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Theory:

In the previous section, we looked into the definition of work. In this section, we are going to see
• Whether force is unidirectional or not.
• Whether work done by a force is positive or negative.
Consider the case in which the force and the displacement are in the same direction:
• A girl pulls a toy car that is parallel to the ground, as shown in figure.
• The baby applies a force in the direction of displacement of the vehicle.

In these cases, the work done will be equal to the product of the force and displacement.

$\mathit{Work}\phantom{\rule{0.147em}{0ex}}\mathit{done}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\mathit{Force}\phantom{\rule{0.147em}{0ex}}×\phantom{\rule{0.147em}{0ex}}\mathit{Displacement}$

In such cases, the work done by the force is taken as positive.

Consider the scenario in which an object moves with a uniform velocity along a particular direction.

Retarding force, $$F$$, is applied in the opposite direction to the object's movement. That is, the angle between the two directions is $$180º$$. Let the object stops after a displacement $$s$$.

In such cases, the work done by the force, $$F$$, is taken as negative and denoted by the minus sign.

$\begin{array}{l}\mathit{Work}\phantom{\rule{0.147em}{0ex}}\mathit{done}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\mathit{Force}\phantom{\rule{0.147em}{0ex}}×\phantom{\rule{0.147em}{0ex}}\mathit{Displacement}\\ \\ \mathit{Workdone}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}F\phantom{\rule{0.147em}{0ex}}×\phantom{\rule{0.147em}{0ex}}\left(-s\right)\\ \\ \mathit{or}\\ \\ \mathit{Workdone}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\left(-F\right)\phantom{\rule{0.147em}{0ex}}×\phantom{\rule{0.147em}{0ex}}s\end{array}$

From the above discussion, the work done by a force can be either positive or negative.