Potential Energy and Conservation Energy.

AI Thread Summary
The discussion revolves around a physics problem involving a pendulum bob with a mass of 0.33 kg moving at a speed of 2.4 m/s at point B, where its kinetic energy is calculated to be 0.95 J. The key question is about the change in gravitational potential energy as the bob comes to a momentary rest, which is determined to also be 0.95 J. Participants emphasize using the conservation of energy principle, where the sum of kinetic and potential energy remains constant. The second question regarding the maximum angle of the swing is left unanswered, with users seeking hints to approach the problem. Overall, the discussion highlights the application of energy conservation in solving pendulum-related physics problems.
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Homework Statement



Suppose the pendulum bob in the figure has a mass of 0.33 Kg and is moving to the right at point B with a speed of 2.4 m/s . Air resistance is negligible.

The diagram is given here :
http://session.masteringphysics.com/prob…

I calculated the bob's kinetic energy at point B which is 0.95 J.

Can anyone help me out with following two questions?

At some point the bob will come to rest momentarily. Without doing an additional calculation, determine the change in the system's gravitational potential energy between point B and the point where the bob comes to rest.

I got 0.95 J for the potential Energy for this..

1. Find the maximum angle the string makes with the vertical as the bob swings back and forth. Ignore air resistance.

Homework Equations



Kinetic Energy = 1/2 mv^2
Potential Energy is = mgh

The Attempt at a Solution



The bob comes to momentary rest at point A which has largest P.E.
For the first question, I think I have to find P.E between point A and B. So I solved it now and got it right...

Second question.. I am completely clueless.. Can anyone please give me any hints?

Can anyone please explain it and help me out?

Thanks a lot before hand!

I solved the whole thing
 
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Hint: Use conservation of energy. K.E.+P.E. = constant
 
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