Work done by a variable force

[tashizzz]

TL;DR: Please point out any mistakes in solution . If there isnt why the work done by force come out to be equal to work done my gravity , and why when not moving it slowly there is an addittional term . If the work done is equal to Wd by gravity , I have read that work done by gravity always remain mgh if it has disp through height h , why extra kinetic erg . Why diff answers in both methods . Thank You

 

[tashizzz]

TL;DR: Please point out any mistakes in solution . If there isnt why the work done by force come out to be equal to work done my gravity , and why when not moving it slowly there is an addittional term . If the work done is equal to Wd by gravity , I have read that work done by gravity always remain mgh if it has disp through height h , why extra kinetic erg . Why diff answers in both methods . Thank You

View attachment 374708

 

[osilmag]

Hi. There doesn't seem to be any mistakes. It comes out this way because of the work-energy principle. The work required to keep it in place or for net zero force is equal to the gravity working against it, if that makes sense.

For the second part, since there is movement too, there is the work against gravity plus the work done to give it a velocity. Whenever there is a velocity, there is kinetic energy.