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Phy 231
Your 'cq_1_21.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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A ball is tossed vertically upward and caught at the position from which it was released.
Ignoring air resistance will the ball at the instant it reaches its original position be traveling faster, slower, or at the same speed as it was when released?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
Thinking about this using energy considerations, I believe the speed will be the same when it reaches its original position as it was when released. It starts at a certain speed (and accordingly a certain KE), gradually loses KE as it goes higher and slows down due to gravity. Meanwhile, the gain in altitude brings a corresponding gain in PE. As I understand it, this is the definition of a conservative force. The KE the ball loses as it does work on gravity is regained as PE it acquires through increasing its altitude.
Then it begins to fall, and if there are no other forces present, all of that just-acquired PE turns back into KE, little by little, as it falls and gains speed. So the energy level returns to its original quantity, and since the mass is constant, that means the speed must return to its original rate as well.
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What, if anything, is different in your answer if air resistance is present? Give your best explanation.
answer/question/discussion: ->->->->->->->->->->->-> scussion:
If air resistance is present, then in addition to doing work against the conservative force of gravity as described above, the ball is also doing work against the nonconservative force of air resistance. So when it reaches the top, it has regained some of the lost KE as gravitational PE (but not all-- some is just lost for good). In visual terms, this means it hasn't climbed as high as it otherwise would have.
Then as it falls, air resistance continues to oppose its motion, meaning that the net force on the ball is less than it would be if gravity alone were working on the ball. So the lower net force (applied to a constant mass) will result in a lower acceleration, and a lower final velocity. This means the ball will be travelling slower than it would have, and slower than it began when originally released.
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Very well done.
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