A92-kg skier is sliding down a ski slope that makes an angle of 30 degrees above the horizontal direction. the coefficient of kinetic friction between the skis and the snow is 0.10. neglecting any air resistance, what is the acceleration of the skier?

a = 4.05 m/s²

Explanation:

Known data

m= 92 kg  : mass of the  skier

θ =30°  :angle θ of the ski slope  with respect to the horizontal direction

μk= 0.10 : coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Newton's second law:

∑F = m*a Formula (1)

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

m : mass s (kg)

a : acceleration  (m/s²)

We define the x-axis in the direction parallel to the movement of the block on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the skier

W: Weight of the skier : In vertical direction

N : Normal force : perpendicular to the ski slope

f : Friction force: parallel to the ski slope

Calculated of the W

W= m*g

W=  92kg* 9.8 m/s² = 901,6 N

x-y weight components

Wx= Wsin θ= 901,6 N *sin 30° = 450.8 N

Wy= Wcos θ = 901,6 N *cos 30° =780.8 N

Calculated of the N

We apply the formula (1)

∑Fy = m*ay    ay = 0

N - Wy = 0

N = Wy

N = 780.8 N

Calculated of the f

f = μk* N=  0.10*780.8 N  

f = 78.08 N

We apply the formula (1) to calculated acceleration of the skier:

∑Fx = m*ax  ,  ax= a  : acceleration of the block

Wx - f = m*a

450.8- 78.08 = ( 92)*a

372.72 =  (92)*a

a = (372.72)/ (92)

a = 4.05 m/s²