Question

A skier of mass 103 kg comes down a slope of constant angle 21◦ with the horizontal.
What is the force on the skier parallel to the slope? Neglect friction. The acceleration
of gravity is 9.8m/s . Answer in units of N.

Answer 1

The force on the skier parallel to the slope is 362 N

Explanation:

Since we are neglecting friction, there are only two forces acting on the skier:

• The force of gravity, $mg$, acting downward and vertically
• The normal reaction of the slope on the skier, perpendicular to the slope

Here we have to find the force on the skier parallel to the slope. Since the normal reaction is perpendicular to the slope, it does not have any component parallel to it, so we just need to find the component of the force of gravity parallel to the slope.

The component of the force of gravity parallel to the slope is given by

$W_p = mg sin \theta$

where

m = 103 kg is the mass of the skier

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

$\theta=21^{\circ}$ is the angle of the slope

Substituting these numbers, we find:

$W=(103)(9.8)(sin 21^{\circ})=362 N$

Learn more about inclined planes here:

brainly.com/question/5884009

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