Answer:
Option B
Explanation:
Forces acting on the skier-
F1 [tex]= -mg sin(30)[/tex] down the slope
F2 [tex]= -mg cos(30)[/tex]
F3 = friction force [tex]= 0.74 mg cos(30)[/tex]
Net force, down the slope
[tex]= -mg sin(30)+ 0.74 mg cos(30)\\= mg(.74 cos(30)-sin(30))\\= 0.14mg\\= 1.38m[/tex]
Acceleration [tex]= F/m= 1.38[/tex] m/s2
Acceleration remains constant, initial speed is 20 m/s
Speed at time t is [tex]1.38t- 20[/tex] m/s
Distance down the slope at time t is [tex]0.69t^2- 20t[/tex]
When the skier stops, her speed is 0. Thus,
, [tex]1.38t- 20= 0\\t= 20/1.38\\= 14.5[/tex] seconds
Distance travelled in 14.5 seconds [tex]= (0.69)(14.52- 20(14.5)= -145 m[/tex](negative because it is down the slope).
Option B is correct
