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The work is done by a constant force on a particle during a straight-line displacement that is in one direction.

Work done is the dot product of force and displacement in the direction of the applied force. Using this concept of the work done, the value work for each case can be calculated using the given formula.

Formula:

The work done by a constant force on a body,

$$\begin{gathered} \mathbf{W}\mathbf{=}\stackrel{\mathbf{鈫�}}{\mathbf{F}}\mathbf{.}\stackrel{\mathbf{鈫�}}{\mathbf{d}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{贵诲肠辞蝉胃} \end{gathered}$$ (1)

When the angle between force and displacement is then the work done on the particle can be given using equation (1) as follows:

$$\begin{gathered} \mathrm{W}=\mathrm{Fdcos}(30掳) \\ =\frac{\sqrt{3}}{2}\mathrm{Fd} \end{gathered}$$

So, the work done is positive.

When the angle between force and displacement is , then the work done using equation (2) is given as:

$$\begin{gathered} \mathrm{W}=\mathrm{Fdcos}(100掳) \\ =-\left(\mathrm{Fdcos}10掳\right) \end{gathered}$$

So, the work done is negative.

Now when force is $2\hat{\mathrm{i}}-3\hat{\mathrm{j}}$and displacement is localid="1657168357089" $-4\hat{\mathrm{i}}$, , then the work done using equation (1) is given as:

$$\begin{gathered} \mathrm{W}=\left(2\hat{\mathrm{i}}-3\hat{\mathrm{j}}\right).\left(-4\hat{\mathrm{i}}\right) \\ =-8 \end{gathered}$$

So, the work done is negative.