If a 2 kg object moving at 5 m/s slows down to a halt after moving 3 m, what is the friction coefficient of the surface that the object was moving over?

Feb 21, 2016

${u}_{k} \cong 0 , 425$

Explanation:

${E}_{k} = \frac{1}{2} \cdot m \cdot {v}^{2} \text{ The kinetic energy of object}$
$W = {F}_{f} \cdot \Delta x \text{ Work doing by friction force}$
${F}_{f} = {u}_{k} \cdot N \text{ N: normal force to contacting surfaces}$
$\text{The kinetic energy turns work}$
$\frac{1}{2} \cdot \cancel{m} \cdot {v}^{2} = {u}_{k} . \cancel{m} \cdot g \cdot \Delta x$
$\frac{1}{2} \cdot {5}^{2} = {u}_{k} \cdot 9 , 81 \cdot 3$
$25 = {u}_{k} \cdot 2 \cdot 3 \cdot 9 , 81$
${u}_{k} = \frac{25}{2.3 .9 , 81}$
${u}_{k} \cong 0 , 425$