# If a 13 kg object moving at 4 m/s slows down to a halt after moving 80 m, what is the friction coefficient of the surface that the object was moving over?

Mar 29, 2018

The coefficient of friction is $= 0.01$

#### Explanation:

Apply the equation of motion to find the acceleration

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

The initial velocity is $u = 4 m {s}^{-} 1$

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

The distance is $s = 80 m$

The acceleration is

$a = \frac{{v}^{2} - {u}^{2}}{2 s} = \frac{0 - {4}^{2}}{2 \cdot 80} = - \frac{16}{160} = - 0.1 m {s}^{-} 2$

The mass of the object is $m = 13 k g$

According to Newton's Second Law of Motion

$F = m a$

The force of friction is ${F}_{r} = 13 \cdot 0.1 = 1.3 N$

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

The normal reaction is $N = m g = 13 \cdot 9.8 = 127.4 N$

The coefficient of kinetic friction is

${\mu}_{k} = {F}_{r} / N = \frac{1.3}{127.4} = 0.01$