Question

In an $XY$ horizontal plane a force field $\vec{F}=-(40N/m)(y\hat{i}+x\hat{j})$ is present where $x$ and $y$ the coordinates of any point on the plane. A smooth rod $AB$ is fixed in the plane as shown in the figure. A particle of mass $5kg$ is to be released with a velocity in this force field such that it reaches to point $B$. Find the minimum velocity (in $m/s$ ) that must be imparted along the rod at $A$ such that it reaches to $B$.

Answer 1

Answer: 2.00

$W$ by $F=\int \vec{F}\cdot \overrightarrow{dr}$
$=-\int 40[ydx+xdy]$
$=-40\int d(xy)$
$=-40[xy{]}_{x_i,y_i}^{x_f,y_f}$

if particle will able to reach point $O$ it will go by itself to $B$.
$WET$ between $A$ and $O$
$W$ by $F=0-\frac{1}{2}m{V}_{\max }^2$
$-40\left[\frac{1}{2}\times \frac{1}{2}-0\times 1\right]=-\frac{1}{2}m{V}_{\min }^2$
$-10=-\frac{1}{2}\times 5V_{\min }^2$
$V_{\min }=2m/s$