Respuesta :
L = length of the incline = 75 m
θ = angle of incline = 22 deg
h = height of skier at the top of incline = L Sinθ = (75) Sin22 = 28.1 m
μ = Coefficient of friction = 0.090
N = normal force by the surface of incline
mg Cosθ = Component of weight of skier normal to the surface of incline opposite to normal force N
normal force "N" balances the component of weight opposite to it hence we get
N = mg Cosθ
frictional force acting on the skier is given as
f = μN
f = μmg Cosθ
v = speed of skier at the bottom of incline
Using conservation of energy
potential energy at the top of incline = kinetic energy at the bottom + work done by frictional force
mgh = f L + (0.5) m v²
mgh = μmg Cosθ L + (0.5) m v²
gh = μg Cosθ L + (0.5) v²
(9.8 x 28.1) = (0.09 x 9.8 x 75) Cos22 + (0.5) v²
v = 20.7 m/s
