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 origin, then the speed of particle at $t=2s$ is

• A. 5 m / s
• B. 14 m / s
• C. 17.5 m / s
• D. 10 m / s

Answer 1

Answer: (D)

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