The work done by a force $\vec{F} = -5x^4 \hat{i} \text{ N}$ in displacing a body from $x = 2 \text{ m}$ to $x = -2 \text{ m}$ is: (in $\text{ J}$)

If a force of $(6x^2 - 4x) \ N$ acts on a body of mass $10 \ kg$,then the work done by the force in displacing the body from $x = 2 \ m$ to $x = 4 \ m$ is: (in $J$)

Obtain the equation of work done by a variable force in one dimension.

The acceleration-displacement graph of a $10 \ kg$ particle is given. Find the work done as the particle moves from $x = 0 \ cm$ to $x = 8 \ cm$.

$A$ bar of mass $m$ resting on a smooth horizontal plane starts moving due to force $|\vec{F}|=\frac{mg}{9}$. The magnitude of the force remains constant with time. The force vector makes an angle $\theta$ with the horizontal which varies with the distance covered as $\theta= Cx$. If the constant $C=10 \text{ degree/meter}$, then the speed of the bar, when $\theta$ becomes equal to $30^{\circ}$ for the first time, is (Assume $g = 10 \text{ m s}^{-2}$) (in $\text{ m s}^{-1}$)

$A$ 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