$A$ gardener pushes a lawn roller through a distance of $20 \,m$. If he applies a force of $30 \,kg-wt$ in a direction inclined at $60^{\circ}$ to the ground, the work done by the gardener in pushing the roller is (Given: $g=9.8 \,m/s^2$) (in $\,J$)

$A$ force $\vec{F} = 6\hat{i} + 2\hat{j} - 3\hat{k}$ acts on a particle,causing a displacement $\vec{S} = 2\hat{i} - 3\hat{j} + x\hat{k}$. If the work done during this displacement is zero,find the value of $x$.

$A$ force $\vec{F}=(3 \hat{i}+4 \hat{j}) \;N$ acts on a particle moving in the $x-y$ plane. Starting from the origin,the particle first moves along the $x$-axis to the point $(4,0) \;m$ and then parallel to the $y$-axis to the point $(4,3) \;m$. The total work done by the force on the particle is ............. $J$.

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 $\text{ J}$)

$A$ force of $(4\hat i + \hat j + 3\hat k) \, N$ acts on a body,causing its displacement from $(3\hat i + 2\hat j - 6\hat k) \, m$ to $(14\hat i + 13\hat j + 9\hat k) \, m$. Calculate the work done in $J$.

The kinetic energy acquired by a body of mass $M$ in travelling a certain distance $d$,starting from rest,under the action of a constant force is: