If an object with a mass of #10 kg # is moving on a surface at #45 m/s# and slows to a halt after # 4 s#, what is the friction coefficient of the surface?

1 Answer
Jun 6, 2018

Answer:

The coefficient of friction is #=1.15#

Explanation:

Calculate the acceleration with the equation

#v=u+at#

The initial velocity is #u=45ms^-1#

The final velocity is #v=0ms^-1#

The time is #t=4s#

The acceleration is

#a=(v-u)/t=(0-45)/4=-11.25ms^-2#

The mass of the object is #m=10kg#

Apply Newton's Second Law to calculate the frictional force

#F_r=m*|a|=10*11.25=112.5N#

The acceleration due to gravity is #g=9.8ms^-2#

The normal force (reaction) is #N=mg=10*9.8=98N#

The coefficient of friction is

#mu_k=F_r/N=112.5/98=1.15#