Homework Help: physics

Posted by **chocolatekisses on Thursday, February 8, 2007 at 5:14pm.

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 - JT, Wednesday, February 9, 2011 at 2:20pm
this is kinda late answer, but still...

35*sin(11)-Fx=0
This works for a = 0

35*sin(11)+ma-Fx=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
physics - JT, Wednesday, February 9, 2011 at 2:22pm
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

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