Opel GT on Incline: How Much HP Needed?

[VoyagerOne]

Homework Statement

Ok this question came up on a car forum and i tried to answer it, but I think my answer is way to simple and I solved the problem incorrectly
1. A 2072lb Opel GT goes up 20 degree incline how much horsepower will the car need to get up the hill at the same speed as the same car on the ground. The car on the ground has 102 HP
2. I didn't solve this one: Both cars now start from rest and travel a quarter mile. Assuming the car on the ground completes the distance in 18 seconds how much HP will be needed for the car on the incline to finish in the same amount of time. (accel is constant)
Dont worry about friction or air resistance all things are equal.

Homework Equations

mgcosΘ and mgsinθ

The Attempt at a Solution

The force against a 9210.46 Newton (2072lb) Opel GT on an incline of 20° would be equal to mass*gravity(sin20°) which rounds to about 3153 Newtons (708.8 lbs)
So without any air resistance, friction, wheels falling off, fuel leaks, drivers, passengers, or dead bodies in rear console, your Opel would "weigh" 2780.8lbs.
Let's say you have the top spec stock high comp GT with 102HP. Your HP to weight ratio would be .03668009 hehe (very roughly 1 hp per 27.8 lbs)
On the ground your HP to weight ratio is 0.04922779! (Very roughly 1 HP per 20.7 lbs)
Xhp/2780.8=.04922779
Hp up hill would have to be about 137hp to equal the car on the level surface!

 

[haruspex]

Doesn't seem to me to be enough information. I don't think you can simply take the ratio of the forces like that.
Suppose the rolling resistance is F, the extra downslope force due to gravity is F', the speed is v, the power on the level is P, and the extra power up the slope is P'.
Then you have
Fv = P
(F+F')v = P+P'
You know P and F', but that leaves three unknowns and only two equations. You can deduce F'v = P', but that still doesn't help. In the attempt posted, you have effectively assumed F = weight of car, which is not true.

 

[dean barry]

Give your car some arbitrary drag co-efficients :
Cd = air drag co-efficient (sea level) = 0.45
Crr = wheel rolling resistance co-efficient = 0.03
Level ground drive wheel power = 102 hp = 76,061 Watts
Mass = 2,072 lbs = 934 kg
Top speed of this vehicle = 115.45 mph
Power required by this vehicle up a 20 degree incline at the same speed = 317.72 hp

 

[CWatters]

As above.

To work out the extra power needed going up hill you need to know or calculate the vertical speed. There isn't enough info in the original problem to do that. For example if the speed was slow (or the hill shallow) then the power required to overcome drag and rolling resistance would dominate and the same power would be required going up hill as on the flat.

 

[HalfLostAndSearching]

I would approach this problem by first considering the forces acting on the car on the incline. We know that the force of gravity acting on the car will be the same as on the level surface, but there will also be an additional component of the force acting against the car due to the incline. This can be calculated using the equation F = mg sinθ, where m is the mass of the car, g is the acceleration due to gravity, and θ is the angle of the incline.

In order to maintain the same speed on the incline as on the level surface, the car will need to have enough horsepower to overcome this additional force. This can be calculated using the equation P = Fv, where P is power, F is force, and v is velocity. Since we are assuming constant acceleration, we can use the equation v = at, where a is acceleration and t is time.

For the first part of the question, we can use the given information of the car's weight and horsepower to calculate its acceleration on the incline. From there, we can use the equation v = at to calculate the time it takes for the car to travel a quarter mile. Then, using the equation P = Fv, we can determine the horsepower needed to achieve this time on the incline.

For the second part of the question, we can use the same approach but with the added information that both cars start from rest and travel a quarter mile in the same time. This allows us to set the equations for the two cars equal to each other and solve for the horsepower needed on the incline.

Overall, this problem requires a more in-depth analysis of the forces and equations involved, and cannot be solved simply by comparing the horsepower to weight ratios of the two cars.