Stumped, Help!

  • #1
This is supposed to be a hard question...

A 1900kg car experiences a combinded force of air resistance and friction that has the same magnitude whether the car goes up or down a hill at 27m/s. Going up a hill, the car's engine needs to produce 47 hp (1horsepower=745.7 Watts) more power to sustain the constant velocity than it does going down the same hill. At what angle is the hill inclined above the horizontal?

Ok, where do I even start?
 

Answers and Replies

  • #2
Tide
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The applied force times the speed is the power delivered by the vehicle. Simply write those out for both cases in terms of the gravitational force and the frictional force. Then simply subtract the equations! :-)
 
  • #3
but doesn't it have to be the average speed. P= F x Vave?
 
  • #4
Where does the 47 hp come into play?
 
  • #5
Tide
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You said the speed is fixed and is the same for both the trip up and and the trip down the hill. You need to figure out what force the engine must exert for both cases and for each case the power delivered will be the respective force times that speed. Hint: Going up the hill the force is [itex]mg \cos \theta + F_{friction}[/itex]
 
  • #6
Hmm, are you sure it isn't mgsin(theta), the x-components right?. Going up hill, wouldn't friction oppose the x-force moving in another direction?
 
  • #7
Tide
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Yes, of course it's sine and not cosine! Good catch!
 
  • #8
ok force of friction, I don't have the coeffecient of friction. Once I find the force to go uphill that's also equal to the force downhill with exception to the extra 47hp. Then I can separate into x and y compents. All I need is the force uphill or downhill.!!!
 
  • #9
krab
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you don't need to know anything about friction, since its force is the same in either direction and so will cancel. You don't know the total hp either, just the difference.
 
  • #10
ok I'm stuck here at this point: I have the F going down is equal to the one going up, right so let's see: (Engine force+1298N (the 47 hp)=the x component [email protected] Here's where I'm stuck I don't know Engine force, and I don't konw @.
 
  • #12
Pyrrhus
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I read the problem wrong again, let me fix it.

[tex] P_{uphill} - P_{downhill} = 47hp [/tex]

Yes read krab's post and the original problem to understand.
 
Last edited:
  • #15
ah nuts, I think I did something wrong. Ok I think it should be [email protected]=Wx/W? that'll make it 2.0 degrees! humm. either too steep or too flat. What's wrong?
 
  • #16
Doc Al
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NotaPhysicsMan said:
Ok so let me see:

P=FV

P(uphill)[email protected] +F(friction)
P(downhill)[email protected](friction)
What you are calling "P" is really the applied force, not the power. (Since you didn't multiply by V.) You need to rethink what force is needed in going up and down the hill.

Since the speed is constant, the applied force must just overcome friction and gravity. Going up, gravity acts against you (thus the applied force must be greater); going down, gravity acts in your favor (so the applied force is less). Friction always acts against you.
 
  • #17
ah dang I forgot about V! Ack I'm on it.
 
  • #18
k wait so all my other stuff are correct right? but I just didn't multiply by V? Adding V makes my angle pretty much 0!
 
  • #19
Doc Al
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NotaPhysicsMan said:
k wait so all my other stuff are correct right? but I just didn't multiply by V? Adding V makes my angle pretty much 0!
Nope, not correct. You have the forces wrong. Reread my post, get the forces right, then worry about multiplying by V.
 
  • #20
P(uphill)=(F+1298)[email protected] +F(friction)
P(downhill)[email protected](friction)

The only thing I see is that maybe I didn't add the force needed to get up there, the extra 47 hp or 1298N. Going downhill, I have friction against so that's neg.

a little help lol.
 
  • #21
hmm, well is my approach at least correct?
 
  • #22
I got it!! the answer is 2.0 degrees?
Simple mistake really, didnt' take account of the 1298N and mixed up some of the negative signs.
 

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