If a #2 kg# object moving at #13 m/s# slows to a halt after moving #19 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

Question

1310 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-24T16:27:48

#u_k~=0.4#

#v_f^2=v_i^2-2*a*Delta x#

#v_f=13/2" "m/s" final velocity"#

#v_i=13" "m/s " initial velocity"#

#Delta x=19" " m" displacement of object"#

#(13/2)^2=13^2-2*a* 19#

#2*a*19=169-169/4#

#38*a=507/4#

#a=597/(38*4)#

#a=597/152 =3.93 m/s^2#

#F_f=u_k*m*g#

#a=F_f/m#

#3.93=(u_k*cancel(m)*g)/cancel(m)#

#u_k=(3.93)/(9.81)#

#g=9.81 N/(kg)#

#u_k~=0.4#