$A$ body of mass $6\,kg$ is under a force which causes displacement in it given by $S = \frac{t^2}{4}$ metres, where $t$ is time. The work done by the force in $2\,seconds$ is ............... $J$.

$A$ $1\text{ kg}$ block subjected to two simultaneous forces $\vec{F}_1 = (2\hat{i} + 3\hat{j} + 4\hat{k})\text{ N}$ and $\vec{F}_2 = (3\hat{i} - \hat{j} - 2\hat{k})\text{ N}$ is moved a distance of $25\text{ m}$ along the $(3\hat{i} - 4\hat{j})$ direction. The work done in this process is . . . . . . $J$.

If a force $(3 \hat{i} + 2 \hat{j} + 5 \hat{k}) \text{ N}$ acting on a body displaces it through $(2 \hat{i} + 2 \hat{j} + 1 \hat{k}) \text{ m}$,then the work done by the force on the body is (in $J$)

$A$ force of magnitude $30\, N$ acting along $\hat{i} + \hat{j} + \hat{k}$ displaces a particle from point $(2, 4, 1)$ to $(3, 5, 2)$. The work done during this displacement is:

$A$ block of mass $M$ is pulled a distance $x$ on a horizontal table. The work done by the weight (gravitational force) is:

$A$ horizontal force of $10 \ N$ is applied on a block of mass $1.5 \ kg$ which is initially at rest on a rough horizontal surface. The work done by the applied force in a time of $6 \ s$ from the beginning of the motion is (Acceleration due to gravity $= 10 \ m/s^2$; the coefficient of kinetic friction between the block and the surface is $0.2$). (in $J$)