# If an object is moving at 8 m/s over a surface with a kinetic friction coefficient of u_k=9 /g, how far will the object continue to move?

May 5, 2018

The distance is $= 3.56 m$

#### Explanation:

The coeffcient of kinetic friction is ${\mu}_{k} = \frac{9}{g}$

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

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

Then,

The normal reaction is $N = m g N$

The frictional force is

${F}_{r} = {\mu}_{k} N = {\mu}_{k} m g = \frac{9}{g} \cdot m g = 9 m N$

According to Newton's Second Law

$F = m a$

The acceleration is $a = \frac{F}{m} = {F}_{r} / m = - 9 \frac{m}{m} = - 9 m {s}^{-} 2$

The initial velocity is $u = 8 m {s}^{-} 1$

Apply the equation of motion

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

The final velocity $v = 0$

$0 = 64 - 2 \times 9 \times s$

The distance is

$s = \frac{64}{18} = 3.56 m$