# If a 4 kg object moving at 5/2 m/s slows to a halt after moving 12 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

Apr 15, 2018

#### Answer:

The coefficient of kinetic friction is $= 0.027$

#### Explanation:

Apply the equation of motion

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

To calculate the acceleration

The initial velocity is $u = \frac{5}{2} m {s}^{-} 1$

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

The distance is $s = 12 m$

The acceleration is

$a = \frac{{v}^{2} - {u}^{2}}{2 s} = \frac{0 - {\left(\frac{5}{2}\right)}^{2}}{2 \cdot 12} = - 0.26 m {s}^{-} 2$

According to Newton's Second Law, the force of friction is

${F}_{r} = m a = 4 \cdot 0.26 = 1.042 N$

The normal force is

$N = m g = 4 \cdot 9.8 = 39.2 N$

The coefficient of kinetic friction is

${\mu}_{k} = {F}_{r} / N = \frac{1.042}{39.2} = 0.027$