# If a 5 kg object moving at 2 m/s slows to a halt after moving 50 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

Dec 14, 2015

${\mu}_{k} = 0 , 004$

#### Explanation:

If frictional force is the only retarding force then it is the resultant force and so by Newton's 2nd Law of Motion (assuming motion along a flat surface in 1 direction)
$\sum F = m a$
$\therefore - {f}_{k} = m a$
$\therefore - {\mu}_{k} N = m a$
$\therefore - {\mu}_{k} m g = m a$

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

We compute the acceleration $a$ from the equations of motion for uniform acceleration in 1 direction as follows :

${v}^{2} = {u}^{2} + 2 a x$

$\therefore a = \frac{{v}^{2} - {u}^{2}}{2 \times x} = - \frac{4}{100} = - 0 , 04 m / {s}^{2}$

Substituting back for ${\mu}_{k}$, we get

${\mu}_{k} = 0 , \frac{04}{9} , 8 = 0 , 004$