Question

The potential energy (in joules) of a particle of mass $1\text{kg}$ moving in a plane is given by $U=3x+4y$. The position coordinates of the point being $x$ and $y$, measured in metres. If the particle is at rest at $(6,4)$; then

• A. its acceleration is of magnitude $5{\text{m/s}}^2$
• B. its speed when it crosses the $y-$axis is $10\text{m/s}$
• C. it crosses the $y-$ axis $(x=0)$ at $y=-4$
• D. it moves in a straight line passing through the origin $(0,0)$

Answer 1

Answer: (A), (B), (C)

it crosses the $y-$ axis $(x=0)$ at $y=-4$

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$=\frac{F}{m}=(-3\hat{i}-4\hat{j}){\text{m/s}}^2$
$\Rightarrow \left|\overrightarrow{a}\right|=5{\text{m/s}}^2$
Let at time ${}^{\prime }{t}^{\prime }$ the particle cross $y-$ axis
$S_x=\frac{1}{2}\left(a_x\right)t^2-6=\frac{1}{2}(-3)t^2\Rightarrow t=2\text{sec}$
Along $y-$ direction:
$S_y=\frac{1}{2}(-4)(2{)}^2=-8$
$y-4=-8$
$\Rightarrow$ Particle crosses $y$-axis at $y=-4$
At $(6,4):U=34\text{J}\&KE=0$
At $(0,-4):U=-16\text{J}\Rightarrow KE-16=34\text{J}$
$\Rightarrow \frac{1}{2}m{v}^2=50\Rightarrow v=10\text{m/s}$