How much work does a pulling force of $40\, N$ do on a $20\, kg$ box in pulling it $8\, m$ across a smooth floor at a constant speed? The pulling force is directed at $60^{\circ}$ above the horizontal. (Answer in $J$)

$A$ $1\text{ kg}$ 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 . . . . . . $J$.

$A$ block of mass $2 \,kg$ is pulled at a constant speed with a taut rope along a frictionless plane that is inclined at $30^{\circ}$. Then the work done by the tension in the rope in pulling it a distance $4 \,m$ along the inclined plane in joule is (acceleration due to gravity $= 10 \,ms^{-2}$)

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

If a force $\vec{F}=(3 \hat{i}-2 \hat{j}) \text{ N}$ acting on a body displaces it from point $(1 \text{ m}, 2 \text{ m})$ to point $(2 \text{ m}, 0 \text{ m})$,then the work done by the force is (in $\text{ J}$)

$A$ particle moves from a point $(-2\hat i + 5\hat j) \ m$ to $(4\hat j + 3\hat k) \ m$ when a force of $(4\hat i + 3\hat j) \ N$ is applied. How much work has been done by the force? $J$