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## Homework Statement

A 21.0 kg box is released on a 38.0° incline and accelerates down the incline at 0.267 m/s

^{2}. Find the friction force impeding its motion.

Friction force is correctly calculated as 121N

## Homework Equations

F

_{fr}= [tex]\mu[/tex]F

_{N}(Friction force equals mu times normal force, in case the symbols are not in its commonly used form)

F

_{N}=mg (Normal force equals mass times gravity)

## The Attempt at a Solution

F

_{fr}= [tex]\mu[/tex]F

_{N}

121=[tex]\mu[/tex]F

_{N}

121=[tex]\mu[/tex]mg

121=[tex]\mu[/tex](21)(9.8)

5.7619=tex]\mu[/tex](9.8)

0.588=[tex]\mu[/tex]

I see this as the correct method, but the answer came out as incorrect, so that means I'm wrong. Is this formula incorrect for finding mu? If so, what is supposed to be used/what am I supposed to do in order to find mu?

note: numbers are obtained using longer, more accurate numbers of 121.09613162202047395584132290722 / ((21)(9.8)) = 0.5879494655004859086491739552964 but shortened down to look better