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 =

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.

To determine the magnitude of the applied force (F), we need to consider the forces acting on the block.

First, let's consider the gravitational force, which is the force pulling the block downward. The gravitational force can be calculated using the formula: F_gravity = mass × acceleration due to gravity.

Given that the mass of the block is 126 kg and the acceleration due to gravity is approximately 9.8 m/s^2, we can calculate the gravitational force as follows:

F_gravity = 126 kg × 9.8 m/s^2 = 1234.8 N

Next, let's consider the frictional force, which opposes the motion of the block. The frictional force can be calculated using the formula: F_friction = coefficient of friction × normal force.

The normal force is the force exerted by the floor on the block in the vertical direction. Since the block is on a horizontal surface, the normal force is equal to the gravitational force (F_gravity) acting on it.

F_friction = 0.7 × F_gravity = 0.7 × 1234.8 N = 864.36 N

Since the block is moving at a constant velocity, the applied force (F) must balance the frictional force (F_friction). Therefore, the magnitude of the applied force is equal to the magnitude of the frictional force.

F_applied = F_friction = 864.36 N

Hence, the magnitude of the applied force is 864.36 N.