Question

A 60.0-kg skier starts from rest at a height of 10.0 m on a 5.0° inclined snow-covered slope. The coefficient of kinetic friction between her waxed skis and the snow is 0.050. If the skier goes straight down the slope, what is the speed of the skier when she reaches the bottom?

Answer 1

9.2 m/s

Step-by-step explanation:

There are three forces acting on the skier:

Weight force mg pulling down,

Normal force N pushing normal to the incline,

Friction force Nμ pushing parallel up the incline.

Sum the forces in the normal direction:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

Sum the forces in the parallel direction:

∑F = ma

mg sin θ − Nμ = ma

mg sin θ − mg cos θ μ = ma

g (sin θ − μ cos θ) = a

Plug in values:

a = 9.8 m/s² (sin 5.0° − 0.050 cos 5.0°)

a = 0.366 m/s²

Given:

s = 10.0 m / sin 5.0° = 114.7 m

u = 0 m/s

a = 0.366 m/s²

Find: v

v² = u² + 2as

v² = (0)² + 2 (0.366) (114.7)

v = 9.16 m/s

Rounded to two significant figures, the final speed is 9.2 m/s.