If a #4 kg# object moving at #6 m/s# slows to a halt after moving #35 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

1 Answer
Feb 27, 2016

Answer:

#\mu_k=0.0525#

Explanation:

Since the object is moving on a plane surface, the kinetic friction experienced by the object will be #\mu_kmg# implying that the de-acceleration due to friction will be #\mu_kg#

So, using the third kinematic equation
#v^2=v_o^2+2ax#

The object is said to have come to a halt so that means #v=0ms^-1#
So, #v_o^2=2ax# (we ignored the negative sign since we know why that's there).

So, we know it travels a distance of #35m# before it stops so #x=35m#. We know it initially had a speed of #6ms^-1# so #v_o=6ms^-1#.

So, #a=v_o^2/(2x)=6^2/(2*35)=36/70=0.51#
So, #a=0.51#. But #a=\mu_kg# So, #0.51=\mu_k*9.8\implies\mu_k=0.51/9.8#

Solving it yourself, you now understand that things look odd even if they are right.