# If an object is moving at 2 ms^-1 over a surface with a kinetic friction coefficient of mu_k=8/g, how far will the object continue to move?

Feb 20, 2016

The object will move $\frac{1}{4}$ (or 0.25) $m$ before it stops.

#### Explanation:

The frictional force will be given by:

${F}_{\text{frict}} = \mu \cdot {F}_{N}$ where ${F}_{N}$ is the normal force, which is the weight force of the object: ${F}_{N} = m g$.

${F}_{\text{frict}} = \mu \cdot {F}_{N} = \frac{8}{g} \cdot m g = 8 m$

The acceleration of the object will be described by Newton's Second Law :

$a = \frac{F}{m} = \frac{8 m}{m} = 8$ $m {s}^{-} 2$

From its initial velocity, $u = 2.0$ $m {s}^{-} 1$, to $v = 0$ $m {s}^{-} 1$, the distance traveled will be given by:

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

Rearranging:

$d = \frac{{v}^{2} - {u}^{2}}{2 a} = \frac{0 - {2}^{2}}{2 \cdot 8} = \frac{4}{16} = \frac{1}{4}$ $m$.