$A$ force of $(6x^2 - 4x + 3) \text{ N}$ acts on a body of mass $0.75 \text{ kg}$ and displaces it from $x = 2 \text{ m}$ to $x = 5 \text{ m}$. The work done by the force is (in $\text{ J}$)

$A$ particle of mass $0.1 \, kg$ is subjected to a force which varies with distance as shown in the figure. If it starts its journey from rest at $x = 0$,its velocity at $x = 12 \, m$ is .......... $m/s$.

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 $\text{ J}$)

$A$ 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

What will be the area under the graph of variable force versus position?

$A$ 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