If a #6kg# object moving at #4 m/s# slows down to a halt after moving 12 m, what is the friction coefficient of the surface that the object was moving over?

1 Answer
Jan 30, 2016

Answer:

#mu_s~~6.5#

Explanation:

Given that the object was moving over a friction surface at a speed of #4ms^-1# initially and finally came to rest after travelling a distance of #12m#.

First we need to find the de-acceleration the object suffered. We'll use the third kinematic equation to solve this.
So, #v^2-v_o^2=2ax#
#x=12m#, #v_o=4ms^-1#, #v=0ms^-1#
So, #0^2-4^2=2a*12\implies-cancel{16}^2=cancel{24}^3a\impliesa=-2/3ms^-2#
The negative sign makes sense, since the object is decelerating.

Now, we know that the object is moving over the friction surface is being acted upon by a frictional force that is proportional to the force the object acts on the surface of friction.
i.e #F_f=\mu_sN=\mu_smg#
We also know that this force is the same force that causes the de-acceleration on the object. So, that means #F_f=ma#

Equating the two, we get #\cancel{m}a=\mu_scancel{m}g#
So, #2/3=\mu_sg# (we ignored the negative sign because both forces here were acting in the same direction, while in the case of the kinematic equation it was negative because velocity and acceleration of the object were in opposite sides).

So substituting the values you'll see what #\mu_s# is.