$A$ force is applied to a particle of mass $0.1 \ kg$ as a function of distance as shown in the figure. If it starts from rest at $x = 0$,its velocity at $x = 12 \ m$ is ....... $m/s$.

For a particle moving under the action of a variable force,the kinetic energy-position $(K-x)$ graph is given. Then:

$A$ particle of mass $10\,g$ moves in a straight line with retardation $2x$,where $x$ is the displacement in $SI$ units. Its loss of kinetic energy for the above displacement is $\left(\frac{10}{x}\right)^{-n}\,J$. The value of $n$ will be $............$.

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 figure shows the frictional force versus displacement for a particle in motion. The loss of kinetic energy in travelling over $s = 0$ to $20\, m$ will be......$J$.

The force $F$ acting on a particle moving in a straight line is shown below. What is the work done by the force on the particle in the $1^{\text{st}}$ meter of the trajectory (in $\,J$)?