I need to use this to find a motor that will be able to pull at most 2250 kg at a 1.5 m/s pace. At peak acceleration, I would like this to accelerate at 0.5 m/s2. The area that the load will be pulled is a concrete sidewalk with, at its most extreme, a 20° incline.
Ignoring friction, because you have not provided any details regarding this in your question, and assuming that the object starts from rest on a $20°$ incline, and accelerates at a peak rate of $0.5m/s^2$ on it's way to a max speed of $1.5m/s$
The force required to stop a static object from rolling down a hill is the same as that required to keep it at a constant speed on that hill, $F_1=mg\sin\theta$.
The additional resultant force required to provide an acceleration of $0.5m/s^2$ can then simply be added using $F=ma$, therefore,
Which equates to a peak force in your case of:
Your question contradicts itself. Acceleration will mean infinitely increasing speed.
If we disregard that and assume a steady speed of 1.5m/s up a ramp of 20 degrees for a mass of 2250kg, you need:
$P= mgh/s= 2250(9.8)sin(20)1.5$
And force = P/vertical velocity= P/sin(20)1.5
This is the force needed. From here depending on the diameter of the winch's pulley you can calculate the torque needed. Just divide the power by the diameter.