If an object is moving at #3 m/s# over a surface with a kinetic friction coefficient of #u_k=5 /g#, how far will the object continue to move?

1 Answer
May 8, 2016

Here KE of the object = Work done against the force of kinetic friction
So #u_kxxmxxgxxd=1/2*m*v^2#
Where #u_k= 5/g#
#m="mass",v="velocity"#
#g="acceleration due to gravity=9.8m/s^2#
#d= "distance traversed"#

#u_kxxmxxgxxd=1/2*m*v^2#
#5/gxxcancel(m)xxgxxd=1/2*cancel(m)*v^2#
#d=v^2/10=3^2/10=0.9m#