# If an object is moving at 150 m/s over a surface with a kinetic friction coefficient of u_k=15 /g, how far will the object continue to move?

Jun 7, 2017

The distance is $= 150 m$

#### Explanation:

The initial velocity is $u = 150 m {s}^{-} 1$

The final velocity is $v = 0 m {s}^{-} 1$

The distance is s=?m

The coefficient of friction is

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

Let the mass be $= m k g$

Then

the normal reaction is $N = m g N$

So,

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

According to Newton's Second Law

$F = m a$

Therefore,

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

The acceleration is $a = - {\mu}_{k} g$ , the negative sign indicates a deceleration

The coefficient of kinetic friction is ${\mu}_{k} = \frac{15}{8}$

We apply the equation of motion

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

$0 = {150}^{2} - 2 \cdot \frac{15}{g} \cdot g \cdot s$

$s = {150}^{2} / 30 = 750 m$