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

Physics

1 answer:

Sonja [21]2 years ago

8 0

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

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