An object starts from rest and is acted upon by a variable force $F$ as shown in the figure. If $-F_0$ is the initial value of the force at $x=0$,then the position of the object where it again comes to rest will be:

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:

The slope of the kinetic energy-displacement curve of a particle in motion is

The work done by a force $\vec{F} = -5x^4 \hat{i} \text{ N}$ in displacing a body from $x = 2 \text{ m}$ to $x = -2 \text{ m}$ is: (in $\text{ J}$)

$A$ 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 relationship between force $F$ and displacement $x$ is shown in the figure. The work done by the object for a displacement from $x = 1 \ m$ to $x = 5 \ m$ is equal to ... $J$.