Question

A 50.0 kg piece of rock slides down the side of Mt. Etna. That part of the volcano slopes up at 30.0° (from the horizontal). If the rock accelerates at 3.0 m/s2, what is the coefficient of friction between the rock and the volcano?

Answer 1

Answer:

0.224

Explanation:

There are two forces acting on the rock along the direction parallel to the slope:

- The component of the weight parallel to the slope, down, given by

$mg sin \theta$

where

m = 50.0 kg is the mass of the rock

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

$\theta=30^{\circ}$ is the angle of the slope

- The force of friction, up along the slope, given by

$F_f=\mu R$

where

$\mu$ is the coefficient of friction

R is the normal reaction

So the equation of the forces along the direction parallel to the slope is

$mg sin \theta - \mu R = ma$ (1)

where

$a=3.0 m/s^2$ is the acceleration

The normal reaction R can be found by looking at the equation of the forces along the direction perpendicular to the slope: in fact, we have that R balances the component of the weight perpendicular to the slope, so

$R=mg cos \theta$

And substituting into (1)

$mg sin \theta - \mu mg cos \theta = ma$

And solving for $\mu$, we find the coefficient of friction:

$\mu = \frac{g sin \theta -a}{g cos \theta}=\frac{(9.8)(sin 30)-3.0}{(9.8)(cos 30)}=0.224$