Question

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

coefficient of friction μk = 0.740. How far down the slope does she go before stopping?


a.119 m

b.145 m

c.170 m

d.199 m

Answer 1

Option B

Step-by-step 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