If a 1/2 kg object moving at 1/4 m/s slows to a halt after moving 1/8 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

Feb 9, 2016

$k \cong 0 , 408$

Explanation:

$k : \text{coefficient of friction}$
${V}^{2} = {V}_{i}^{2} - 2 \cdot a \cdot x$
$V = 0 \text{ if object stops.}$
$0 = {\left(\frac{1}{4}\right)}^{2} - \cancel{2} \cdot a \cdot \frac{1}{\cancel{8}}$
$\frac{1}{16} = \frac{a}{4}$
$a = 4 \frac{m}{s} ^ 2 \text{ acceleration}$
$k \cdot N = m \cdot a$
$\text{N: normal force contacting surface}$
$N = m \cdot g$
$k \cdot \cancel{m} \cdot g = \cancel{m} \cdot a$
$k = \frac{a}{g}$
$k = \frac{4}{9 , 81}$
$k \cong 0 , 408$