If a 2 kg object moving at 80 m/s slows to a halt after moving 9 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

Apr 19, 2017

The coefficient of kinetic friction is $= 36.28$

Explanation:

We apply the equation of motion to find the deceleration

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

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

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

The distance is $s = 9 m$

$0 = {80}^{2} + 2 \cdot a \cdot 9$

$a = - {80}^{2} / \left(2 \cdot 9\right)$

$a = - 355.6 m {s}^{-} 2$

The frictional force is

${F}_{r} = m a = 2 \cdot 355.6 = 711.1 N$

$N = 2 g$

The coefficient of kinetic friction is

${\mu}_{k} = {F}_{r} / N = \frac{711.1}{2 g} = 36.28$