$A$ particle moves from position $3\hat i + 2\hat j - 6\hat k$ to $14\hat i + 13\hat j + 9\hat k$ due to a uniform force of $(4\hat i + \hat j + 3\hat k) \, N$. If the displacement is in meters,the work done will be ......... $J$.

$A$ force $\vec{F} = (-2\hat{i} + 15\hat{j} + 6\hat{k}) \, N$ acts on an object,causing it to move along the $Y$-axis. If the displacement along the $Y$-axis is $10 \, m$,find the work done by the force in $J$.

$A$ force $F = (5\hat i + 3\hat j) \, N$ is applied to a particle,which displaces it from its origin to the point $r = (2\hat i - 1\hat j) \, m$. The work done on the particle is:

$A$ 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?

$A$ 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 $\text{ J}$)

Define work exerted by force.