If a #1 kg# object moving at #16 m/s# slows down to a halt after moving #10 m#, what is the friction coefficient of the surface that the object was moving over?

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Answer 1 · 2016-04-22T17:27:00

#mu=0,98#

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#v^2=v_i^2-2*a*s#

#"where ; " v=8 m/s" "v_i=16m/s#
#s=10m" area of under graph" #

#"a=acceleration"#

#8^2=16^2-2*a*10#

#64=256-20a#

#20a=256-64#

#20a=192#

#a=192/20#

#a=9,6 m/s^2" acceleration"#

#F=m*a#

#F_f=mu*N" "N=mg" friction force"#

#mu*cancel(m)*g=cancel(m)*a#

#mu=a/g#

#g=9,81 m/s^2#

#mu=(9,6)/(9,81)#

#mu=0,98#