The potential energy$\phi$ , in joule, of a particle of mass 1 kg, moving in the x - y plane, obeys the law $\phi =3x+4y$ , where (x,y) are the coordinates of the particle in metre. If the particle is at rest at (6, 4) at time t = 0, then

a) ${\text{F}}_x=\frac{-d\phi }{dx},{\text{F}}_y=-\frac{d\phi }{dt}$

Force acting on particle $\text{F}=-\left(3i+4j\right)\text{N}$

$\text{F}=\text{constant}\Rightarrow \text{a}=\text{constant}$

b) $\text{a}=\frac{\text{F}}{\text{m}}=\frac{-\left(3i+4j\right)}{1},\left|\text{a}\right|=\sqrt{9+18}=5{\text{m/s}}^{\text{2}}$

$x={\text{u}}_x\text{t}+\frac{1}{2}{\text{a}}_x{\text{t}}^2=6-\frac{1}{2}\times 3{\text{t}}^2$

$\text{y}={\text{u}}_{\text{y}}\text{t}+\frac{1}{2}{\text{a}}_{\text{y}}{\text{t}}^2=4-\frac{1}{2}\times 4{\text{t}}^2=4-2{\text{t}}^2$

when particle crosses x – axis,

$\text{y}=0\Rightarrow {\text{t}}_1=\sqrt{2}\text{S}$

displacement${\text{S}}_{\text{1}}=\frac{1}{2}{\text{at}}^2=5\text{m}$

work$\text{W}=\text{FS}=5\times 5=25$

c) particle crosses y - axis$x=0\Rightarrow {\text{t}}_2=2\sec$

speed $\text{v}={\text{at}}_2=5\times 2=10$

d) $\text{at t}=4\sec ,x=6-\frac{3}{2}{\left(4\right)}^2=-18\text{m}$

$\text{y}=4-2{\left(2\right)}^2=-28\text{m}$