A block with a mass of 126 kg is pulled with a horizontal force of F applied across a rough floor. The coefficent of friction between the floor and the block is 0.7. If the block is moving at a constant velocity, what is the magnitude of F applied?

F applied =

1 answer

Wt. of block=m*g = 126kg * 9.8N/kg=1235N
= Fn = Normal force.

Fk = u*Fn = 0.7 * 1235 = 864.5 N.= Force
of kinetic friction.

Fap-Fk = m*a = m*0 = 0
Fap = Fk = 864.5 N = Force applied.