Same old block on the floor problem

To find the minimum time we have to first find the maximum acceleration that this force induces in the body. Assume the force to be acting at an angle $\theta$ as shown in the figure.

Vertical Force balance equation:

$N=mg-Fsin\theta$

Horizontal force balance equation:

$Fcos\theta -\mu N=ma$

$Fcos\theta -\mu (mg-Fsin\theta )=ma$

$a=\frac { F(cos\theta +\mu sin\theta ) }{ m } -\mu g$

To find maximum acceleration, differentiate and equate to 0

$\frac { da }{ d\theta } =\frac { F(-sin\theta +\mu cos\theta ) }{ m } =0$

$tan\theta =\mu$

${ a }_{ max }=\frac { F }{ m } (\frac { 1 }{ \sqrt { 1+{ \mu }^{ 2 } } } +\frac { { \mu }^{ 2 } }{ \sqrt { 1+{ \mu }^{ 2 } } } )\quad -\mu g$

$a_{ max }=\frac { F }{ m } \frac { 1+{ \mu }^{ 2 } }{ \sqrt { 1+{ \mu }^{ 2 } } } -\mu g$

${ a }_{ max }=\frac { F }{ m } (\sqrt { 1+{ \mu }^{ 2 } } )-\mu g$

Substituting the values we get

${ a }_{ max }=17.36ms^{ -2 }$

Now let the distance moved by the block while the force is acting on it be $x$ and the velocity at that displacement be $v$ and time taken to reach there be $t$ .

Therefore,

$2{ a }_{ max }x={ v }^{ 2 }$ ...............................( $1$ )

For the 2nd part of the journey when only frictional force is acting

$2(\mu g)(l-x)={ v }^{ 2 }$ ....................................( $2$ )

From ( $1$ ) and ( $2$ ) we get

$x=\frac { \mu gl }{ { a }_{ max }+\mu g }$

Also for region 1,

$x=\frac { 1 }{ 2 } { a }_{ max }{ t }^{ 2 }$

$t=\sqrt { \frac { 2x }{ { a }_{ max } } }$

$t=\sqrt { \frac { 2\mu gl }{ { a }_{ max }({ a }_{ max }+\mu g) } }$

Substituting the values we get

$t=7s$

I tried to solve it by energy conservation...

If AB=l , the work done by force F in time 't' be Fx ,if the block moves a distance 'x' in time 't'; So the work done Fx =work done by frictional force in the remaining distance...

Fx=umg(l-x);

Also acc. 'a' during which force was applied..... a=(F-umg)/m;

and 0.5at^2=x;

Combining these equations ;solve 't' t=[7.11]=7 sec