# If a 1/2 kg object moving at 7/4 m/s slows to a halt after moving 3/8 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

Jan 21, 2017

The answer is ${\mu}_{k} = 1.25$

#### Explanation:

We solve in the horizontal direction ${\rightarrow}^{+}$

$u = \frac{7}{4} m {s}^{-} 2$

$v = 0$

$s = \frac{3}{8} m$

$a =$ acceleration

We use the equation

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

$0 = \frac{49}{16} + 2 \cdot \frac{3}{8} \cdot a$

$a = - \frac{49}{16} \cdot \frac{4}{3} = - \frac{49}{12} m {s}^{-} 2$

By Newton`s second Law

${F}_{r} = m a = - \frac{49}{4} \cdot \frac{1}{2} = - \frac{49}{8} N$

The reaction is $N = \frac{1}{2} g = \frac{1}{2} \cdot 9.8 = 4.9 N$

The coefficient of kinetic friction is ${\mu}_{k} = {F}_{r} / N = \frac{49}{8} \cdot \frac{1}{4.9} = 1.25$