If a force of constant magnitude acts in a direction perpendicular to the motion of a particle,then its

The work done against gravity in taking a $10 \, kg$ mass to a $1 \, m$ height in $1 \, s$ will be .... $J$.

$A$ block of mass $m$ is pulled along a circular arc by means of a constant horizontal force $F$ as shown. Work done by this force in pulling the block from $A$ to $B$ is ............

$A$ 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 ............. $J$.

$A$ particle is made to move from the origin in three spells of equal distances,first along the $x$-axis,second parallel to the $y$-axis,and third parallel to the $z$-axis. One of the forces acting on it has a constant magnitude of $50 \ N$ and always acts along the direction of motion. Work done by this force in the three spells of motion are equal,and the total work done in all three spells is $300 \ J$. The final coordinates of the particle will be:

$A$ rope is used to lower a block of mass $M$ vertically downwards by a distance $x$ with a constant downward acceleration of $g/2$. What is the work done by the rope on the block?