# If an object is moving at 7 ms^-1 over a surface with a kinetic friction coefficient of u_k=2/g, how far will the object continue to move?

Aug 29, 2016

The distance traveled is $24.5$ $m$.

The mass cancels out and so does the value of 'g', and we can use the calculated acceleration and the fact that the final velocity is 0 to find the distance traveled.

#### Explanation:

The force causing the deceleration will be the frictional force, given by ${F}_{\text{frict"=muF_"norm}}$, where the normal force ${F}_{\text{norm}} = m g$, so ${F}_{\text{frict}} = \mu m g$.

The acceleration of the object will be given by $a = {F}_{\text{frict}} / m = \frac{\mu m g}{m}$

The mass cancels, which is convenient since we have not been told its value: $a = \mu g = \frac{2}{g} \times g$

The g also cancels, so that the acceleration is just equal to $2$ $m {s}^{-} 2$. Since it is in the opposite direction to the initial velocity, we can write this as $- 2$ $m {s}^{-} 2$.

Now we can use ${v}^{2} = {u}^{2} + 2 a s$. The object comes to rest, so the final velocity, $v = 0$ $m {s}^{-} 1$.

${0}^{2} = {7}^{2} + \left(- 2\right) s$

Rearranging:

$s = \frac{49}{2} = 24.5$ $m$