# If a 4 kg object moving at 1 m/s slows down to a halt after moving 90 m, what is the friction coefficient of the surface that the object was moving over?

Feb 16, 2016

$5 \cdot {10}^{-} 4$

#### Explanation:

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

$a = \frac{\Delta {v}^{2}}{2 s} = \frac{1}{180}$

$f = \mu N$

$N = m g = 40 N$

${D}_{f} = \mu \frac{N}{m} = \mu g$

$\mu g = \frac{1}{180}$

$\mu = \frac{1}{1800} = 0.00055555555 \approx 5 \cdot {10}^{-} 4$