If a force of $(6x^2 - 4x) \ N$ acts on a body of mass $10 \ kg$,then the work done by the force in displacing the body from $x = 2 \ m$ to $x = 4 \ m$ is: (in $J$)

$A$ particle is projected with velocity $v_{0}$ along the $x-$axis. $A$ damping force is acting on the particle which is proportional to the square of the distance from the origin,i.e.,$ma = -\alpha x^{2}$. The distance at which the particle stops is:

The figure shows the $F-x$ graph,where $F$ is the force applied and $x$ is the distance covered by the body along a straight-line path. Given that $F$ is in newton $(N)$ and $x$ is in metre $(m)$,calculate the total work done in joules $(J)$.

The relation between the displacement $X$ of an object produced by the application of the variable force $F$ is represented by the graph shown in the figure. If the object undergoes a displacement from $X = 0.5 \, m$ to $X = 2.5 \, m$,the work done will be approximately equal to .............. $J$.

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 .....

$A$ force $F = 20 + 10y$ acts on a particle in the $y$ direction,where $F$ is in newton $(N)$ and $y$ is in meter $(m)$. The work done by this force to move the particle from $y = 0$ to $y = 1 \; m$ is ...... $J$.