Given that a force $\vec{F}$ acts on a body for time $t_1$ and displaces the body by $\vec{d}$. In which of the following cases must the velocity of the body increase?

$A$ particle is moved from $(0, 0)$ to $(a, a)$ under a force $\vec{F} = (3\hat{i} + 4\hat{j})$ through two paths. Path $1$ is $OP$ and path $2$ is $OQP$. Let $W_1$ and $W_2$ be the work done by this force in these two paths respectively. Then

The kinetic energy $(KE)$ acquired by a mass $m$ in travelling a certain distance $d$,starting from rest,under the action of a constant force $F$ is ..........

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 ............... $J$.

$A$ 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

$A$ $1\text{ kg}$ block subjected to two simultaneous forces $\vec{F}_1 = (2\hat{i} + 3\hat{j} + 4\hat{k})\text{ N}$ and $\vec{F}_2 = (3\hat{i} - \hat{j} - 2\hat{k})\text{ N}$ is moved a distance of $25\text{ m}$ along the $(3\hat{i} - 4\hat{j})$ direction. The work done in this process is . . . . . . $J$.