Answer:
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
