# If a 3 kg object moving at 20 m/s slows to a halt after moving 500 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

Feb 6, 2017

I got $0.04$

#### Explanation:

The only horizontal force acting on the object is kinetic friction:

${f}_{k} = {\mu}_{k} N$

where:
${\mu}_{k}$ is the coefficient of kinetic friction;
$N$ is the modulus of the Normal Reaction (of the surface); in this purely horizontal situation (no inclination or sloped surface) the normal will be equal to the weight of the object or:
$N = W = m g$

We know that the object is travelling at $20 \frac{m}{s}$ when friction kicks in and slows it down to zero in $d = 500 m$ so we can write (from Kinematics):

${v}_{f}^{2} = {v}_{i}^{2} + 2 a d$
or:
$0 = {20}^{2} + 2 a \cdot 500$

giving an acceleration of:

$a = - \frac{400}{2 \cdot 500} = - 0.4 \frac{m}{s} ^ 2$ negative to indicate a deceleration that will slow down our object (opposite to the direction of motion).

Finally we can use Newton's second law to equate the resultant of the forces acting on the object to mass and acceleration:

$\Sigma \vec{F} = m \vec{a}$

BUT: horizontally, the only force acting on the object is friction (IN OPPOSITE DIRECTION TO THE MOTION ) so that we can write:

$- {f}_{k} = m a$
$- {\mu}_{k} \cdot m g = m a$

in numbers:

$- {\mu}_{k} \cancel{3} \cdot 9.8 = \cancel{3} \left(- 0.4\right)$

${\mu}_{k} = \frac{0.4}{9.8} = 0.04$