# If an object is moving at 7 ms^-1 over a surface with a kinetic friction coefficient of u_k=14 /g, how far will the object continue to move?

Feb 17, 2016

Finding the frictional force and the acceleration due to it allows us to find the distance traveled, $1.75$ $m$, before the object stops.

#### Explanation:

The force accelerating (decelerating) the mass will be the frictional force, which is the frictional coefficient times the normal force:

${F}_{\text{fric}} = \mu {F}_{N}$

In this case the normal force is the weight force of the object:

${F}_{N} = m g$

So

${F}_{\text{fric}} = \mu m g = \frac{14}{g} \cdot m \cdot g$

$g$ cancels out to leave a frictional force of $14 m$ where $m$ is the mass. We don't know it, but be patient. ;-)

The acceleration (deceleration) of the object with this force acting on it is given by Newton's Second Law :

$a = \frac{F}{m} = \frac{14 m}{m} = 14$ $m {s}^{-} 2$ ($m$ cancels)

This should be given a minus sign, because it is an acceleration in the opposite direction to the object's velocity - a deceleration.

$a = - 14$ $m {s}^{-} 2$

We know the initial velocity $u = 7$ $m {s}^{-} 1$ and the final velocity $v = 0$ $m {s}^{-} 1$ and the acceleration, and we are asked for the distance the object moves before stopping.

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

Rearranging:

$d = \frac{{v}^{2} - {u}^{2}}{2 a} = \frac{{0}^{2} - {7}^{2}}{2 \cdot - 14} = \frac{49}{28} = 1.75 m$