## Precalculus (6th Edition) Blitzer

The magnitude of the work done is $1966\text{ foot-pounds}$.
The wagon handle makes an angle of $35{}^\circ$ with the road. Hence, the work-done is: $W=\left\| \mathbf{F} \right\|\left\| \mathbf{d} \right\|\cos \theta$ ……(1) Where, $\left\| \mathbf{F} \right\|=40\text{ pounds}$, $\left\| \mathbf{d} \right\|=60\text{ feet}$ and $\theta =35{}^\circ$. Now put it in equation (1), \begin{align} & W=\left( 40 \right)\left( 60 \right)\cos 35{}^\circ \\ & =2400\left( 0.8191 \right) \\ & =1965.84 \\ & \approx 1966\text{ foot-pounds} \end{align}