# If an object is moving at 2 m/s over a surface with a kinetic friction coefficient of u_k=6 /g, how far will the object continue to move?

Feb 26, 2016

$x = \frac{1}{3} m$

#### Explanation:

We can use the Energy Work Theorem to solve this equation.
${K}_{2} - {K}_{1} = \setminus \int \setminus \vec{F} \setminus \cdot \setminus \hat{\mathrm{dx}}$

Since the force and the distance travelled are in opposite directions, then $\setminus \vec{F} \setminus \cdot \setminus \hat{\mathrm{dx}} = - F \mathrm{dx}$
Also, since it force works uniformly over the entire movement until rest of object (i,e till ${K}_{2} = 0$), the integral will equate as such
$- \int F \mathrm{dx} = - F x$

So, all in all, it reduces to ${K}_{1} = F x$

Now, force acting on the object is $F = \setminus {\mu}_{k} m g$ (assuming it is a proper horizontal planar surface )
So, $\frac{1}{2} \cancel{m} {v}_{1}^{2} = \setminus {\mu}_{k} \cancel{m} g x \setminus \implies \frac{1}{2} {v}^{2} = \setminus {\mu}_{k} g x$
So, that means x=v^2/(2\mu_kg

Given $\setminus {\mu}_{k} = \frac{6}{g} \setminus \implies \setminus {\mu}_{k} g = 6$, $v = 2 m {s}^{-} 1 \setminus \implies {v}^{2} = 4 {\left(m {s}^{-} 1\right)}^{2}$

Solving this gives the given answer above.