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

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)