The work done in displacing a particle from $y=a$ to $y=2a$ by a force $F = -\frac{K}{y^2}$ acting along the $y$-axis is:

Consider a force $F = Kx^3$, which acts on a particle at rest. The work done by the force for a displacement of $2 \,m$ is given that $K = 2 \,N \cdot m^{-3}$. (in $\,J$)

$A$ 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

$A$ body of mass $8\,kg$ is moved by a force $F = 3x\,N$,where $x$ is the distance covered. The initial position is $x = 2\,m$ and the final position is $x = 10\,m$. The initial speed is $0.0\,m/s$. The final speed is ........... $m/s$.

$A$ body moves from $x = 0$ to $x = x_1$ under the influence of a force $F = cx$. The work done during this process is:

$A$ 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?