The acceleration of the skier is [tex]4.98 m/s^2[/tex] down along the incline
Explanation:
We can solve this problem by writing the equations of motion of the skier along the incline.
Along the direction perpendicular to the incline, we have:
[tex]N-mg cos \theta=0[/tex] (1)
where
N is the normal reaction of the plane on the skier
[tex]mgcos \theta[/tex] is the component of the weight perpendicular to the plane, with
m = 60 kg is the mass of the skier
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
[tex]\theta=35^{\circ}[/tex] is the angle of the incline
Along the direction parallel to the incline, we have
[tex]mg sin \theta - \mu N = ma[/tex] (2)
where
[tex]mg sin \theta[/tex] is the component of the weight parallel to the incline
[tex]\mu N[/tex] is the force of friction, with
[tex]\mu=0.08[/tex] is the coefficient of friction
a is the acceleration of the skier
From (1) we find
[tex]N=mg cos \theta[/tex]
And substituting into (2),
[tex]mg sin \theta - \mu (mg cos \theta) = ma\\a=g sin \theta - \mu g cos \theta = g(sin \theta-\mu cos \theta) = (9.8)(sin 35^{\circ}-0.08cos 35^{\circ})=4.98 m/s^2[/tex]
Learn more about inclined plane here:
brainly.com/question/5884009
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