If a force $(3 \hat{i} + 2 \hat{j} + 5 \hat{k}) \text{ N}$ acting on a body displaces it through $(2 \hat{i} + 2 \hat{j} + 1 \hat{k}) \text{ m}$,then the work done by the force on the body is (in $J$)

If a force $\vec F = 4\hat i + 5\hat j$ causes a displacement $\vec s = 3\hat i + 6\hat k$,the work done is:

$A$ force $\overrightarrow{F} = 6\hat{i} + 2\hat{j} - 3\hat{k}$ acts on a particle and produces a displacement of $\overrightarrow{s} = 2\hat{i} - 3\hat{j} + x\hat{k}$. If the work done is zero,the value of $x$ is

$A$ body of mass $m$ starts from rest and acquires a velocity $v$ in time $t_1$. The work done in time $t$ is proportional to:

Show that a force that does no work must be a velocity-dependent force.

$A$ 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 ......... $J$.