phy121
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
Assuming we are including friction and air resistance (typical), that automobile will be moving slower than it was when it started up.
The KE of the automobile is being transformed into PE as it goes up the hill. However, some of the KE is being used up to over come friction and air resistance. When the vehicle has stopped at the top, that PE will then start being trasformed back into KE.
As the vehicle drifts down, it will have to overcome the same friction and air resistance. This will give you a KE = KEI - 2(work done against friction and air resistance).
KEI > KEF so vI > vF
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&#This looks very good. Let me know if you have any questions. &#