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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?
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0m=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: \(\sf W=\Delta E\)

korosh23
 one year ago
Best ResponseYou've already chosen the best response.0Thank 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?

korosh23
 one year ago
Best ResponseYou've already chosen the best response.0I do not know how to do C.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I'm pondering...

korosh23
 one year ago
Best ResponseYou've already chosen the best response.0Ok. 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?

korosh23
 one year ago
Best ResponseYou've already chosen the best response.046) a) 36J b) 150J c) 180J

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1part a) \[KE = \frac{1}{2}m{v^2} = \frac{1}{2} \times \frac{1}{2} \times 144 = 36Joules\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1part b) \[PE = mgh = \frac{1}{2} \times 10 \times 30 = 150Joules\] here g=10 m/sec^2

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! Lol! for part c) we can write this equation: \[\Large \frac{1}{2}mv_{INITIAL}^2  \frac{1}{2}mv_{FINAL}^2 = mgh\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1where v_FINAL= 12 m/sec

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1and v_INITIAL is the starting speed of your ball

korosh23
 one year ago
Best ResponseYou've already chosen the best response.0Starting speed is o m/s

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1it is impossible, since v_INITIAL =0 then the ball can not go away

korosh23
 one year ago
Best ResponseYou've already chosen the best response.0Then how the answer is 180J? You are tight it is impossible.

korosh23
 one year ago
Best ResponseYou've already chosen the best response.0One more question? it is important for me to know.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I think that it is an approximated value, since the work done by Bob has to be equal to the initial kinetic energy

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1namely: \[W = \frac{1}{2}mv_{FINAL}^2 + mgh = 36 + 150\]

korosh23
 one year ago
Best ResponseYou've already chosen the best response.0W is change in kinetic energy, but not mechanical energy. It is fine, we can move to another question.

korosh23
 one year ago
Best ResponseYou've already chosen the best response.0May I ask you my another question?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1W 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1the same reasoning goes if we replace the spring with Bob

korosh23
 one year ago
Best ResponseYou've already chosen the best response.0Can I ask you one more question. It is important for me to know for my exam

korosh23
 one year ago
Best ResponseYou've already chosen the best response.0I open a new question
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