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

Apr 12, 2016

$0.20$, rounded to second decimal place.

#### Explanation:

To find out the retardation due to kinetic friction let us use the equation
${v}^{2} - {u}^{2} = 2 a s$,

where $v , u , a , s$ are the final velocity, initial velocity, acceleration and distance moved respectively.
${0}^{2} - {1}^{2} = 2 \times a \times \frac{1}{4}$
or $a = - 2 m {s}^{-} 2$

Now Force $F = m . a$
or Retarding force which is the force $| {F}_{f} | = 2 \times 2 = 4 N$
We know that ths ${F}_{f} = \mu \times \text{Normal Reaction}$, where $\mu$ is the coefficient of kinetic friction.

Assuming that the surface is a level surface, $\therefore \text{Normal Reaction} = m g$
Taking $g = 9.81 m {s}^{-} 2$, we obtain
$4 = \mu \times 2 \times 9.81$, solving for $\mu$
$\mu = \frac{4}{2 \times 9.81} = 0.20$, rounded to second decimal place.