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

$A$ bar of mass $m$ resting on a smooth horizontal plane starts moving due to force $|\vec{F}|=\frac{mg}{9}$. The magnitude of the force remains constant with time. The force vector makes an angle $\theta$ with the horizontal which varies with the distance covered as $\theta= Cx$. If the constant $C=10 \text{ degree/meter}$, then the speed of the bar, when $\theta$ becomes equal to $30^{\circ}$ for the first time, is (Assume $g = 10 \text{ m s}^{-2}$) (in $\text{ m s}^{-1}$)

$A$ 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:

$A$ sphere of mass $m$ is set in motion with initial velocity $v_0$ on a surface where $kx^n$ is the frictional force,with $k$ and $n$ as constants and $x$ as the distance from the starting point. Find the distance at which the sphere will stop.

$A$ force $(3x^2 + 2x - 5) \text{ N}$ displaces a body from $x = 2 \text{ m}$ to $x = 4 \text{ m}$. The work done by this force is .......... $J$.

$A$ 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 ................. $\,ms^{-1}$.