# If an object is moving at 15 m/s over a surface with a kinetic friction coefficient of u_k=5 /g, how far will the object continue to move?

May 24, 2017

#### Answer:

The distance is $= 22.5 m$

#### Explanation:

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

Then,

the coefficient of friction is

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

$N = m g$

So,

${F}_{r} = {\mu}_{k} \cdot N = \frac{5}{g} \cdot m g = 5 m$

We apply Newton's Second Law

$F = m a = 5 m$

The deceleration is $= 5 m {s}^{-} 2$

We apply the equation of motion

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

$v = 0$

$u = 15 m {s}^{-} 1$

$a = - 5 m {s}^{-} 2$

$0 = {15}^{2} - 2 \cdot 5 \cdot s$

$s = {15}^{2} / 10 = 22.5 m$