Askier is moving down a snowy hill with an acceleration of 0.40 m/s^2. the angle of the slope is 5.0° to the horizontal. what is the acceleration of the same skier when she is moving down a hill with a slope of 19°? assume the coefficient of kinetic friction is the same in both cases.

ax=2.76  m/s²

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:

a=0.40 m/s²

β =5°

g = 9.81 m/s² : acceleration due to gravity

W= m*g

x-y weight components

Wx= Wsinβ= m*g*sin5°

Wy= Wcosβ=m*g*cos5°

Friction force : Ff

Ff= μk*N

μk: coefficient of kinetic friction

N : Normal force (N)  

Problem development

look at the skier's free body diagram in the attached graphic

We apply the formula (1)

∑Fy = m*ay    , ay=0

N-Wy = 0

N = Wy

N= m*g*cos5°

∑Fx = m*ax

Wx-Ff= m*ax

m*g*sin5° -μk*m*g*cos5° = m*0.4 : we divide by m on both sides of the equation:

g*sin5°- μk*g*cos5°= 0.4

g*sin5° - 0.4 = μk*g*cos5°

μk = (9.8*sin5° - 0.4 ) /( 9.8*cos5°)

μk = 0.0465

Acceleration of the same skier when she is moving down a hill with a slope of 19°, μk = 0.0465

∑Fx = m*ax

Wx-Ff= m*ax     Wx=m*g*sin 19°, N=*m*g*cos19°, Ff=μk*N

m*g*sin 19°-(0.0465)*(m*g*cos19°) =  m*ax   we divide by m , and g= 9.8

9.8*sin 19°-(0.0465)*(9.8*cos19°) = ax

ax=2.76  m/s²

from

explanation:

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