# If a 5 kg object moving at 6 m/s slows to a halt after moving 2 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

Jan 25, 2018

The coefficient of kinetic friction is $= 0.92$

#### Explanation:

The mass of the object is $m = 5 k g$

The initial speed of the object is $u = 6 m {s}^{-} 1$

The final speed of the object is $v = 0 m {s}^{-} 1$

The distance is $s = 2 m$

Apply the equation of motion

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

to calculate the acceleration

The acceleration is

$a = \frac{{v}^{2} - {u}^{2}}{2 s} = \frac{{0}^{2} - {6}^{2}}{2 \cdot 2} = - \frac{36}{4} = - 9 m {s}^{-} 2$

According to Newton's Second Law

$F = m a$

The frictional force is ${F}_{r} = 5 \cdot 9 = 45 N$

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

The normal force is $N = m g = 5 \cdot 9.8 = 49 N$

The coefficient of kinetic friction is

${\mu}_{k} = {F}_{r} / N = \frac{45}{49} = 0.92$