posted by mary on .
A still block slides over a still surface,if the mass of the pot is 2.00kg what is the force of friction opposing the motion?coefficient of kinetic friction is taken as 0.42.Gravity is 9.8
Wb = mg = 2.0kg * 9.8N/kg = 19.6N. =
Weight of block.
Fb = 19.6N @ 0deg.
Fv = 19.6cos(0) = 19.6N. = Force perpendicular to the surface.
Fk = u*Fv = 0.42*19.6 = 8.23N. = Force of kinetic friction.