If an object is moving at #2# #ms^-1# over a surface with a kinetic friction coefficient of #mu_k=8/g#, how far will the object continue to move?

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Answer 1 · 2016-02-20T10:20:16

The object will move #1/4# (or 0.25) #m# before it stops.

The frictional force will be given by:

#F_"frict"=mu*F_N# where #F_N# is the normal force, which is the weight force of the object: #F_N=mg#.

#F_"frict"=mu*F_N=8/g*mg=8m#

The acceleration of the object will be described by Newton's Second Law :

#a=F/m=(8m)/m=8# #ms^-2#

From its initial velocity, #u=2.0# #ms^-1#, to #v=0# #ms^-1#, the distance traveled will be given by:

#v^2=u^2+2ad#

Rearranging:

#d=(v^2-u^2)/(2a)=(0-2^2)/(2*8)=4/16=1/4# #m#.