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

[Jack_Straw]

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:
F_θ=(121,000/49)/cos(sin^-1(1/20))=2472.48N

find F_N at angle θ:
F_Nθ=(1490N)(sin(sin^-1(1/20)=74.5N

find F_netθ:
F_netθ=F_θ-F_Nθ=2472.48N-74.5N=2397.98

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

This answer is reported as wrong. Any ideas? Thanks

 

[collinsmark]

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:
F_θ=(121,000/49)/cos(sin^-1(1/20))=2472.48N

I'm not sure I follow what the purpose of this F_θ 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.