A44 kg skier skis directly down a frictionless slope angled at 15° to the horizontal. choose the positive direction of the x axis to be downhill along the slope. a wind force with component fx acts on the skier. what is fx if the magnitude of the skier's velocity is (a) constant, (b) increasing at a rate of 1.2 m/s2, and (c) increasing at a rate of 2.4 m/s2. use g = 9.81 m/s2?

a) Fx= 111.72 N

b) Fx= 58.92 N

c)  Fx = 6.12 N

Explanation:

Conceptual analysis

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Data:

m=44 kg

β = 15°

g = 9.81 m/s²

W= m*g = 44 kg*9.81 m/s²= 431.64 N : skier skis weight

x-y weight components

Wx= Wsinβ=431.64 N*sin15° = 111.72 N

Wy= Wcosβ=431.64 N*sin15° =416.9 N

Problem development

We apply the formula (1)

a)  velocity is constant, then, ax=0

∑Fx = m*ax

Wx-Fx= m*0

Wx=Fx

b) ax= 1.2m/s²

∑Fx = m*ax

Wx-Fx= m*ax

111.72 -Fx= 44*1.2

111.72-44*1.2 = Fx

Fx= 58.92 N

c) ax= 2.4m/s²

∑Fx = m*ax

Wx-Fx= m*ax

111.72 -Fx= 44*2.4

111.72-44*2.4 = Fx

Fx = 6.12 N

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