# If an object is moving at 120 m/s over a surface with a kinetic friction coefficient of u_k=8 /g, how far will the object continue to move?

Mar 23, 2016

$\Delta x = 900 \text{ m}$

#### Explanation:

$\text{Object has a kinetic energy due to motion}$
$\text{The kinetic energy of object is changed to heat energy by}$
$\text{the friction force}$
${F}_{f} = \mu \cdot m \cdot g \text{ } \mu = \frac{8}{g}$
${F}_{f} = \frac{8}{\cancel{g}} \cdot m \cdot \cancel{g}$

$\frac{1}{2} \cdot m \cdot {v}^{2} = {F}_{f} \cdot \Delta x$
$\frac{1}{2} \cdot \cancel{m} \cdot {v}^{2} = 8 \cdot \cancel{m} \cdot \Delta x$
$\frac{1}{2} \cdot {120}^{2} = 8 \cdot \Delta x$
$\Delta x = {120}^{2} / 16$
$\Delta x = \frac{14400}{16}$
$\Delta x = 900 \text{ m}$