$A$ $50\,kg$ man with a $20\,kg$ load on his head climbs up $20$ steps of $0.25\,m$ height each. The work done in climbing is......$J$

In the figure shown,all the surfaces are frictionless,and the mass of the block $m$ is $1\,kg$. The block and wedge are held initially at rest. Now,the wedge is given a horizontal acceleration of $5\,m/s^2$ by applying a force on the wedge,such that the block does not slip on the wedge. Then,the work done by the normal force in the ground frame on the block in $2\,sec$ 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 $\text{ J}$)

$A$ body of mass $50 \ kg$ is lifted to a height of $20 \ m$ from the ground in two different ways as shown in the figure. The ratio of work done against gravity in both the respective cases will be:

$A$ force $\vec F = 5\hat i + 6\hat j - 4\hat k$ acting on a body produces a displacement $\vec s = 6\hat i + 5\hat k$. The work done by the force is $......$ units.

$A$ force $F = (5\hat i + 3\hat j) \, N$ is applied to a particle,which displaces it from its origin to the point $r = (2\hat i - 1\hat j) \, m$. The work done on the particle is: