Question

Calculate the acceleration of a skier heading down a 10.0º slope, assuming the coefficient of friction for waxed wood on wet snow. (b) Find the angle of the slope down which this skier could coast at a constant velocity. You can neglect air resistance in both parts. Explicitly show how you follow the steps in the Problem-Solving Strategies.

Answer 1

Solution:

[Note: Waxed Wood on Wet Snow Kinetic friction is 0.1]

Normal force of the slope on the skier is
Fn = mgcosθ

Friction force is
Ff = μFn
Ff = μmgcosθ

Which is the force restricting acceleration down hill

The force accelerating the skier down hill is

Fa = mgsinθ

The net force is

F = Fa - Ff
F = mgsinθ - μmgcosθ
F = mg(sinθ - μcosθ)

a)From Newton

F = ma
we can then see that the net acceleration is
a = g(sinθ - μcosθ)

a = 9.81(sin10 - 0.1cos10)

a = 0.74 m/s²

b) at constant velocity the acceleration equals zero

a = g(sinθ - μcosθ)
0 = g(sinθ - μcosθ)
as g is assumed to be not zero, then
0 = sinθ - μcosθ
sinθ = μcosθ
sinθ/cosθ = μcosθ/cosθ
tanθ = μ
tanθ = 0.1
θ = 5.71°

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