Answer
The magnitude of the work done is $1966\text{ foot-pounds}$.
Work Step by Step
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),
$\begin{align}
& W=\left( 40 \right)\left( 60 \right)\cos 35{}^\circ \\
& =2400\left( 0.8191 \right) \\
& =1965.84 \\
& \approx 1966\text{ foot-pounds}
\end{align}$
