$A$ force acts on a $3.0 \ g$ particle in such a way that the position of the particle as a function of time is given by:
$x = 3t - 4t^2 + t^3$
Where $x$ is in metres and $t$ is in seconds. The work done during the first $4 \ s$ is ................. $mJ$

When a force $\vec{F} = (17 - 2x + 6x^2) \text{ N}$ acts on a body of mass $2 \text{ kg}$ and displaces it from $x = 0 \text{ m}$ to $x = 8 \text{ m}$,the work done is: (in $\text{ J}$)

$A$ force $\vec F = k(y\hat i + x\hat j)$ where $k$ is a positive constant acts on a particle moving in the $x-y$ plane. The particle moves along a straight line from the point $(3, 5)$ to $(5, 7)$. The work done by the force is: ............... $k$

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 $\,J$)?

$A$ 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

$A$ 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 .......... $J$.