elementary particle dynamics (1 Viewer)

[shsshs]

hi could someone please help out here im quite confused about this..

the total air+road resistance of a car is proportional to v^2 (v is velocity). When the engine is disengaged the car moves down as inclined plane, angle arcsin (1/30), with a velocity 30ms.

Find the force necessary to drive the car up the plane with a steady speed of 24ms. The mass of the car is 1200 and gravity is 10.



I dont understand how the a force can produce a "steady speed", since F=ma.
Could someone please post the solution?

Also does this type of question come out often in the HSC? Because some books dont have this, they just start on resisted motion. thanks

 

[SeDaTeD]

I presume when it states that it moves down the incline plane with v = 30m/s, that would be its terminal velocity, which occurs when the resistive force is equal to the component of gravity along the slope. (Draw a free body diagram and resolve forces to be along the slope and perpendicular to the slope).
Setting them equal yields: mkv^2 = mgsin(theta), with v=30, theta being arcsin (1/30). This should give k = 1/2700.

Now consider when there is a driving force up the slope. In this case the resistive force would be pointing down the slope, as the car is travelling up. Since it is moving at a steady speed up the slope, the net force along the slope is equal to zero, so this yields F = mkv^2 + mgsin(theta), where F is the driving force. Subbing should yield 656N if I am not mistaken.