If an object is moving at #7# #ms^-1# over a surface with a kinetic friction coefficient of #u_k=2/g#, how far will the object continue to move?

1 Answer
David G. · Stefan V.
Aug 29, 2016

The distance traveled is #24.5# #m#.

The mass cancels out and so does the value of 'g', and we can use the calculated acceleration and the fact that the final velocity is 0 to find the distance traveled.

Explanation:

The force causing the deceleration will be the frictional force, given by #F_"frict"=muF_"norm"#, where the normal force #F_"norm"=mg#, so #F_"frict"=mumg#.

The acceleration of the object will be given by #a=F_"frict"/m=(mumg)/m#

The mass cancels, which is convenient since we have not been told its value: #a=mug=2/g xx g#

The g also cancels, so that the acceleration is just equal to #2# #ms^-2#. Since it is in the opposite direction to the initial velocity, we can write this as #-2# #ms^-2#.

Now we can use #v^2=u^2+2as#. The object comes to rest, so the final velocity, #v=0# #ms^-1#.

#0^2 = 7^2 +(-2)s#

Rearranging:

#s=49/2=24.5# #m#

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