# 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?

Jul 18, 2017

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 $N = m g$

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