$A$ body of mass $2.4 \, kg$ is subjected to a force which varies with distance as shown in the figure. The body starts from rest at $x=0$. Its velocity at $x=9 \, m$ is

$A$ 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:

$A$ particle moves under the effect of a force $F = cx$ from $x = 0$ to $x = x_1$. The work done in the process is

$A$ force $F = \alpha + \beta x^2$ acts on an object in the $x$-direction. The work done by the force is $5 \ J$ when the object is displaced by $1 \ m$. If the constant $\alpha = 1 \ N$,then $\beta$ will be: (in $N/m^2$)

The figure shows the variation of a force $F$ acting on a particle along the $x$-axis. If the particle begins at rest at $x=0$,what is the particle's coordinate when it again has zero speed?

$A$ force of $F=(5 y+20) \hat{j} \, N$ acts on a particle. The work done by this force when the particle is moved from $y=0 \, m$ to $y=10 \, m$ is $... \, J.$