$A$ force $\vec{F} = -5 \hat{i} - 7 \hat{j} + 3 \hat{k}$ acting on a particle causes a displacement $\vec{s} = 3 \hat{i} - 2 \hat{j} + a \hat{k}$. If the work done is $14 \ J$,then the value of '$a$' is:

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

$A$ 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

$A$ person draws water from a $5\,m$ deep well using a bucket of mass $2\,kg$ and capacity $8\,litre$ with a rope of mass $1\,kg.$ What is the total work done by the person? (Assume $g = 10\,m/s^2$ and the density of water is $1\,kg/litre$.)

$A$ particle of mass $m$ moves along a circular path of radius $r$ under the influence of a centripetal force $F$. What is the work done when the particle moves along a semi-circular path?

$A$ body of mass $6 \,kg$ is under a force which causes displacement in it given by $s = \frac{t^2}{4} \,m$, where $t$ is time in seconds. The work done by the force in $2 \,s$ is: (in $\,J$)