Question

A 400-n block is dragged along a horizontal surface by an applied force as shown. the coefficient of kinetic friction is uk = 0.4 and the block moves at constant velocity. the magnitude of is:

Answer 1

The block moves with constant velocity: for Newton's second law, this means that the resultant of the forces acting on the block is zero, because the acceleration is zero.

We are only concerned about the horizontal direction, and there are only two forces acting along this direction: the force F pushing the block and the frictional force $F_f$ acting against the motion. Since their resultant must be zero, we have:
$F-F_f = 0$
The frictional force is
$F_f = \mu mg$
where
$\mu=0.4$ is the coefficient of kinetic friction
$mg=400 N$ is the weight of the block. 

Substituting these values, we find the magnitude of the force F:
$F=F_f = \mu mg=(0.4 )(400 N)=160 N$