Figure 43 shows a man pushing on a lawn mower handle with a force of 30 lb. How much work is done in moving the lawn mower a distance of 75 ft if the handle makes an angle of $60^{\circ }$ with the ground?

Figure 43 $\skew5\overrightarrow{\it AB}=75\,\mathbf{i}$

Solution We set up the coordinate system so that the lawn mower is moved from $(0,0)$ to $(75,0)$. Then the motion occurs along $\skew5\overrightarrow{\it AB}=75\,\mathbf{i}$. The force vector $\mathbf{F}$, as shown in Figure 44, makes an angle of $300^\circ$ to the positive $x$-axis. Since $\Vert \mathbf{F} \Vert = 30$, $\mathbf{F}$ is given by $$\begin{equation*} \mathbf{F}=30[ (\cos 300^{\circ })\mathbf{i}+(\sin 300^{\circ })\mathbf{ j}] =30\left[ \dfrac{1}{2}\mathbf{i}-\dfrac{\sqrt{3}}{2}\mathbf{j} \right] =15\mathbf{i}-15\sqrt{3}\mathbf{j} \end{equation*}$$

Figure 44 $\Vert \mathbf{F} \Vert = 30$

Then the work $W$ done is $$W=\mathbf{F}\,{\cdot}\, \skew5\overrightarrow{\it AB}=(15\mathbf{i}-15\sqrt{3}\mathbf{j} )\,{\cdot}\, 75\mathbf{i}=1125 \hbox{ft-lb}$$