Question

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 meter. If the particle is at rest at $(6,4)$ at time $t=0$ , then

• A. the particle has constant acceleration.
• B. the work done by the external forces on the particle till the instant it crosses the $x$ -axis is 25 $J$
• C. the speed of the particle when it crosses the $y$ -axis is
  15 $m{s}^{-1}$
• D. the coordinates of the particle at time $t=4$ s are $(-18,-28).$

Answer 1

Answer: (A), (B), (D)

the coordinates of the particle at time $t=4$ s are $(-18,-28).$

Since potential energy of the particle is equal to $\phi$ this means
$x$ - component of the force acting on the particle is equal to
$F_x=-\frac{d\varphi }{dx}$
and $y$ - component of the force on the particle is equal to
$F_y=-\frac{d\varphi }{dy}$
Hence, force acting on the particle is $\overrightarrow{F}=-(3\hat{i}+4\hat{j})$ $N$. It means, force acting on the particle is constant. Hence, the particle moves with constant acceleration. So option (a) is correct.
Acceleration of the particle will be $\overrightarrow{a}=\frac{\overrightarrow{F}}{m}=-(3\hat{i}+4\hat{j}){ms}^{-2}$
Its magnitude is $a=\sqrt{3^2+4^2}=5{ms}^{-2}$
Since, the particle was initially at rest at $(6,4),$ position of the particle at any time $t$ is given by
$x=6+\frac{1}{2}{a}_x{t}^2=(6-\frac{3}{2}{t}^2)m$
and $y=4+\frac{1}{2}{a}_y{t}^2=(4-2t^2)m$
When the particle crosses the $x$ -axis , $y=0$
$⟹$ $t_1=\sqrt{2}s$
$\therefore$ Displacement of the particle during this time is
$s_1=\frac{1}{2}a{t}^2=5m$
Hence, work done by the force upto this instant will be
$F_{s_1}=5\times 5J=25J$
Hence, option (b) is also correct.
The particle crosses $y$ -axis when $x=0$
Hence, $6-\frac{3}{2}{t}_2^2=0$ or $t_2=2s$
Speed of the particle at this instant will be
$v=a{t}_2=5\times 2=10{ms}^{-1}$
So, option (c) is wrong
At $t=4s,x=6-\frac{3}{2}(4{)}^2=-18m$
and $y=4-2(4{)}^2=-28m$
Hence, option $\left(d\right)$ is also correct.