The force $F$ acting on a body varies with its position $x$ as shown in the figure. At which point is the body in unstable equilibrium?

The figure shows the $F-x$ graph,where $F$ is the force applied and $x$ is the distance covered by the body along a straight-line path. Given that $F$ is in newton $(N)$ and $x$ is in metre $(m)$,calculate the total work done in joules $(J)$.

$A$ force acts on a particle moving along the $x$-axis,and its variation with position is shown in the figure. At which point will the particle be in a stable equilibrium position?

$A$ force of $(6x^2 - 4x + 3) \text{ N}$ acts on a body of mass $0.75 \text{ kg}$ and displaces it from $x = 2 \text{ m}$ to $x = 5 \text{ m}$. The work done by the force is (in $\text{ J}$)

The components of a force acting on a particle vary according to the graphs shown. To reach point $B\, (8, 20, 0)$ from point $A\, (0, 5, 12)$,the particle moves on paths parallel to the $x$-axis,then the $y$-axis,and then the $z$-axis. The work done by this force is ............ $J$.

$A$ particle of mass $0.1 \, kg$ is subjected to a force which varies with distance as shown in the figure. If it starts its journey from rest at $x = 0$,its velocity at $x = 12 \, m$ is .......... $m/s$.