Question

In an $XY$ horizontal plane a force field $\vec{F}=-(40 \,N / m )(y \hat{i}+x \hat{j})$ is present where $x$ and $y$ the coordinates of any point on the plane. A smooth rod $A B$ is fixed in the plane as shown in the figure. A particle of mass $5 \,kg$ 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

$W$ by $F =\int \vec{ F } \cdot \vec{ dr }$
$=-\int 40[y d x+x d y]$
$=-40 \int d(x y)$
$=-40[x y]_{x_{i}, y_{i}}^{x_{f}, y_{f}}$

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 5 V _{\min }^{2} $
$V_{\min } =2 \,m / s $