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**Refujoi** · June 30th 2006, 12:19 PM

A wooden block slide directly down an inclined plane velocity of 6.0 m/s How large is the coefficient of kinetic friction, if the plane makes an angle of 25 deegre with the horizontal?

**CaptainBlack** · July 1st 2006, 01:29 AM

Originally Posted by Refujoi

A wooden block slide directly down an inclined plane velocity of 6.0 m/s How large is the coefficient of kinetic friction, if the plane makes an angle of 25 deegre with the horizontal?

As the block slides with constant velocity down the plane the Normal Reaction
$N$ , the Frictional force $F$ , and the Gravitational force $mg$ on the block are in equilibrium.

Resolve the gravitational force into components down the plane and normal to
the plane and set these equal to the frictional force and the normal reaction
(allowing for the appropriate signs):

$F=mg \sin(\theta)$ ,
$N=mg \cos(\theta)$

where $\theta$ is the inclination of the plane.

Now $F=\mu N$ , where $\mu$ is the coefficient of
kinetic/sliding ... friction. So:

$\mu\ mg \cos(\theta)=mg \sin(\theta)$

rearranging:

$\mu=\tan(\theta)$ .

In this problem $\theat=25^{\circ}$ , so:

$\mu=\tan(25^{\circ}) \approx 0.466$ .

RonL

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