If an object is moving at #16# #ms^-1# over a surface with a kinetic friction coefficient of #u_k=80/g#, how far will the object continue to move?

1 Answer
May 10, 2016

Answer:

The distance taken for the object to stop will be #1.6# #m#.

Explanation:

The frictional force will be given by:

#F_"frict"=mu_kF_"norm"#

The normal force #F_"norm"=mg#, so:

#F_"frict"=mu_kmg=80/g*mg=80m#

The acceleration of the object will be given by #a=F_"frict"/m=(80m)/m = 80# #ms^-2#. Given that it is a deceleration (acceleration in the opposite direction to the velocity) give it a minus sign: #a=-80# #ms^-2#.

The initial velocity, #u=16# #ms^-1# and the final velocity when the object stops will be #v=0# #ms^-1#.

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

Rearranging:

#d=-(u^2)/(2d)=-(16^2)/(2*-80)=1.6# #m#