# Stumped, Help!

1. Oct 16, 2004

### NotaPhysicsMan

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?

2. Oct 16, 2004

### Tide

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. Oct 16, 2004

### NotaPhysicsMan

but doesn't it have to be the average speed. P= F x Vave?

4. Oct 16, 2004

### NotaPhysicsMan

Where does the 47 hp come into play?

5. Oct 16, 2004

### Tide

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 $mg \cos \theta + F_{friction}$

6. Oct 16, 2004

### NotaPhysicsMan

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. Oct 16, 2004

### Tide

Yes, of course it's sine and not cosine! Good catch!

8. Oct 16, 2004

### NotaPhysicsMan

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. Oct 16, 2004

### krab

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. Oct 16, 2004

### NotaPhysicsMan

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 mgsin@. Here's where I'm stuck I don't know Engine force, and I don't konw @.

11. Oct 16, 2004

### NotaPhysicsMan

ok according to Tide, The force going uphill is mgsin@+Ffriction

so the one going down hill is mgsin@-Ffriction

He said subtract the two, which doesn't make sense to me.

12. Oct 16, 2004

### Pyrrhus

I read the problem wrong again, let me fix it.

$$P_{uphill} - P_{downhill} = 47hp$$

Yes read krab's post and the original problem to understand.

Last edited: Oct 16, 2004
13. Oct 16, 2004

### NotaPhysicsMan

Ok so let me see:

P=FV

P(uphill)=mgsin@ +F(friction)
P(downhill)=mgsin@-F(friction)

ok so I subtract them:

47hp=mgsin@+F - (mgsin@-F(friction)
47hp = 35047 Watts or 1298 N.

1298N=mgsin@-mgsin@+F+F
1298N=2F(friction)
F= 649 N. Funny Friction doesn't cancel out!

hmm,

Cos@=Wx/W
=649N/mg
=649N/18639N
=cos inverse(0.0348)
=88 degrees!

sounds kinda funny, that would be too steep.

14. Oct 16, 2004

### NotaPhysicsMan

anyone want to verify?

15. Oct 16, 2004

### NotaPhysicsMan

ah nuts, I think I did something wrong. Ok I think it should be sin@=Wx/W? that'll make it 2.0 degrees! humm. either too steep or too flat. What's wrong?

16. Oct 16, 2004

### Staff: Mentor

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. Oct 16, 2004

### NotaPhysicsMan

ah dang I forgot about V! Ack I'm on it.

18. Oct 16, 2004

### NotaPhysicsMan

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. Oct 16, 2004

### Staff: Mentor

Nope, not correct. You have the forces wrong. Reread my post, get the forces right, then worry about multiplying by V.

20. Oct 16, 2004

### NotaPhysicsMan

P(uphill)=(F+1298)sin@ +F(friction)
P(downhill)=mgsin@-F(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.

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