If an object is moving at #12 m/s# over a surface with a kinetic friction coefficient of #u_k=24 /g#, how far will the object continue to move?

1 Answer
Feb 8, 2016

Answer:

#x=3 meters#

Explanation:

#v^2=v_i^2-2*a*x#
#v=0 " when object is stops"#
#v_i^²=2*a*x#
#k*m*g=m*a#
#24/cancel(g)*cancel(m)*cancel(g)=cancel(m)*a#
#a=24 m/sec^2#
#12^2=2*24*x#
#144=2*24*x#
#144=48.x#
#x=144/48#
#x=3 meters#