Answer
The work done is 56.21 foot-pounds.
Work Step by Step
Here, the force can be resolved into two vectors, such that:
$\mathbf{F}=\left( 6\cos 40{}^\circ,6\sin 40{}^\circ \right)$ …… (1)
As the object is moved from point $\left( 5,9 \right)$ to $\left( 8,20 \right)$, therefore, the point to which the object has traveled is:
$\begin{align}
& \mathbf{AB}=\left[ \left( 8,20 \right)-\left( 5,9 \right) \right] \\
& =\left[ \left( 8-5 \right),\left( 20-9 \right) \right] \\
& =\left( 3,11 \right)
\end{align}$
$\mathbf{AB}=3\mathbf{i}+11\mathbf{j}$ …… (2)
Now, as it is known that:
$\mathbf{W}=\mathbf{F}\cdot \mathbf{AB}$ …… (3)
Substituting equation (1) and (2) in equation (3),
$\begin{align}
& \mathbf{W}=\left( 6\cos 40{}^\circ,6\sin 40{}^\circ \right)\cdot \left( 3,11 \right) \\
& =\left[ \left( 6\cos 40{}^\circ \times 3 \right)+\left( 6\sin 40{}^\circ \times 11 \right) \right] \\
& =18\cos 40{}^\circ +66\sin 40{}^\circ \\
& \approx 56.21
\end{align}$
