text: # physics
  
  posted by .
  
  A 35-kg 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*sinTheta-Windfriction=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?
  
  • physics -
  
  this is kinda late answer, but still...
  
  35*sin(11)-Fx=0
  This works for a = 0
  
  35*sin(11)+ma-Fx=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
  
  • physics -
  
  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
  
  • physics -
  
  ur supposed to multiply the 35sintheta with the acceleration due to gravity (9.8)
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