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 move at constant velocity is equal to the gravity working against it, if that makes sense.

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

 

[kuruman]

It would be easier for us to help you if you posted the statement of the problem as was given to you. We need to know what question(s) you are trying to answer before we can explain to you what you did right, what you did wrong and what you do not understand.