$A$ 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 $, J$)

$A$ particle moves from $x = 0$ to $x = x_1$ under the influence of a force $F = Cx$. Find the work done in this process.

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 $J$)

$A$ 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?

When a rubber band is stretched by a distance $x$,it exerts a restoring force of magnitude $F = ax + bx^2$,where $a$ and $b$ are constants. The work done in stretching the unstretched rubber band by $L$ is

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 .............. $J$.