# If a 2 kg object moving at 80 m/s slows to a halt after moving 4 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

Nov 5, 2017

The coefficient of kinetic friction is $= 81.6$

#### Explanation:

Apply the equation of motion

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

to determine the acceleration $a$

The initial speed is $u = 80 m {s}^{-} 1$

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

The distance is $s = 4 m$

The acceleration is

$a = \frac{{v}^{2} - {u}^{2}}{2 s} = \frac{{0}^{2} - {80}^{2}}{2 \times 4} = - 80 \cdot \frac{80}{8} = - 800 m {s}^{-} 2$

The frictional force is ${F}_{r} = m a = 2 \times 800 = 1600 N$

Let the acceleration due to gravity be $g = 9.8 m {s}^{-} 2$

The normal force is $N = m g = 2 \cdot 9.8 = 19.6 N$

The coefficient of kinetic friction is

${\mu}_{k} = {F}_{r} / N = \frac{1600}{19.6} = 81.6$