Respuesta :
Answer:
Net acceleration, [tex]a=3.55\ m/s^2[/tex]
Explanation:
It is given that,
mass of the person, m = 56 kg
Inclination with the horizontal, [tex]\theta=30^{\circ}[/tex]
The coefficient of friction between the skis and the snow is 0.16
Let [tex]a_g[/tex] is the acceleration due to gravity on the object,
[tex]a_g=g\ sin\theta[/tex]
[tex]a_g=9.8\ sin(30)[/tex]
[tex]a_g=4.9\ m/s^2[/tex]
Let [tex]a_f[/tex] is the acceleration of the normal reaction. It is given by:
[tex]a_f=\mu g\ cos\theta[/tex]
[tex]a_f=0.16\times 9.8\times cos(30)[/tex]
[tex]a_f=1.35\ m/s^2[/tex]
Let a is the net acceleration of the person. It is equal to,
[tex]a=a_g-a_f[/tex]
[tex]a=4.9-1.35[/tex]
[tex]a=3.55\ m/s^2[/tex]
So, the magnitude of net acceleration is [tex]3.55\ m/s^2[/tex]. Hence, this is the required solution.
Answer:
3.54 m/s²
Explanation:
mass, m = 56 kg
inclination, θ = 30°
coefficient of friction, μ = 0.16
As the person moves down
So, the acceleration is
a = g Sinθ - μ g Cos θ
a = 9.8 ( Sin 30 - 0.16 Cos 30)
a = 9.8 (0.5 - 0.139)
a = 3.54 m/s²
Thus, the acceleration is 3.54 m/s².
