A65 kg skier starts at rest at the top of a 150 m long hill that has an incline of 28 degrees. how fast will she be going at the bottom of there is 50n of friction?

34 m/s

Explanation:

Potential energy at top = kinetic energy at bottom + work done by friction

PE = KE + W

mgh = ½ mv² + Fd

mg (d sin θ) = ½ mv² + Fd

Solving for v:

½ mv² = mg (d sin θ) − Fd

mv² = 2mg (d sin θ) − 2Fd

v² = 2g (d sin θ) − 2Fd/m

v = √(2g (d sin θ) − 2Fd/m)

Given g = 9.8 m/s², d = 150 m, θ = 28°, F = 50 N, and m = 65 kg:

v = √(2 (9.8 m/s²) (150 m sin 28°) − 2 (50 N) (150 m) / (65 kg))

v = 33.9 m/s

Rounded to two significant figures, her velocity at the bottom of the hill is 34 m/s.

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