Question

An Olympic skier moving at 20.0 m/s down a 30.0o slope encounters a region of wet snow, of

Answer 1

Answer:

Option B

Explanation:

Forces acting on the skier-

F1 $= -mg sin(30)$ down the slope  

F2 $= -mg cos(30)$

F3 = friction force  $= 0.74 mg cos(30)$  

Net force, down the slope  

$= -mg sin(30)+ 0.74 mg cos(30)\\= mg(.74 cos(30)-sin(30))\\= 0.14mg\\= 1.38m$

Acceleration $= F/m= 1.38$ m/s2  

Acceleration remains constant, initial speed is 20 m/s

Speed at time t is $1.38t- 20$ m/s

Distance down the slope at time t is $0.69t^2- 20t$

When the skier stops, her speed is 0. Thus,  

, $1.38t- 20= 0\\t= 20/1.38\\= 14.5$ seconds

Distance travelled in 14.5 seconds $= (0.69)(14.52- 20(14.5)= -145 m$(negative because it is down the slope).

Option B is correct