Elementary Particle Dynamics Q (1 Viewer)

[Mathematician]

Plz solve this q for me.

The combined air and road resistance of a car in motion is proportional to v^2 , where v is its speed. When the engine is disengaged the car moves down an incline making an angle InverseSine(1/30) with the horizontal, with a velocity of 30m/s . Find the force required to drive the car up the incline with a steady speed of 24m/s , given that the mass of the car is 1200kg.

 

[Affinity]

necessary working may have been omitted:

consider the component of gravity parallel to the inclined plane.
that would have a magnitude of (mg/30) and direction down the plane.

the total acceleration when the car is rolling down with the engines off would be :

a = F/m = (kv^2 - mg/30)/m for some positive k

at terminal velocity U, a=0

0=(k/m)U^2 - g/30

and we know that U=30

900(k/m) -g/30 = 0

k = 2g/45


now consider the car driving up.

resistance +gravity in this case is

F= -kv^2 - mg/30

(because the resistance is acting in the same direction as gravity ie both opposing the motion.)

to drive the car up with a steady speed at 24 m/s, the resultant acceleration and hence force must be both 0.

let E be the applied force we want to find.

E + (-k(24)^2 - mg/30) = 0

E = 576k + mg/30

E = 576*2g/45 + 40g

E =642.88 Newtons taking g as 9.8 ms^-2

 

[Harimau]

Is this examinable in the HSC... because it looks damn hard. Can i ask how you got mg/30?

 

[Affinity]

force due to gravity is mg downwards

the component parallel to the plane equals mg*sin(t) where t is the angle of inclination of the plane, show this fact with a diagram. we're told that sin(t) = 1/30 so.. that makes mg/30

yes this is examinable, it's not that hard considering no calculus is involved.