goes if we replace the spring with Bob

  31. korosh23
    • one year ago
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    Ok. Makes sense.

  32. korosh23
    • one year ago
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    Can I ask you one more question. It is important for me to know for my exam

  33. Michele_Laino
    • one year ago
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    ok!

  34. korosh23
    • one year ago
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    I open a new question

  35. Michele_Laino
    • one year ago
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    ok!

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