If a #2 kg# object moving at #12 m/s# slows to a halt after moving #144 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

1 Answer
Jul 2, 2016

Answer:

#\mu_s=0.05#

Explanation:

So, given than the object was moving with a speed of #12m/s#

We'll use the third kinematic equation to solve this.

The third kinematic equation is
#v_f^2=v_o^2+2ax#
So, #v_o=12m/s# and #v_f=0m/s#, #x=144m#

So, #0^2=12^2+2*a*144\implies-144=144*2a#

Simple cancellation and rearranging gives us #-1/2m/s^2=a#
It's reasonable to realize why it's negative since the acceleration is decreasing rather than increasing the velocity of the object. But we'll drop the symbol knowing that friction is invariant of direction for further solving.

Now, we know the effective acceleration of the object. We also know that the effective acceleration of an object while travelling over a friction surface is #\mu_s*g#

So, equating the two, and taking #g=10m/s^2#, we get #\mu_s*10=1/2#.

Rearranging gives you what you now see in the answer box.