The potential energy U in joule of a particle of mass 2 kg moving in X-Y plane follows the law $U=2x+4y$  , where (x, y) are the co-ordinates (in metre) of the particle. If the particle is at rest at $(4,4)$ at time t=0 then, 
 

$$\begin{gathered} U=2x+4y F=-\frac{\partial U}{\partial x}\widehat{i}-\frac{\partial U}{\partial y}\widehat{j}-\frac{\partial U}{\partial z}\widehat{k}=(-2\widehat{i}-4\widehat{j})N \\ Att=0,at(4,4),P=24J \end{gathered}$$

Kinetic energy K=0   Total energy, $E=P+K+24J$ and acceleration $\overrightarrow{a}=(-\widehat{i}-2\widehat{j})m/s^2$

To cross x-axis, along y,  $4=\frac{1}{2}{a}_y{t}_1^2=\frac{1}{2}2t_1^2\Rightarrow t_1=2\sec$

To cross x-axis, along y,  $4=\frac{1}{2}{a}_x{t}_2^2=\frac{1}{2}\mathrm{.1.}{t}_2^2\Rightarrow t_2=2\sqrt{2}\sec$

Also, Crossing the x-axis  co-ordinates are  $(4-\frac{1}{2}{a}_x{t}_1^2,0)=(2,0)$

$\therefore P=4J\Rightarrow K=E-P=20J$ 

Crossing the y-axis co-ordinates are $(0,4-\frac{1}{2}{a}_y{t}_2^2)=(0,-4)$

 $\therefore P=-16J,K=E-P=40J$