# If you use a horizontal force of 30.0 N to slide a 12.0 kg wooden crate across a ﬂoor at a constant velocity, what is the coefficient of friction between the crate and the floor?

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Nov 4, 2015

#### Answer:

I found: ${\mu}_{k} = 0.25$

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Newton's law is split into two components along $x$ and $y$, that, using our data, give:
$30 - {\mu}_{k} {R}_{n} = 0$
${R}_{n} = m g = 12 \cdot 9.81 = 117.7 \approx 118 N$
So:
${\mu}_{k} = \frac{30}{118} = 0.25$

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