In this situation, the mass and weight are proportional and irrelevant to the problem. Other than proper respect of units, why would it really matter? I would agree that using mass+kg would remain correct and be less unusual, but it doesn't matter a lot.
It does not matter, it is just a pet peeve of mine. Might be due psychological trauma from when I was a kid and arguing with an older cousin about how pounds and kilograms are not units of the same thing, and the older cousin "winning" the argument in the eyes of the elders because the cousin was quite a few years older than me and considered to be smart in school.
As a European I learned in metric. When I first learned pounds, it was as the imperial system's equivalent of grams and a conversion factor was given. Force in physics class was taught in Newtons (kg*m/s^2).
Pounds as a mass unit are perverse enough. Things like pound force and psi (pounds per square inch) were used only to make fun of old mechanics papers and textbooks. Also, btu. It is quite amazing actually that someone would see the SI and think “no, too simple; I’ll keep my pounds, ounces, inches, and feet”.
Anyway, yes, the proper unit of force is the Newton.
I think there is some weird dual usage that makes them either mass or weight depending on the context. For example, torque is in ft•lbs or N•m so the pounds there are lbs force.
Though checking wikipedia again, it actually specifies that torque is measured in as lbf•ft. I take that to mean that 1 ft•lb is the torque of 1/2 oz (1/32 of an lb) at 1 foot. I expect that's a test question almost everyone would get wrong, myself included.
Think like an ordinary person. You know, an ordinary person who would say that they weigh about 80 kilograms. Only science nerds would say that they mass about 80 kilograms, or that they weigh about 785 Newtons. Similarly, anyone who's used to living with US customary (or Imperial) units understands that a pound of force is what a pound of mass weighs and would see no reason, under any circumstances, why anyone would want to divide the gravitational acceleration out of a pound-foot to arrive at a "real" torque value. When the pound value is expressing a weight-equivalent force, that force is the force of a pound under normal gravitational acceleration at or near the surface of the Earth.
