I don't think this was ever submitted properly to the wiki. There are three ends to this. On the lowest end, it would take 10 kN (1019.72 kg of force; Class 5) according to someone who has been to high school physics: https://www.quora.com/How-much-force-is-necessary-to-rip-a-fully-grown-humans-arm-off
Granted high school studies aren't exactly as good as university studies or trusted governmental resources, especially if recalled rather than gotten straight from the source, but it serves as a decent enough minimum requirement. As the feat can be done with a quick movement, we'll get the distance. The deltoids has a mean length of 16.2 cm (0.162 meters).: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC7854847/
Muscular extension can vary, ranging from 150% (https://wexnermedical.osu.edu/blog/benefits-of-stretching ) to 200% their original length (https://yinyoga.com/you-shouldnt-stretch-ligaments-tendons-joints-connective-tissue-or-fascia/ ). I looked into the context on elasticity. Apparently it's more of a wording thing than a definable thing, since this article says that ropes can be stretched 50% of their original length with context that the ropes can become 1.5x their original length before breaking: https://apps.dtic.mil/sti/tr/pdf/ADA084622.pdf
The 150% statement also includes a statement that claims that tendons can only be stretched 4% their original length, so based on the wording, the stretching here is more like the change in length of the muscle (aka the distance) than the actual length of the muscle. This should equate to a distance of 0.243 meters as that is the furthest the deltoids can go before snapping.
Let's see how much work is exerted.:
10000*.243=2430 joules (Street level)
A Stackexchange discussion puts the force needed to tear the arm off at 30 to 200 kN (about 3.06 to 20.39 tonnes of force, or Class 5 to Class 25). This would serve as a medium end. Let's see how much work this exerts.
30000*0.243=7290 joules (Street level)
200000*0.243=48600 joules (Wall level)
The ballpark figure taken is 85000 newtons, so may as well account for that as well.
85000*.243=20655 joules (Wall level)
This range seems closer to the real-life example of the drawn-and-quartered execution, where horses are made to tear a person apart. The peak power of a horse, exerted over a few seconds, is 14.88 horsepower according to a 1926 study (https://lib.dr.iastate.edu/bulletin/vol20/iss240/1 ), which is equivalent to 11096.014 watts (11096.014 joules per second; Street Level+). Four horses are typically used, so this should put full-body dismemberment at 44384.056 joules (Wall level). This is equivalent to the following force values:
11096.014/0.243=45662.60905 newtons (4656.290278 kgf; Class 5) (one limb)
44384.056/.243=182650.4362 newtons (18625.16111 kgf; Class 25) (full-body dismemberment)
At the highest, we have this journal: https://journals.le.ac.uk/ojs1/index.php/jist/article/download/2694/2511
I managed to rediscover the journal in this calculation or misplaced CRT of FNaF: User blog:Sharousisback1/Freddy Fazbear tore my arm
The journal mistook 10 cms for 1 cm, making the cross-sectional area 100x smaller than it actually is. As such, the force figure the study calculated is actually 0.0314 m² and the force is actually 200646 newtons. Let's determine the work there.
200646*0.243=48756.978 joules (Wall level)
So yeah, there you have it. Personally I think the StackExchange range is closest to the real-life example of dismemberment we have (the drawn-and-quartered execution), so I believe that's the pick to go for.