$A$ block of mass $m$,kept on a horizontal surface,is moved through a distance $s$ by applying a horizontal force $F$ to it. What is the work done by the normal reaction?

$A$ body of mass $3\, kg$ is under a constant force which causes a displacement $s$ in metres in it,given by the relation $s = \frac{1}{3}t^2$,where $t$ is in seconds. Work done by the force in $2$ seconds is

The kinetic energy $K$ of a particle moving in a straight line depends upon the distance $s$ as $K = as^2$. The force acting on the particle is

$A$ uniform force of $(3\hat i + \hat j) \text{ N}$ acts on a particle of mass $2 \text{ kg}$. The particle is displaced from position $(2\hat i + \hat k) \text{ m}$ to position $(4\hat i + 3\hat j - \hat k) \text{ m}$. The work done by the force on the particle is ............. $J$.

$A$ $1\text{ kg}$ 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 . . . . . . $J$.

$A$ particle moves from $(1, 0, 3)$ to the point $(-3, 4, 5)$,when a force $F = \hat{i} + 5 \hat{k}$ acts on it. The amount of work done in joules is: