# If a 1 kg object moving at 10 m/s slows down to a halt after moving 50 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

Dec 8, 2015

$\mu = \frac{1}{9.8}$

#### Explanation:

First we must find the acceleration experienced by the object.
Here we will use the kinematic equation:
${\left(v f\right)}^{2} = {\left(v i\right)}^{2} + 2 a d$
plug in our given data...
$0 = 100 + 100 a$
$a = - 1 \frac{m}{s} ^ 2$
We can now use this acceleration to calculate force using $F = m a$
$F = 1 k g \cdot - 1 \frac{m}{s} ^ 2$
$F = - 1 N$ this must be the force of friction because no other resistive forces are given.

Kinetic friction is $f k = \mu N$
The normal force in this case is the weight of the object. $N = m g$
$N = 1 k g \cdot 9.8 \frac{m}{s} ^ 2$
$N = 9.8$

Plugging in our values for $f k$ and $N$....
$1 = \mu \cdot 9.8$
$\mu = \frac{1}{9.8}$