If an object is moving at #15 m/s# over a surface with a kinetic friction coefficient of #u_k=15 /g#, how far will the object continue to move?

1 Answer
Apr 23, 2016

Answer:

#x=7,5m#

Explanation:

#"you can use the formula "v^2=v_i^2-2*a*x#
#"v=0 ,if object stops"#
#0=v_i^2-2*a*x#
#v_i^2=2*a*x" (1)"#
#"where "v_i=15m/s" ; "a:"acceleration ; x:displacement" #

#F_f=15/g*m*g=15*m" friction force"#
#F=m*a" the Newton's law"#
#F=F_f#
#m*a=m*15#
#a=15m/s^2#

#"now use equation (1)"#
#15^2=2*15*x#
#15=2*x#
#x=7,5m#