If a #5 kg# object moving at #2 m/s# slows to a halt after moving #50 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

1 Answer
Dec 14, 2015

Answer:

#mu_k=0,004#

Explanation:

If frictional force is the only retarding force then it is the resultant force and so by Newton's 2nd Law of Motion (assuming motion along a flat surface in 1 direction)
#sumF=ma#
#therefore -f_k=ma#
#therefore -mu_kN=ma#
#therefore -mu_kmg=ma#

#therefore mu_k=-a/g#

We compute the acceleration #a# from the equations of motion for uniform acceleration in 1 direction as follows :

#v^2=u^2+2ax#

#therefore a=(v^2-u^2)/(2xxx)=-4/100=-0,04m//s^2#

Substituting back for #mu_k#, we get

#mu_k=0,04/9,8=0,004#