A force is given \[F=-\widehat{i}+2\widehat{j}-3\widehat{k}\]along \[z\]-direction. What amount of work is done by this force to move a point size object by a distance of\[4m\] along the \[z\]-direction?

Hint : The work by a force in displacing a body to some distance is equal to the product of force and displacement. The component of displacement in the direction of force is to be considered. Then substitute the values in the equation to get the required work done by the force to move the object in the same direction.

Complete step-by-step solution:
Force is defined as a push and pull done on a body. Force will do some effect on the body like change in shape and size of body, changing the state of motion of the body and it is also defined as vector quantity. S.I. Unit of force is Newton and C.G.S. The unit for force is dyne. In calculating the force we have to take the angle between direction of force and direction of displacement. S.I. unit of displacement is metre and C.G.S. The unit of displacement is centimetre.
 The value of force given in the question is as follows:-
 \[F=-\widehat{i}+2\widehat{j}-3\widehat{k}\]
According to the given question, force is acting along \[z\]direction and direction of displacement is also along \[z\]direction.
Since work done by a force is expressed as the product of the component of force in the direction of motion and the displacement covered by the object under this force. It is expressed mathematically as follows:
\[\] \[W=\overrightarrow{F}.\overrightarrow{S}\]
Here the symbol \[S\]is representing the displacement.
Work done is defined as a dot product of force and displacement. It is a scalar quantity.
Here the given data is:-
\[\overrightarrow{F}=-\widehat{i}+2\widehat{j}-3\widehat{k}\]
\[\overrightarrow{S}=4\widehat{k}\]
Because in the question it is mentioned that displacement is along the\[z\]-direction. It means direction of force and direction of displacement are the same.
Now substituting these values in the equation of work done ,we get
\[W=(-\widehat{i}+2\widehat{j}+3\widehat{k}).(4\widehat{k})\]
Using product rule,we get the following results,
\[W=(0+0+12)J\]
\[W=12J\]
S.I unit of Work is Joule and C.G.S unit of erg.
So work done of \[12J\] is done in moving a point size object by a distance of\[4m\] along the \[z\]-direction.

Note: Newton’s Law of motion is defined to study forces, and Newton’s second of motion states that rate of change in momentum is directly proportional to external force applied and derived from this law is defined as force is equal to product of mass and acceleration. Newton’s third law states that every action has an equal to opposite reaction.