If an object with a mass of #10 kg # is moving on a surface at #4 m/s# and slows to a halt after # 6 s#, what is the friction coefficient of the surface?

1 Answer
Aug 10, 2016

Answer:

#mu = 1/15#

Explanation:

The initial kinetic energy is lost against the friction losses.

#1/2mv_0^2 = mg mu Delta x#.

Here #mu# is the kinetic friction coefficent and #Deltax# is the path until repos.

Also the movement along the path #Delta x# is dictated by

#Delta x = v_0 t - 1/2(mu g)t^2#

Solving for #Delta x# and #mu# the system

#{ (1/2mv_0^2 = mg mu Delta x), (Delta x = v_0 t - 1/2(mu g)t^2) :}#

we obtain

#Delta x = (v_0 t)/2# and
#mu = v_0/(g t)#

Assuming #g = 10# we have

#mu = 4/60 = 1/15#