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PHY 201
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: ->->->->->->->->->->->-> :
Same speed
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• What, if anything, is different in your answer if air resistance is present? Give your best explanation.
answer/question/discussion: ->->->->->->->->->->->-> :
Same speed. When air resistance is present it will act in both directions equally. It is treated like friction, which is a coefficient times *m * g. M and g will not change if it is going up or going down. So that means that ‘dW_nc will be equal and opposite, up and down. ‘dKE and ‘dPE will also change signs for up and down. When rising, -‘dW_nc = ‘dPE - ‘dKE. When falling, ‘dW_nc = -‘dPE + ‘dKE.
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Air friction does negative work in both directions, since the air resistance is always in the direction opposite motion.
So `dW_nonconservative_ON will always be negative, so `dKE + `dPE will be negative.
Since `dPE is not influenced by nonconservative forces, `dKE will be less at the end than at the beginning.
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