# If a 3 kg object moving at 5 m/s slows to a halt after moving 10 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

Dec 15, 2015

${\mu}_{k} = 0 , 12755$

#### Explanation:

I will show you 2 different methods to do this question :

Method 1 - Using Newton's Laws and equations of motion

The acceleration of the object may be found from the equations of motion for constant linear acceleration as follows :
${v}^{2} = {u}^{2} + 2 a x$
$\therefore a = \frac{{v}^{2} - {u}^{2}}{2 x} = \frac{{0}^{2} - {5}^{2}}{2 \times 10} = - 1 , 25 m / {s}^{2}$

This acceleration is caused by the resultant force of friction acting on the object and by Newton 2 given as :
$\sum F = m a$
$\therefore - {f}_{k} = m a$
$\therefore - {\mu}_{k} N = m a$
$\therefore - {\mu}_{k} m g = m a$

$\therefore {\mu}_{k} = \frac{m a}{- m g} = \frac{- 1 , 25}{- 9 , 8} = 0 , 12755$.

Method 2 - Using energy considerations

From conservation of energy, work done by friction equals change in kinetic energy brought about.

$\therefore {W}_{f k} = \Delta {E}_{K}$

$\therefore {f}_{k} \cdot x = \frac{1}{2} m \left({v}^{2} - {u}^{2}\right)$

$\therefore {\mu}_{k} m g \cdot x = \frac{1}{2} m {v}^{2}$

$\therefore {\mu}_{k} = \frac{\frac{1}{2} m {v}^{2}}{m g x} = \frac{\frac{1}{2} \times {5}^{2}}{9 , 8 \times 10} = 0 , 12755$.