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Phy 241
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: ->->->->->->->->->->->-> :
the only forces on the ball are the intial force of the ball thrown up in the air and the gravitational force. There is no air resistance, friction or any other forces.
And the conservation of energy shows us that no energy is lost in the act of throwing the ball up and down. At the peak when the ball begins to come back down, KE is used all up and become PE, and when it comes back down, the same amount of energy is created
So if there is no energy loss and the only forces on the ball is gravity, than the ball neither gains or loses 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: ->->->->->->->->->->->-> :
If air resistance is present, it acts as another force on the ball and energy is lost. So the amount of energy the ball starts off with is not the same amount of energy
is ends up with, so therefore the ball would be traveling at a slower speed, because it has less energy to climb in to air and on the way back down has less distance for gravity to accelerate it.
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Then please compare your old and new solutions with the expanded discussion at the link
Solution
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Phy 241
Your 'cq_1_21.2' 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 typical automobile coasts up a typically paved incline, stops, and coasts back down to the same position.
When it reaches this position, is it moving faster, slower or at the same speed as when it began? Explain
answer/question/discussion: ->->->->->->->->->->->-> :
Theres two answers:
The automobile is moving slower due to all the forces on the car. Assuming air resistance, friction force of the tires on the road and gravity are all forces acting on the car
Therefore the is acually less KE when the car comes back down the incline because of air resistance and frictional forces are both nonconservative forces.
2nd answer
You could also assume that gravity was the only force acting on the car, and therefore the speed of the car would be the same, because gravity is a conservative force, because
the amount of energy lost on the way up is gained by the same gravitational force and no actual energy is lost.
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A real auto on a real incline would lose energy to rolling friction.
&#See any notes I might have inserted into your document, and before looking at the link below see if you can modify your solutions. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your old and new solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified.If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#
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