$A$ position-dependent force $F$ acts on a particle,and its force-position curve is shown in the figure. The work done on the particle for a displacement from $0$ to $5\,m$ is ............. $J$.

The work done in displacing a particle from $y=a$ to $y=2a$ by a force $F = -\frac{K}{y^2}$ acting along the $y$-axis is:

An object starts from rest and is acted upon by a variable force $F$ as shown in the figure. If $-F_0$ is the initial value of the force at $x=0$,then the position of the object where it again comes to rest will be:

$A$ body of mass $3 \ kg$ moves under the influence of a force such that its displacement is given by $s = t^3/3 \ m$. What is the work done by the force in the first $2 \ s$ in $J$?

The displacement $s$ of a body of mass $3 \,kg$ under the action of a force is given by $s = \frac{t^3}{3}$, where $s$ is in metres and $t$ is in seconds. The work done by the force in the first two seconds is (in $\,J$)

The figure shows the $F-x$ graph,where $F$ is the force applied and $x$ is the distance covered by the body along a straight-line path. Given that $F$ is in newton $(N)$ and $x$ is in metre $(m)$,calculate the total work done in joules $(J)$.