Question

The potential energy of a particle of mass $5 \, kg$ moving in the $x-y$ plane is given by $U=\left(\right.-7x+24y\left.\right) \, J$ , $x$ 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 $t=2 \, s$ ?

Answer 1

$\overset{ \rightarrow }{F}=-\frac{\partial U}{\partial x}\hat{i}-\frac{\partial U}{\partial y}\hat{j}=7\hat{i}-24\hat{j}$
$\therefore \, \, \, a_{x}=\frac{F_{x}}{m}=\frac{7}{5}=1.4 \, ms^{- 2}$ along positive $x$ -axis
$a_{y}=\frac{F_{y}}{m}=-\frac{24}{5}$
$=4.8ms^{- 2}$ along negative $y$ -axis
$\therefore \, v_{x}=a_{x}t=1.4\times 2$
$=2.8 \, ms^{- 2}$
and $v_{y}=4.8\times 2=9.6 \, ms^{- 1}$
$\therefore \, v=\sqrt{v_{x}^{2} + v_{y}^{2}}=10 \, ms^{- 1}$