What force is needed to slide a 250-kg crate across the floor at constant velocity if the coefficient of sliding friction between a crate and a horizontal floor is 0.25?

Hello!
Newton's Second law will help us. It states that the (vector) sum of all forces acting on a body is the same as its mass multiplied by its acceleration, vecF = m veca.
There are four forces here: gravitational force mg downwards, reaction force N upwards, traction force F_T horizontally and friction F_f also horizontally but in reverse direction.
The acceleration is zero, because velocity is constant. So we obtain
m vec(g) + vecN + vec(F_T) + vec(F_f) = vec0.
Vertical forces must balance each other and horizontal, too, so in magnitudes we obtain
N = mg  and  F_T = F_f.
Also we know that F_f = mu N, where mu is the coefficient of sliding friction. Hence F_T = F_f = mu N = mu m g. This is the final formula, and numerically F_T approx 0.25*250*9.8 =612.5 (N).
The answer: the force of about 612.5 N is needed.