The force $F$ acting on a body moving in a circle of radius $r$ is always perpendicular to the instantaneous velocity $v$. The work done by the force on the body in one complete rotation is:

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

$A$ uniform rope of length $4 \,m$ and mass $0.4 \,kg$ is held on a frictionless table in such a way that $0.6 \,m$ of the rope is hanging over the edge. The work done to pull the hanging part of the rope onto the table is, (Assume $g = 10 \,m/s^2$) (in $\,J$)

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

$A$ man of mass $50 \ kg$ carries a load of mass $20 \ kg$ and climbs $20$ steps,each of height $0.25 \ m$. Calculate the work done in $J$.

$A$ horizontal force of $5 \,N$ is required to maintain a velocity of $2 \,m/s$ for a block of $10 \,kg$ mass sliding over a rough surface. The work done by this force in one minute is $J$.