# If an object is moving at 12 m/s over a surface with a kinetic friction coefficient of u_k=2 /g, how far will the object continue to move?

Mar 9, 2016

$\Delta x = 36 m$

#### Explanation:

$\text{Changing of energy of the kinetic is equal work doing}$
$\text{ by the Friction Force}$
$\Delta {E}_{k} = W$
${E}_{k 1} = \frac{1}{2} \cdot m \cdot {v}^{2}$
${E}_{k 2} = 0 \text{ object stops.}$
$\Delta {E}_{k} = {E}_{k 2} - {E}_{k 1} \text{ } \Delta {E}_{k} = \frac{1}{2} \cdot m \cdot {v}^{2}$
${W}_{f} = {F}_{f} \cdot \Delta x \text{ work doing by the friction force}$
${W}_{f} = {u}_{k} \cdot N \cdot \Delta x$
${W}_{f} = \Delta {E}_{k}$
${u}_{k} \cdot N \cdot \Delta x = \frac{1}{2} \cdot m \cdot {v}^{2} \text{ } \left(1\right)$
$N : \text{Normal Force acting contacting surfaces. } N = m \cdot g$
$\text{rearranging the equation (1)}$
${u}_{k} \cdot \cancel{m} \cdot g \cdot \Delta x = \frac{1}{2} \cdot \cancel{m} \cdot {v}^{2}$
${u}_{k} = \frac{2}{g} \text{ } v = 12 \frac{m}{s}$
$\frac{2}{\cancel{g}} \cdot \cancel{g} \cdot \Delta x = \frac{1}{2} \cdot {12}^{2}$
$\Delta x = \frac{144}{4}$
$\Delta x = 36 m$