If force and displacement of a particle in the direction of force are doubled,the work done would be:

$A$ man of mass $50 \ kg$ carrying a load of $10 \ kg$ reaches the top of a building in $4 \ minutes$. The work done by the man is $6 \times 10^4 \ J$. If he reaches the top in $2 \ minutes$,the work done by the man will be:

$A$ 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

$A$ particle is displaced from point $P(3 \ m, 4 \ m, 5 \ m)$ to a point $Q(2 \ m, 3 \ m, 4 \ m)$ under a constant force $\vec{F}=(3 \hat{i}+4 \hat{j}+5 \hat{k}) \ N$. The work done by the force in this process is

$A$ chain of mass $4 \ kg$ and length $2 \ m$ is hanging such that one-fourth of its length is hanging over the edge of a table. The work done to pull the hanging part onto the table is . . . . . . $J$. (Take $g = 10 \ m/s^2$)