If a #5 kg# object moving at #15 m/s# slows to a halt after moving #12 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

1 Answer
Jul 18, 2017

Answer:

#μ=0.956#

Explanation:

We know that the work done by friction is equal to the change in kinetic energy of the object.

#ΔΚ=W_f =>K_"after"-K_"initial"=F_f*d =>0-1/2*5*15^2=#

#-Nμd# because the force of friction is #F_f=-Nμ#

where #N# is the upward force from the surface applied to the object and it is equal to the gravitational force because there is no movement in the upward or downward direction.

so #N=mg#

and #μ# is the coefficient of friction

#0-1/2*5*15^2=-Nμd=>1125/2=mgμd=>#

#μ=1125/(2mgd)=1125/(2*5*9,81*12)=0.956#