A skier is sliding down a slope. The gravitational force acting on the skier is directed perpendicular to the slope. The normal force is directed perpendicular to the slope, pushing the skier into the slope. The friction force is parallel to the slope, opposing the skier's motion down the slope. For an object on an inclined plane sliding at constant velocity, the force of friction equals the component of gravity parallel to the plane, and the normal force equals the component of gravity perpendicular to the plane.

1. A skier is sliding down a slope. The gravitational force acting upon the skier is directed _____.a. straight upward b. straight downward c. perpendicular to the slope d. parallel to the slope

2. A skier is sliding down a slope. The gravitational force acting upon the skier is directed _____.a. straight upward b. straight downward c. perpendicular to the slope d. parallel to the slopeFg

3. The normal force acting upon the skier is directed _____.a.perpendicular to the slope b. parallel to the slopec. straight downward d. straight upwardFg

4. The normal force acting upon the skier is directed _____.a.perpendicular to the slope b. parallel to the slopec. straight downward d. straight upwardFnFg

5. The friction force acting upon the skier is directed _____.a. parallel to the slope b. straight upward c. straight downward d. perpendicular to the slope FnFg

6. The friction force acting upon the skier is directed _____.a. parallel to the slope b. straight upward c. straight downward d. perpendicular to the slope FnFfrFg

7. An object upon an inclined plane is sliding at a constant speed down the incline. Friction is present. Which one of the following diagrams represent the free-body diagram for such an object?

8. How large is the force of friction?FnFfrFg

9. How large is the force of friction?Construct a coordinate system…+yFnFfrFg+x

10. How large is the force of friction?Construct a coordinate system…+yFnFfrqFg+xq

11. How large is the force of friction?Construct a coordinate system……now break Fg into its components…+yFnFfrFperpqFg+xqFpar

12. A downhill skier has a weight of 500 N,and is skiing down a 30 hill. Determine FperpFpar+yFnFfrFperpqFg+xqFpar

13. A downhill skier has a weight of 500 N,and is skiing down a 30 hill. Determine Fperp = (500 N) cos 30 = 433 NFpar= (500 N) sin 30=250 N+yFnFfrFperp30Fg= 500 N+xFpar

14. A downhill skier has a weight of 500 N,and is skiing down a 30 hill. Determine +yFnFfr433 N = Fperp30Fg= 500 N+x250 N = Fpar

15. If the skier is not accelerating, h how large is the force of friction? How large is the normal force?+yFnFfr250 N = Fpar433 N = Fperp30Fg= 500 N+x

16. A 12 kg box slides down a ramp at a constant velocity.Sketch a free body diagram for the box. 20

17. A 12 kg box slides down a ramp at a constant velocity.Sketch a free body diagram for the box.FnFfrFg 20

18. A 12 kg box slides down a ramp at a constant velocity.b. Make a tilted coordinate system and find Fperp andFpar. FnFfrFg 20

19. A 12 kg box slides down a ramp at a constant velocity.b. Make a tilted coordinate system and find Fperp andFpar. +yFnFfrFg+x 20

20. A 12 kg box slides down a ramp at a constant velocity.b. Make a tilted coordinate system and find Fperp andFpar. Fperp = (117.6 N) cos 20 = 110.5 NFpar = (117.6 N) sin 20=40.2 N+yFnFfr 20Fg+x 20

21. A 12 kg box slides down a ramp at a constant velocity.c. Determine the force of friction and the normal force on the block.Ff = Fn =+yFnFfr110.5 N 20Fg+x 2040.2 N

22. A 12 kg box slides down a ramp at a constant velocity.c. Determine the force of friction and the normal force on the block.Ff = 40.2 NFn = 110.5 N+yFnFfr40.2 N110.5 N+x 20

23. A 12 kg box slides down a ramp at a constant velocity. If there was no friction, what would be the acceleration down the ramp? FNET = m a 40.2 N = (12 kg) aa = 3.35 m/s^2+yFn40.2 N110.5 N+x 20