Question

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?

Answer 1

`x=1/3m`

Explanation:

We can use the Energy Work Theorem to solve this equation.
`K_2-K_1=\int\vec{F}\cdot\hat{dx}`

Since the force and the distance travelled are in opposite directions, then `\vec{F}\cdot\hat{dx}=-Fdx`
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 Fdx=-Fx`

So, all in all, it reduces to `K_1=Fx`

Now, force acting on the object is `F=\mu_kmg` (assuming it is a proper horizontal planar surface )
So, `1/2cancel{m}v_1^2=\mu_kcancel{m}gx\implies1/2v^2=\mu_kgx`
So, that means `x=v^2/(2\mu_kg`

Given `\mu_k=6/g\implies\mu_kg=6`, `v=2ms^-1\impliesv^2=4(ms^-1)^2`

Solving this gives the given answer above.