Question

The 60kg skier is shown below skiing down a 35° incline with a coefficient of friction is 0.08. Determine the acceleration of the skater

Answer 1

The acceleration of the skier is $4.98 m/s^2$ 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:

$N-mg cos \theta=0$ (1)

where

N is the normal reaction of the plane on the skier

$mgcos \theta$ is the component of the weight perpendicular to the plane, with

m = 60 kg is the mass of the skier

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=35^{\circ}$ is the angle of the incline

Along the direction parallel to the incline, we have

$mg sin \theta - \mu N = ma$ (2)

where

$mg sin \theta$ is the component of the weight parallel to the incline

$\mu N$ is the force of friction, with

$\mu=0.08$ is the coefficient of friction

a is the acceleration of the skier

From (1) we find

$N=mg cos \theta$

And substituting into (2),

$mg sin \theta - \mu (mg cos \theta) = ma\\a=g sin \theta - \mu g cos \theta = g(sin \theta-\mu cos \theta) = (9.8)(sin 35^{\circ}-0.08cos 35^{\circ})=4.98 m/s^2$

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