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

[pmd28]

No, Oh is it f=Fw+mgSin(theta)

 

[TSny]

So, Fw is?

 

[pmd28]

Fw=f-mgSin(theta)

 

[TSny]

Good. So, find an expression for the power due to Fw going up and also for coming down. (Don't worry about numbers yet.)

 

[pmd28]

Up the hill: P=Fw(v) = (f+mgSinθ)v
Down the hill: P=Fw(v) = (f-mgSinθ)v

 

[TSny]

Right. How do you now bring in the 44 hp?

 

[pmd28]

Set the equations equal to each other and add 44 to the "up hill" side

 

[TSny]

Do you want to add it to the "up hill" side or the "down hill" side? Also, is hp the SI unit for power?

 

[pmd28]

o right becasue going uphill is 44 more hp you would add it to the downhill side. and no 1hp=745watt

 

[TSny]

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

 

[pmd28]

So just checking my equation:

v(f+mgSinθ)=v(f-mgSinθ) + 32810.8w

Now I can just plug and chug.

 

[TSny]

Yep. Good work!

 

[pmd28]

I got a domain error... [never mind calculator error]

 

[TSny]

What?

 

[pmd28]

I just put my nunmbers in my calculator wrong. No big deal thanks again.

 

[TSny]

ok. No problem.