Question

The potential energy $U$ in joules of a particle of mass $1\text{kg}$ moving in the $x-y$ plane obeys the law $U=3x+4y$, where $(x,y)$ are the co-ordinates of the particle in metres. If the particle is released from rest at $(6,4)$ at time $t=0$, then

• A. the particle has zero acceleration
• B. co-ordinates of the particle at $t=1\text{s}$ is $(2,4.5)$
• C. co-ordinates of the particle at $t=1\text{s}$ is $(4.5,2)$
• D. co-ordinates of the particle at $t=1\text{s}$ is $(2,2)$

Answer 1

The correct option is C co-ordinates of the particle at $t=1\text{s}$ is $(4.5,2)$

Given: $U=3x+4y$
We know that,
$F=-(\frac{dU}{dx}\hat{i}+\frac{dU}{dy}\hat{j}+\frac{dU}{dz}\hat{k})$
$\Rightarrow F=-(3\hat{i}+4\hat{j})\text{N}$
$\Rightarrow F=(-3\hat{i}-4\hat{j})\text{N}$
$\therefore$ Acceleration of particle$=\frac{F}{m}=(-3\hat{i}-4\hat{j}){\text{m/s}}^2$

Since $u=0$ (initial velocity)
$\therefore$ At $t=1\text{sec}$
$s_x=\frac{1}{2}{a}_x{t}^2=\frac{1}{2}(-3)\times 1=-1.5\text{m}$
$s_y=\frac{1}{2}{a}_y{t}^2=\frac{1}{2}(-4)\times 1=-2\text{m}$

Therefore, final position of particle
$(s_{\text{net}}{)}_x=6-1.5=4.5\text{m}$
$(s_{\text{net}}{)}_y=4-2=2\text{m}$
$\therefore$ Position after $t=1\text{sec}$ is
$(4.5,2)\text{m}$