a 1550N box is being pulled across a level floor at a aconstant speed
by a force of 500N at an angle of 15 above the horizontal. what is the
coefficient of kinectic friction between the box and the floor

Sounds like a homework problem to me.

In the absence of friction, what happens to the velocity of an object
subjected to a constant force?  In this problem, you are told that the
box is being pulled by a constant applied force, but its velocity is
constant.  What does that tell you?

The constant applied force is not directed exactly along the direction
of motion.  How might you resolve this force into two components - one
directed along the direction of motion, and the other perpendicular to
it?

0.34

The equations are

Sum(Fx's)==> 0 = N - 500*sin(15) - 1550
Sum(Fy's)==> 0 = MuF*N - 500*cos(15)

where MuF is the coefficient of kinetic friction and N is the normal force.

When you solve this system of eqn's N = 1420.59N and MuF = .3399 ~ .34.

This was the easiest physics question ever.  You should try harder.