If a position-dependent force $F(x) = (3x^2 - 2x + 7) \text{ N}$ acting on a body of mass $2 \text{ kg}$ displaces it from $x = 0 \text{ m}$ to $x = 5 \text{ m}$,then the work done by the force is: (in $J$)

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

$A$ particle moves along the $X$-axis from $x=0$ to $x=1 \, m$ under the influence of a force given by $F=3x^2+2x-10$. The work done in the process is ............. $J$.

The force on a particle as a function of displacement $x$ (in $x$-direction) is given by $F = 10 + 0.5x$. The work done corresponding to the displacement of the particle from $x = 0$ to $x = 2$ units is:

$A$ force $\overrightarrow F = - K(y\hat i + x\hat j)$ (where $K$ is a positive constant) acts on a particle moving in the $x-y$ plane. Starting from the origin,the particle is taken along the positive $x$-axis to the point $(a, 0)$ and then parallel to the $y$-axis to the point $(a, a)$. The total work done by the force $\overrightarrow F$ on the particle is

$A$ 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 $J$?