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Physics 11 Online
OpenStudy (korosh23):

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?

OpenStudy (anonymous):

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: \(\sf W=\Delta E\)

OpenStudy (korosh23):

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?

OpenStudy (korosh23):

@Michele_Laino

OpenStudy (korosh23):

I do not know how to do C.

OpenStudy (michele_laino):

I'm pondering...

OpenStudy (korosh23):

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?

OpenStudy (michele_laino):

ok!

OpenStudy (korosh23):

46) a) 36J b) 150J c) 180J

OpenStudy (michele_laino):

part a) \[KE = \frac{1}{2}m{v^2} = \frac{1}{2} \times \frac{1}{2} \times 144 = 36Joules\]

OpenStudy (korosh23):

Yes

OpenStudy (michele_laino):

part b) \[PE = mgh = \frac{1}{2} \times 10 \times 30 = 150Joules\] here g=10 m/sec^2

OpenStudy (korosh23):

Yes

OpenStudy (korosh23):

It is challengic! :D

OpenStudy (michele_laino):

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

OpenStudy (korosh23):

No,

OpenStudy (michele_laino):

where v_FINAL= 12 m/sec

OpenStudy (korosh23):

ok go on

OpenStudy (michele_laino):

and v_INITIAL is the starting speed of your ball

OpenStudy (korosh23):

Starting speed is o m/s

OpenStudy (korosh23):

I think

OpenStudy (michele_laino):

it is impossible, since v_INITIAL =0 then the ball can not go away

OpenStudy (korosh23):

Then how the answer is 180J? You are tight it is impossible.

OpenStudy (korosh23):

One more question? it is important for me to know.

OpenStudy (korosh23):

easy for you

OpenStudy (michele_laino):

I think that it is an approximated value, since the work done by Bob has to be equal to the initial kinetic energy

OpenStudy (michele_laino):

namely: \[W = \frac{1}{2}mv_{FINAL}^2 + mgh = 36 + 150\]

OpenStudy (korosh23):

W is change in kinetic energy, but not mechanical energy. It is fine, we can move to another question.

OpenStudy (korosh23):

May I ask you my another question?

OpenStudy (michele_laino):

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

OpenStudy (michele_laino):

the same reasoning goes if we replace the spring with Bob

OpenStudy (korosh23):

Ok. Makes sense.

OpenStudy (korosh23):

Can I ask you one more question. It is important for me to know for my exam

OpenStudy (michele_laino):

ok!

OpenStudy (korosh23):

I open a new question

OpenStudy (michele_laino):

ok!

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