$A$ force $F = (5 \hat{i} + 4 \hat{j}) \text{ N}$ acts on a body and produces a displacement $s = (6 \hat{i} - 5 \hat{j} + 3 \hat{k}) \text{ m}$. The work done by the force is: (in $\text{ J}$)

$A$ body of mass $2 \ kg$ is moving along the $x$-direction such that its displacement as a function of time is given by $x(t) = \alpha t^2 + \beta t + \gamma \ m$,where $\alpha = 1 \ m/s^2$,$\beta = 1 \ m/s$,and $\gamma = 1 \ m$. The work done on the body during the time interval $t = 2 \ s$ to $t = 3 \ s$ is . . . . . . $J$.

$A$ horizontal force of $10 \ N$ is applied on a block of mass $1.5 \ kg$ which is initially at rest on a rough horizontal surface. The work done by the applied force in a time of $6 \ s$ from the beginning of the motion is (Acceleration due to gravity $= 10 \ m/s^2$; the coefficient of kinetic friction between the block and the surface is $0.2$). (in $J$)

$A$ ball is dropped from the top of a tower. What is the ratio of the work done by the force of gravity during the $1^{st}$,$2^{nd}$,and $3^{rd}$ seconds?

Two persons $A$ and $B$ perform the same amount of work in moving a body through a certain distance $d$ with the application of forces acting at angles $45^{\circ}$ and $60^{\circ}$ with the direction of displacement,respectively. The ratio of the force applied by person $A$ to the force applied by person $B$ is $\frac{1}{\sqrt{x}}$. The value of $x$ 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 $J$.