Question

A skier moves down a 12° slope at constant speed what can you say about the coefficient of friction assume the speed is low enough that the air resistance can be ignored

Answer 1

The coefficient of friction is 0.213

Explanation:

We start by writing the equation of the forces along the directions parallel and perpendicular to the incline. We have:

Perpendicular direction:

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

where

N is the normal reaction of the plane

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

m is the mass

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

$\theta=12^{\circ}$  is the angle of the incline

Parallel direction:

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

where :

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

$\mu_k N$ is the force of friction, where

$\mu_k$ is the coefficient of friction

N is the normal reaction  

a is the acceleration

In this case, the skier is moving at constant speed, so its acceleration is zero and the equation becomes:

$mg sin \theta - \mu_k N =0$ (2)

From (1) we find

$N=mg cos \theta$

And substituting into (2)

$mg sin \theta - \mu_k mg cos \theta = 0$

Solving for $\mu_k$,

$\mu_k = tan \theta = tan(12)=0.213$

Learn more about slope and force of friction here:

brainly.com/question/5884009

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