# If a 4 kg object moving at 16 m/s slows down to a halt after moving 800 m, what is the friction coefficient of the surface that the object was moving over?

Jun 2, 2016

${\mu}_{k} = 0.013 \text{ has no unit}$

#### Explanation:

${v}_{f}^{2} = {v}_{i}^{2} - 2 \cdot a \cdot \Delta x$

${v}_{f} = 8 \text{ } \frac{m}{s}$

${v}_{i} = 16 \text{ } \frac{m}{s}$

$\Delta x = 800 \text{ } m$

$a : \text{acceleration of the object}$

${8}^{2} = {16}^{2} - 2 \cdot a \cdot 800$

$64 = 256 - a \cdot 1600$

$1600 \cdot a = 256 - 64$

$1600 \cdot a = 192$

$a = \frac{192}{1600} \text{ } \frac{m}{s} ^ 2$

${F}_{f} = {\mu}_{k} \cdot \cancel{m} \cdot g = \cancel{m} \cdot a$

${\mu}_{k} = \text{coefficient of friction}$

${\mu}_{k} \cdot g = a$

${\mu}_{k} = \frac{a}{g}$

${\mu}_{k} = \frac{\frac{192}{1600}}{9.18}$

${\mu}_{k} = \frac{192}{1600 \cdot 9.18}$

${\mu}_{k} = \frac{192}{14688}$

${\mu}_{k} = 0.013$