A force of 28.0 N is required to start a 3.2 kg box moving across a horizontal concrete floor.If the 28.0 N force continues, the box accelerates at 0.50 m/s2. What is the coefficient of kinetic friction?
2006-11-18 05:21:36 · 3 answers · asked by tjmierzwa 2 in Science & Mathematics ➔ Physics
from the other guys F fric = 26.4
F fric = Coeficiant x F normal = 26.4
F normal = mg
so divide 26.4 by (3.2)(9.8) and you got the coefficient of kinetic friction
2006-11-18 07:12:07 · answer #1 · answered by adklsjfklsdj 6 · 0⤊ 0⤋
F=ma
F=(3.2)(.05)
F=1.6 n (this is the net force)
Fnet= Friction + Apply
1.6= Friction + 28
Friction= -26.4 N
2006-11-18 05:41:45 · answer #2 · answered by 7 · 0⤊ 0⤋
force=mass * acceleration
28N - frictional force= 3.2*0.5
=1.6
frictional force=28-1.6
=26.4 N
well I think...
2006-11-18 05:33:21 · answer #3 · answered by Christophe 2 · 0⤊ 0⤋