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

$A$ particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement $x$ is proportional to

The kinetic energy $(E_k)$ of a particle moving along the $X$-axis varies with its position $(X)$ as shown in the figure. The force acting on the particle at $X = 10 \ m$ is

$A$ 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 force $\vec F = F\hat i$ on a particle of mass $2\, kg$,moving along the $x$-axis is given in the figure as a function of its position $x$. The particle is moving with a velocity of $5\, m/s$ along the $x$-axis at $x = 0$. What is the kinetic energy of the particle at $x = 8\, m$?

The work done on a particle of mass $m$ by a force,$F = K \left[ \frac{x}{(x^2+y^2)^{3/2}} \hat{i} + \frac{y}{(x^2+y^2)^{3/2}} \hat{j} \right]$ (where $K$ is a constant of appropriate dimensions),when the particle is moved from the point $(a, 0)$ to the point $(0, a)$ along a circular path of radius $a$ about the origin in the $x-y$ plane is: