A 35kg skier skis directly down a frictionless slope angled at 11° to the horizontal. Assume the skier moves in the negative direction of an x axis along the slope. A wind force with component Fx acts on the skier. What is Fx (in N) if the magnitude of the skier's velocity is.. constant? ..increasing at a rate of 1.0 m/s^2? ..increasing at a rate of 2.0 m/s^2?
Writing the sum of the forces down the slope....
Weight*sinThetaWindfriction=mass*acceleration.
In the first part, a=0
so..
35*sin(11)Fx=0
and solvee for Fx?
.. as for the increasing rate.. is it telling me that the 1.0m/s2 is added to the gravity of 9.8m/s2?
3 answers

this is kinda late answer, but still...
35*sin(11)Fx=0
This works for a = 0
35*sin(11)+maFx=0
More complete answer... (a!=0)
So, for increasing rate of a = 1m/s^2
This is going in negative direction, so
35*sin(11)+(35)(1)=Fx
And for a = 22/s^2
35*sin(11)+(35)(2)=Fx
Crunch in calculator for Fx in Newtons

Actually, just noticed that was for in x direction, so change sign on acceleration
35*sin(11)+(35)(1)=Fx
35*sin(11)+(35)(2)=Fx

ur supposed to multiply the 35sintheta with the acceleration due to gravity (9.8)
