$A$ body of $0.5 \,kg$ moves along the positive $X$-axis under the influence of a varying force $F$ (in newton) as shown below. If the speed of the object at $x=4 \,m$ is $3.16 \,ms^{-1}$,then its speed at $x=8 \,m$ is ................. $\,ms^{-1}$.

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

$A$ $1 \,kg$ box placed at the origin starts sliding along the $x$-axis under the action of a force $\vec{F} = F \hat{i}$. Its acceleration as a function of $x$ is given by $a(x) = \beta x$,where $\beta = 5 \,s^{-2}$. The work done by $\vec{F}$ in moving the box from $x = 2 \,cm$ to $x = 5 \,cm$ in joules is:

Consider a force $F = Kx^3$, which acts on a particle at rest. The work done by the force for a displacement of $2 \,m$ is given that $K = 2 \,N \cdot m^{-3}$. (in $\,J$)

When a force $\vec{F} = (17 - 2x + 6x^2) \text{ N}$ acts on a body of mass $2 \text{ kg}$ and displaces it from $x = 0 \text{ m}$ to $x = 8 \text{ m}$,the work done is: (in $\text{ J}$)

$A$ 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 $\text{ J}$)