If a force of constant magnitude acts in a direction perpendicular to the motion of a particle,then its

At time $t=0$,a particle starts moving along the $x$-axis. If its kinetic energy increases uniformly with time $t$,the net force acting on it must be proportional to

$A$ 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?

$A$ force of $(4 \hat{i}+2 \hat{j}+\hat{k}) \text{ N}$ is acting on a particle of mass $2 \text{ kg}$,displacing the particle from a position of $(2 \hat{i}+2 \hat{j}+\hat{k}) \text{ m}$ to a position of $(4 \hat{i}+3 \hat{j}+2 \hat{k}) \text{ m}$. The work done by the force on the particle in joules is: (in $\text{ J}$)

Given that $W = \vec F \cdot \vec s = 0$ and $F \neq 0, s \neq 0$,then

$A$ force $\vec{F}=(3 \hat{i}+4 \hat{j}) \;N$ acts on a particle moving in the $x-y$ plane. Starting from the origin,the particle first moves along the $x$-axis to the point $(4,0) \;m$ and then parallel to the $y$-axis to the point $(4,3) \;m$. The total work done by the force on the particle is ............. $J$.