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

Jun 16, 2018

${\mu}_{k} = 0.272$

#### Explanation:

The original kinetic energy was $K E = \frac{1}{2} \cdot \frac{3}{2} k g \cdot {\left(\frac{4}{3} \frac{m}{s}\right)}^{2} = \frac{4}{3} J$.

Work of 4/3 J needs to be done by the friction to stop it.

$\text{work} = \frac{4}{3} J = F \cdot \frac{1}{3} m$

$F = \frac{\frac{4}{\cancel{3}} J}{\frac{1}{\cancel{3}} m} = 4 \frac{J}{m} = 4 N$

To find the value of ${\mu}_{k}$, we solve

${F}_{\text{fk}} = 4 N = {\mu}_{k} \cdot m \cdot g = {\mu}_{k} \cdot \frac{3}{2} k g \cdot 9.8 \frac{m}{s} ^ 2 = {\mu}_{k} \cdot 14.7 N$

${\mu}_{k} = \frac{4 N}{14.7 N} = 0.272$

I hope this helps,
Steve