# If a 1/24 kg object moving at 1/3 m/s slows to a halt after moving 10/27 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

Apr 15, 2017

The coefficient of kinetic friction is $= 0.015$

#### Explanation:

The coefficient of kinetic friction is

${\mu}_{k} = {F}_{r} / N$

${F}_{f} = m a$

We apply the equation of motion

${v}^{2} = {u}^{2} + 2 a s$

$0 = {\left(\frac{1}{3}\right)}^{2} + 2 \cdot a \cdot \frac{10}{27}$

$\frac{20}{27} a = - \frac{1}{9}$

$a = - \frac{27}{9} \cdot \frac{1}{20} = - \frac{3}{20} m {s}^{-} 2$

${F}_{r} = \frac{1}{24} \cdot - \frac{3}{20} = - \frac{1}{160} N$
${\mu}_{k} = \frac{1}{160} \cdot \frac{1}{\frac{1}{24} g}$
$= \frac{1}{65.3} = 0.015$