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:

Obtain the equation of work done by a variable force in one dimension.

An object of mass '$m$' initially at rest on a smooth horizontal plane starts moving under the action of a force $F = 2 \text{ N}$. In the process of its linear motion,the angle $\theta$ (as shown in the figure) between the direction of the force and the horizontal varies as $\theta = kx$,where $k$ is a constant and $x$ is the distance covered by the object from its initial position. The expression for the kinetic energy of the object is $E = \frac{n}{k} \sin \theta$. The value of $n$ is .....

$A$ block of mass $m=1 \; kg$ moving on a horizontal surface with speed $v_{i}=2 \; m \; s^{-1}$ enters a rough patch ranging from $x=0.10 \; m$ to $x=2.01 \; m$. The retarding force $F$ on the block in this range is inversely proportional to $x$ over this range,$F_{r} = -k/x$ for $0.1 < x < 2.01 \; m$,and $F_{r} = 0$ for $x < 0.1 \; m$ and $x > 2.01 \; m$,where $k=0.5 \; J$. What is the final kinetic energy $K_{f}$ and speed $v_{f}$ of the block as it crosses this patch?

$A$ 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?

For a particle moving under the action of a variable force,the kinetic energy-position $(K-x)$ graph is given. Then: