# If an object is moving at 120 m/s over a surface with a kinetic friction coefficient of u_k=7 /g, how much time will it take for the object to stop moving?

Dec 21, 2015

I found $17.1 \sec$

#### Explanation:

Kinetic friction ${f}_{k}$ is responsible for introducing a deceleration given by (Newton's Second law):
$- {f}_{k} = m a$
$- {u}_{k} \cdot N = m a$
$- {u}_{k} \cdot \cancel{m} g = \cancel{m} a$
So:
$a = - \frac{7}{g} \cdot g = - 7$
We can now use:
$\textcolor{red}{{v}_{f} = {v}_{i} + a t}$
To get:
$0 = 120 - 7 t$
$t = \frac{120}{7} = 17.1 \sec$