Question

The potential energy of a particle of mass $5kg$ moving in the $x-y$ plane is given by $U=(-7x+24y)J$ , $x$ and $y$ being in metre . If the particle starts from rest from origin, then what is the speed (in $m{s}^{-1}$ ) of particle at $t=2s$ ?

Answer 1

Answer: 10

$\overrightarrow{F}=-\frac{\partial U}{\partial x}\hat{i}-\frac{\partial U}{\partial y}\hat{j}=7\hat{i}-24\hat{j}$
$\therefore a_x=\frac{F_x}{m}=\frac{7}{5}=1.4m{s}^{-2}$ along positive $x$ -axis
$a_y=\frac{F_y}{m}=-\frac{24}{5}$
$=4.8m{s}^{-2}$ along negative $y$ -axis
$\therefore v_x=a_{x}t=1.4\times 2$
$=2.8m{s}^{-2}$
and $v_y=4.8\times 2=9.6m{s}^{-1}$
$\therefore v=\sqrt{v_x^2+v_y^2}=10m{s}^{-1}$