$A$ 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 $J$.

$A$ body of mass $m$ is moving in a circle of radius $r$ with a constant speed $u$. The force on the body is $mv^2/r$ and is directed towards the centre. What is the work done by this force in moving the body over half the circumference of the circle?

Work done in time $t$ on a body of mass $m$ which is accelerated from rest to a speed $v$ in time $t_1$ as a function of time $t$ is given by

$A$ man starts walking from a point on the surface of the Earth (assumed smooth) and reaches the diagonally opposite point. What is the work done by him?

$A$ particle moves from position $3\hat i + 2\hat j - 6\hat k$ to $14\hat i + 13\hat j + 9\hat k$ due to a uniform force of $(4\hat i + \hat j + 3\hat k) \, N$. If the displacement is in meters,the work done will be ......... $J$.

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 $J$)