Lifting Rope.

Let The at t=t , 'x' length of rope is pulled by us and let it be as our system which is uplift by us in presence of gravity , and at the Lower end of rope there is heap of rope so no Tension act's at the end of heap. Using Laws of motion of Newton Sir (Here $F=ma$ does not Valid, Because it is not an closed system )

$\displaystyle{{ { F }_{ o }-\lambda gx=\cfrac { d(\lambda xv) }{ dt } }\\ { F }_{ o }-\lambda gx=\lambda \cfrac { vd(xv) }{ vdt } \quad (\because vdt=dx)\\ ({ F }_{ o }-\lambda gx)x=\lambda \cfrac { xvd(xv) }{ dx } \quad (multiply\quad by\quad 'x'\quad both\quad side)\\ \int _{ 0 }^{ x }{ ({ F }_{ o }-\lambda gx)xdx } =\lambda \int _{ 0,0 }^{ x,v }{ xvd(xv) } }$ By integrating we should get $(\displaystyle{v=\sqrt { \cfrac { { F }_{ o } }{ \lambda } -\cfrac { 2gx }{ 3 } } \quad \\ \int _{ 0 }^{ l }{ \cfrac { dx }{ \sqrt { \cfrac { { F }_{ o } }{ \lambda } -\cfrac { 2gx }{ 3 } } } } =\int _{ 0 }^{ T }{ dt } }$ Now Evaluate This Elementary integral we should get $\displaystyle{\boxed { T=\cfrac { 3 }{ g } (\sqrt { \cfrac { { F }_{ o } }{ \lambda } } -\sqrt { \cfrac { { F }_{ o } }{ \lambda } -\cfrac { 2gl }{ 3 } } ) } }$

I had added an Image to This Problem , Can you please confirm it ? Is it Ok