# If an object is moving at 16 ms^-1 over a surface with a kinetic friction coefficient of u_k=80/g, how far will the object continue to move?

May 10, 2016

The distance taken for the object to stop will be $1.6$ $m$.

#### Explanation:

The frictional force will be given by:

${F}_{\text{frict"=mu_kF_"norm}}$

The normal force ${F}_{\text{norm}} = m g$, so:

${F}_{\text{frict}} = {\mu}_{k} m g = \frac{80}{g} \cdot m g = 80 m$

The acceleration of the object will be given by $a = {F}_{\text{frict}} / m = \frac{80 m}{m} = 80$ $m {s}^{-} 2$. Given that it is a deceleration (acceleration in the opposite direction to the velocity) give it a minus sign: $a = - 80$ $m {s}^{-} 2$.

The initial velocity, $u = 16$ $m {s}^{-} 1$ and the final velocity when the object stops will be $v = 0$ $m {s}^{-} 1$.

So ${v}^{2} = {u}^{2} + 2 a d$

Rearranging:

$d = - \frac{{u}^{2}}{2 d} = - \frac{{16}^{2}}{2 \cdot - 80} = 1.6$ $m$