If $\vec{F}$ and $\vec{S}$ are force and displacement respectively,and the work done by them is zero,then:

$A$ force of $(4 \hat{i}+2 \hat{j}+\hat{k}) \text{ N}$ is acting on a particle of mass $2 \text{ kg}$,displacing the particle from a position of $(2 \hat{i}+2 \hat{j}+\hat{k}) \text{ m}$ to a position of $(4 \hat{i}+3 \hat{j}+2 \hat{k}) \text{ m}$. The work done by the force on the particle in joules is: (in $\text{ J}$)

$A$ force acts on a $30 \,g$ particle in such a way that the position of the particle as a function of time is given by $x = \alpha t^2$, where $x$ is in metre, $t$ is in seconds and $\alpha = 1 \,m/s^2$. The work done during the first $4 \,s$ is (in $J$)

If a constant force of $(2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \text{ N}$ acting on a body of mass $5 \text{ kg}$ displaces it from $(3 \hat{i} - 4 \hat{k}) \text{ m}$ to $(2 \hat{i} + 2 \hat{j} + 3 \hat{k}) \text{ m}$, then the work done by the force on the body is (in $\text{ J}$)

$A$ particle moves from position $\vec{r_1} = 3\,\hat{i} + 2\,\hat{j} - 6\,\hat{k}$ to position $\vec{r_2} = 14\,\hat{i} + 13\,\hat{j} + 9\,\hat{k}$ under the influence of a force $\vec{F} = 4\,\hat{i} + \hat{j} + 3\,\hat{k}$. Find the work done in $J$.

$A$ particle moves from position $\vec{r}_1 = 3\hat{i} + 2\hat{j} - 6\hat{k}$ to position $\vec{r}_2 = 14\hat{i} + 13\hat{j} + 9\hat{k}$ under the action of force $\vec{F} = 4\hat{i} + \hat{j} + 3\hat{k} \text{ N}$. The work done will be ............ $\text{J}$.