Given Power, Weight, and Vmax find Vmax at angle θ

In summary: And, if you want to calculate the component of that force along the incline, you need to multiply by cosθ, not divide. find FN at angle θ:FNθ=(1490N)(sin(sin-1(1/20)=74.5NSame comment as before, but now you need to multiply by sinθ. find Fnetθ:Fnetθ=Fθ-FNθ=2472.48N-74.5N=2397.98Again, I don't think you need to calculate a new Fθ. And, I think you're using the wrong values for FNθ and Fnetθ. insert Fnetθ into V=W/F:V=121,
  • #1
Jack_Straw
4
0

Homework Statement



A car of weight 1490 N operating at a rate of 121 kW develops a maximum speed of 49 m/s on a level, horizontal road.

Assuming that the resistive force (due to friction and air resistance) remains constant, what is the car’s maximum speed on an incline of 1 in 20; i.e., if θ is the angle of the incline with the horizontal, sin θ = 1/20? Answer in units of m/s.

Homework Equations



P=FV

The Attempt at a Solution



find force:
force=power/velocity
F=121,000/49

find θ:
θ=sin-1(1/20)

find F at angle θ:
Fθ=(121,000/49)/cos(sin-1(1/20))=2472.48N

find FN at angle θ:
F=(1490N)(sin(sin-1(1/20)=74.5N

find Fnetθ:
Fnetθ=Fθ-F=2472.48N-74.5N=2397.98

insert Fnetθ into V=W/F:
V=121,000/2397.98=50.46m/s

This answer is reported as wrong. Any ideas? Thanks
 
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  • #2
Jack_Straw said:

Homework Statement



A car of weight 1490 N operating at a rate of 121 kW develops a maximum speed of 49 m/s on a level, horizontal road.

Assuming that the resistive force (due to friction and air resistance) remains constant, what is the car’s maximum speed on an incline of 1 in 20; i.e., if θ is the angle of the incline with the horizontal, sin θ = 1/20? Answer in units of m/s.
Is the "sin θ = 1/20" actually given to you, or is that your own relation? I ask because typically incline grades (of the form "one part in some other part") are specified in rise/run. If so, that means tanθ = 1/20, not sine.

It turns out that it doesn't really matter too much, since tanθ ≈ sinθ for small θ, and θ is pretty small here. But, I thought I should mention it.

Homework Equations



P=FV

The Attempt at a Solution



find force:
force=power/velocity
F=121,000/49

find θ:
θ=sin-1(1/20)

find F at angle θ:
Fθ=(121,000/49)/cos(sin-1(1/20))=2472.48N
I'm not sure I follow what the purpose of this Fθ is. The problem statement says "Assuming that the resistive force (due to friction and air resistance) remains constant..." That tells me that you don't need to calculate a new frictional force. Just use the same frictional force as before.
 

Related to Given Power, Weight, and Vmax find Vmax at angle θ

1. What is the formula for calculating Vmax at angle θ?

The formula for calculating Vmax at angle θ is Vmax = √(Power / Weight) * sin(θ).

2. How do you determine the angle θ for a given Vmax?

The angle θ can be determined by taking the inverse sine of (Vmax * Weight) / Power.

3. Can Vmax be negative in this equation?

No, Vmax cannot be negative in this equation as it represents the maximum velocity which is always a positive value.

4. What is the unit of measurement for Vmax?

The unit of measurement for Vmax is meters per second (m/s).

5. How does the weight of an object affect the Vmax at angle θ?

The weight of an object directly affects Vmax at angle θ. The heavier the object, the lower the Vmax will be at a given angle θ. This is because a heavier object requires more power to achieve the same velocity as a lighter object.

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