# If an object is moving at 8 m/s over a surface with a kinetic friction coefficient of u_k=64 /g, how far will the object continue to move?

Apr 11, 2018

The distance is $= 0.5 m$

#### Explanation:

The coefficient of kinetic friction is

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

Let the mass of the object be $= m k g$

Then,

The normal force $N = m g$

The acceleration due to gravity is $g = 9.8 m {s}^{-} 2$

Therefore,

${F}_{r} = {\mu}_{k} N = {\mu}_{k} m g$

But according to Newton's Second Law

$F = m a$

Therefore,

$m a = {\mu}_{k} m g$

The decelleration is $a = {\mu}_{k} g$

Apply the equation of motion

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

$0 = {8}^{2} - 2 {\mu}_{k} g s$

${\mu}_{k} = \frac{64}{g}$

The distance is

$s = \frac{64}{2 \cdot \frac{64}{g} \cdot g} = \frac{1}{2} = 0.5 m$