If an object is moving at #12 m/s# over a surface with a kinetic friction coefficient of #u_k=2 /g#, how far will the object continue to move?

1 Answer
Mar 9, 2016

Answer:

#Delta x=36 m #

Explanation:

#"Changing of energy of the kinetic is equal work doing"#
#" by the Friction Force"#
#Delta E_k=W#
#E_(k1)=1/2*m*v^2#
#E_(k2)=0" object stops."#
#Delta E_k=E_(k2)-E_(k1)" "Delta E_k=1/2*m*v^2#
#W_f=F_f*Delta x" work doing by the friction force"#
#W_f=u_k*N*Delta x#
#W_f=Delta E_k#
#u_k*N*Delta x=1/2*m*v^2" "(1)#
#N:"Normal Force acting contacting surfaces. "N=m*g#
#"rearranging the equation (1)"#
#u_k*cancel(m)*g*Delta x=1/2*cancel(m)*v^2#
#u_k=2/g" "v=12 m/s#
#2/cancel(g)*cancel(g)*Delta x=1/2*12^2#
#Delta x=144/4#
#Delta x=36 m #