If a #4 kg# object moving at #16 m/s# slows down to a halt after moving #800 m#, what is the friction coefficient of the surface that the object was moving over?

1 Answer
Jun 2, 2016

Answer:

#mu_k=0.013" has no unit"#

Explanation:

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

#v_f=8" "m/s#

#v_i=16" "m/s#

#Delta x=800" "m#

#a:"acceleration of the object"#

#8^2=16^2-2*a*800#

#64=256-a*1600#

#1600*a=256-64#

#1600*a=192#

#a=192/1600" "m/s^2#

#F_f=mu_k*cancel(m)*g=cancel(m)*a#

#mu_k="coefficient of friction"#

#mu_k*g=a#

#mu_k=a/g#

#mu_k=(192/1600)/(9.18)#

#mu_k=192/(1600*9.18)#

#mu_k=192/14688#

#mu_k=0.013#