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

$A$ force $\vec{F} = (-6x^3 \hat{i}) \ N$ acts on a particle,causing a displacement from $x = 4 \ m$ to $x = -2 \ m$. The work done by the force is .......... $J$.

$A$ 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?

$A$ 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 $\text{ J}$)

$A$ block of mass $m$ moving with a velocity $v_0$ on a smooth horizontal surface strikes and compresses a spring of stiffness $k$ until the mass comes to rest,as shown in the figure. This phenomenon is observed by two observers:
$A$: standing on the horizontal surface
$B$: standing on the block
To an observer $A$,the work done by the normal reaction $N$ between the block and the spring on the block is

$A$ force acts on a body of mass $10 \ kg$,resulting in its displacement given as $x = \left(\frac{t^3}{25}\right) \ m$,where $t$ is the time in seconds. The work done by the force in the first $2 \ s$ is (in $J$)