If a #6 kg# object moving at #1 m/s# slows down to a halt after moving #3 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

1 Answer
Dec 29, 2017

Answer:

#mu=1/(80)#

Explanation:

We know that,
#v^2=u^2+2as#

#1/4 = 1 +2xx3xxa#

#-3/4 = 6xxa#

#a= -1/8#

By Newton's law,

#F=ma#

#f= 6xx1/8#

#mu N=3/4#

#mu=3/(4xx60)#

#mu=1/(80)#