Car going up a hill. Find the Angle of the hill.

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  • #26
TSny
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The question gives you the difference in power of the engine as you switch from going down the slope to going up the slope. So, you'll need to think about how Fw changes when going down compared to up.
 
  • #27
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Would it be right to plug in Fw into P=Fv?
 
  • #28
TSny
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You have found that Fw = f + mgSin(theta) going up.

Would the equation be the same for going back down? If not, how would it change? Just repeat the steps that you used for getting the equation for going up (including drawing a good free-body diagram).
 
  • #29
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Wouldn't it be the same?
 
  • #30
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Do all the forces act in the same direction coming down as they did going up?
 
  • #31
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No, Oh is it f=Fw+mgSin(theta)
 
  • #32
TSny
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So, Fw is?
 
  • #33
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Fw=f-mgSin(theta)
 
  • #34
TSny
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Good. So, find an expression for the power due to Fw going up and also for coming down. (Don't worry about numbers yet.)
 
  • #35
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Up the hill: P=Fw(v) = (f+mgSinθ)v
Down the hill: P=Fw(v) = (f-mgSinθ)v
 
  • #36
TSny
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Right. How do you now bring in the 44 hp?
 
  • #37
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Set the equations equal to each other and add 44 to the "up hill" side
 
  • #38
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Do you want to add it to the "up hill" side or the "down hill" side? Also, is hp the SI unit for power?
 
  • #39
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o right becasue going uphill is 44 more hp you would add it to the downhill side. and no 1hp=745watt
 
  • #40
TSny
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Right. (1 hp = 746 W the last I checked, but who's counting?) [OK, just checked, it's 745.699872 W for 1 "mechanical hp" and exactly 746 W for "electrical hp". Trivia for the day.]

http://en.wikipedia.org/wiki/Horsepower
 
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  • #41
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So just checking my equation:

v(f+mgSinθ)=v(f-mgSinθ) + 32810.8w

Now I can just plug and chug.
 
  • #42
TSny
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Yep. Good work!
 
  • #43
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I got a domain error... [never mind calculator error]
 
  • #44
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What?
 
  • #45
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I just put my nunmbers in my calculator wrong. No big deal thanks again.
 
  • #46
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ok. No problem.
 

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