$A$ 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?

Force acting on a body of mass $1 \ kg$ is related to its position $x$ as $F = x^3 - 3x \ N$. It is at rest at $x = 1$. Its velocity at $x = 3$ can be ......... $m/s$.

$A$ force $\vec{F} = (7 - 2x + 3x^2) \, N$ is applied to an object. Calculate the work done as the object moves from $x = 0$ to $x = 5 \, m$. (in $, J$)

The figure shows the frictional force versus displacement for a particle in motion. The loss of kinetic energy in travelling over $s = 0$ to $20\, m$ will be......$J$.

$A$ force $\overrightarrow F = - K(y\hat i + x\hat j)$ (where $K$ is a positive constant) acts on a particle moving in the $x-y$ plane. Starting from the origin,the particle is taken along the positive $x$-axis to the point $(a, 0)$ and then parallel to the $y$-axis to the point $(a, a)$. The total work done by the force $\overrightarrow F$ on the particle is

$A$ force $F = \alpha + \beta x^2$ acts on an object in the $x$-direction. The work done by the force is $5 \ J$ when the object is displaced by $1 \ m$. If the constant $\alpha = 1 \ N$,then $\beta$ will be: (in $N/m^2$)